Talk:Division by zero/Archive 1

Archive 1

Archive 2

Early discussion

In particular, it is incorrect to say that a ÷ 0 is infinity. The argument that any number a, divided by a very small one, becomes extremely large is unconvincing: a negative number a divided by a small positive number does not become large, and neither does a positive number a divided by a small negative number.

Fine. But how about, a ÷ 0 is infinity if a is positive, and negative infinity if a is negative?

I have no idea what 0 ÷ 0 would be, of course. Evercat 23:19 6 Jul 2003 (UTC)

(Disclaimer: IANAM; poorly remembered high school algebra follows!) Lemme try to show why that doesn't work (aside from the definitional problem already explained in the article)... If you plot y = a/x (where a is positive) you'll see the trouble:

                  |* <- approaching 0 from positive x
                  |*    we get a limit of positive infinity
                  |*
                  | *
                  | *
                  |  **
                  |    **************
------------------+------------------
**************    |
              **  |
                * |
                * |
                 *|
                 *|   approaching 0 from negative x
                 *| <- we get a limit of negative infinity 

If a is negative, the curve is upside-down but has the same split. a/x approaches positive infinity when approaching x=0 from negative x, and negative infinity when approaching from positive x.

If we try to just come straight out with x=0, we can't fake it with the limit, since the curve is discontinuous. There's no reason to favor positive infinity over negative infinity, whatever the sign of a.

Now, depending on what you're doing, positive or negative infinity may be a useful way to treat a divide-by-zero case, but which is appropriate probably depends on which side of 0 you're approaching from in the denominator, as well as the sign of the numerator.

It may be useful to have a pretty graph in the article to illustrate this, as the quoted sentence tries to say it in words and I think is even more confusing than my attempt here. :) --Brion 23:45 6 Jul 2003 (UTC)

What you say may be insightful (and the italicized passage quoted at the beginning of this thread mentions this briefly.) However, I don't think a lengthier discussion than the passage above is warranted, and this why.

The reasoning that this explanation attempts to demolish only applies to a problem which is tangential to the question "what's 1/0?" What the above does say is: "the limit of 1/x as x tends to zero does not exist, even in the extended sense when limits are allowed to be +∞ or -∞."

The remainder of the article makes this clear. Division is the inverse of multiplication. 0*x is 0 for any x, therefore 0*x isn't invertible (because 0*x isn't injective.) This is the core of the matter.

When you ask a politician a question they don't like ("will you provide transmogrifiers to all the citizens?") their reply is sometimes tangential, sometimes even nonsensical ("I will work so that all citizens are treated justly. My opponent doesn't support the fluxification of space, but I will put a flux on the moon if it kills me. Remember the Alamo.") You then have two choices. You can take that long statement, and make a longer statement explaining how that's nonsense and doesn't answer the question, or you can try to give a more correct answer. To the question "what is 1/0?", the answer "if x is small and positive, 1/x is large, thus 1/0 is +∞" is at best tangential, and at worst nonsensical, and I don't think it's worth a lengthy rebuke.

Since it is in fact a widespread fallatious reasoning, it is worth mentioning as it is now, but wasting electrons on a long exposition about what is essentially bollocks (as I'm doing now) isn't ideal.

Ok, I'll shut up now. -- Loisel 05:16 7 Jul 2003 (UTC)

---

Wow. all that over a simple idea like 1/0. Seriously, guys, I always made the assumption that x/0 was either 0 or null since if you have any quantity "x" and you place equal fractions of x in a number of containers equal to 0 then how many pieces of x are there in each containter? Zero. But, you actually don't even have any containers, so really "null". And I never believed in the concept of infinity so the other statements about 1/0 = ∞ were insane at the get-go.

Regardless, the question I always wanted to know was: what's the practical application (if any) of the concept of x/0 anyway? No mathematician has ever been able to answer that, and I doubt they ever would.

Um, integration. Morwen 22:51, Dec 18, 2003 (UTC)

While true, one must wonder if this is just a result of our mathematical system. If x*0 is 0 (defined), why then isn't 0*0=x ?(as it should be if you reverse the process) Yet, x/0=(undefined) because the undefined * 0 also doesn't = x. Almost a little broken at points. Maybe one day (certainly not in my lifetime) it will be fixed.

I'm not sure if I understand what you mean; modifying x*0 = 0 would lead to 0/0 = x, which makes x undefined, which is why one can't divide by zero. Er.. I dunno. Evil saltine 23:01, 18 Dec 2003 (UTC)

---

I look at this question as is being similar to how for example

$x=2\Rightarrow x^{2}=4$

$x=-2\Rightarrow x^{2}=4$

yet

$x^{2}=4\Rightarrow \mathbf {W}$

the bold W meaning whatever or simply "?" indicating unknowledge. Using sets, however, it can be argued that

$x=\{-2,2\}\!$

Should this analogy be applied to division by zero

${0 \over 0}={\mbox{number}}$

such that

$\forall x,\ x\subset {0 \over 0}$

In other words zero divided by zero is the most ambigious number; it could be any number of any type: scalar, vector, or tensor; set or matrix; real or imaginary; differential (Rⁿ); albiet combinations thereof. Zero is an amazing number. As for the division of any nonzero number by zero,

${x \over 0}\ {\mbox{DNE}}\Leftarrow x\neq 0$

this being due in part, if not mainly, to the fact that there is more than one infinity: positive, negative, imaginary, complex; others perhaps.

I recommend avoiding the usage of the words "possible" or "impossible" in formulating conjectures about division by zero. To be a step ahead of human knowledge, one should acknowledge the possiblity of that which we accept as true being false.

--Lindberg G Williams Jr 19:30, 22 May 2004 (UTC)

0/0=0

Try reading http://members.lycos.co.uk/zerobyzero and see how certain logical approach can result in solution to real life problems otherwise un solvable due to your error theory.

0 / 0 = 0

Try this simple formula for average speed.

Total Distance Traveled / Hours Traveled = Average Speed

10 Kilometers / 2 Hours = 5 Kilometers Per Hour

What happens when you're not traveling

0 Kilometers / 0 Hours = 0 km/h

0/0=0

There is no other solution. We have created a world of limited mathematics, eliminated the limit of zero, and told the world it was unlimited, providing you did not divide by zero which limited it. Seems you all forget there are other applications and formulas that ABSOLUTELY NEED IT. Anyhow that is just one of thousands of them.

ZBZ

Well, so much for the theory that 0/0 belongs here rather than at indeterminate form. Charles Matthews 18:25, 24 Sep 2004 (UTC)

world of limited mathematics? CSTAR 19:44, 24 Sep 2004 (UTC)

This is actually a good example of why 0/0 is not defined!

If you go at 0 kilometers for any amount of time (say 1 hour), the 0/1=0, as you would expect. Now imagine you are going at X km/hr, for any X. How far will you go in 0 hours? you will go 0 kilometres. Therefore 0 kilometers / 0 hours = X km/hr for any X!

Think of it this way, - say you ran 20 kilometres. To define the speed, if you ran over it in 2 hours, it would be 10 km/h. If I took a "snapshot" of a zero-second timeframe (nil), and you ran 0 km during this period, it does not mean your speed is 0 km/h. Because the timeframe taken is insufficient/invalid and isn't an accurate reflection of actual speed, and therefore the value given is undefined because it could be anything. -- Natalinasmpf 09:41, 25 Apr 2005 (UTC)

You are partially correct, 0/0 does in fact equal 0 but that isn't the only answer. Let's say a/b=x, that would mean bx=a. So when this is applied to the equation 0/0=x you could truthfully say that 0x=0 and this would mean that x is every number (except for any other a/0 equation of course). - RyanAH

I'm interested in this conversation, but I think that in reality, 0/0 should equal 1 (as long as the 0's are the exact same 0's). What I mean is that, in reality, any 0 is only apporximately 0, and is just a very small number - mostly no useful quantity can be 0. Thus x/x is 1 even when x is infinitely small, ie. 0.

Also, I wanted to point out a flaw in the shpeil about average speed. "0 Kilometers / 0 Hours = 0 km/h" isn't quite right. Average speed requires a change of time. Speed is a change in distance over time, dL/dt. A derivative is by definition a limit, and a limit cannot occur if you don't have a spot from which to approach the limit. Thus 0 Km / ( 0 hrs) is not a valid speed. In reality such a quantity would be impossible to find. But if 0 was instead thought of as an infintesimal amount, then 0 Km / 0 hrs = 1 Km/Hr, as long as (0 Km)/Km = (0 hrs) / hrs. I hope i made some sort of sense. Fresheneesz 00:52, 4 February 2006 (UTC)

Therefore, you still can't divide by zero, because you can't say "I went zero miles per hour for the last zero minutes" It simply doesn't make any kind of sense to determine your speed for a non-existent period of time. How long were you not travelling? An hour? So 0/1=0 (km/hr) But you're not dividing by zero, you're dividing zero by something. The ANSWER is zero, and since if your speed is 0 the answer will always be zero, you chose "0" for your time because we can all agree if you don't travel you must be going 0 km/hr. But, yeah, it's still meaningless.

Zero is not defined as a very small number that is not zero. It is defined as zero. It doesn't matter how you think zero should be thought about nor how small a quantity of anything you can find. If you can't find it, you have, in fact, a quantity of zero of that thing you cannot find -- I can, for example, not find any dragons that spew fire, thus I can find exactly zero -- not just a very small amount of fire spewing dragons. Please stop using personal opinion as argument for mathematics -- that kind of arguments belong in religion or politics. —Preceding unsigned comment added by 80.167.145.223 (talk) 00:13, 17 October 2009 (UTC)

• let me end this dumb attractor

Two two branches converge at ∞. ∞ is nonpositive and nonnegative, just like 0. There is no sense in equivocating its sign with its side; there are only ∞⁺ and ∞^-, not +∞ and -∞. So 1/0 = (1)R∞, where R is any real number. Zero distance over zero time does not produce zero speed; it produces any speed. If any mathematicians disagree with me, and they do because they don't know me, they're wrong. lysdexia 19:44, 13 Oct 2004 (UTC)

I agree, 0/0 equals anything and everything, including zero, any finite number, and infinity. You can place zero into zero as many times as you want, from 1 to infinity as still get zero, so x = 0/0 works for all these numbers and infinity. Also you may choose not to place zero in at all, and you would still get zero, hence the other solution where x = 0. Also, if you subtract zero from zero an infinite amount of times, the answer is still zero, hence negative numbers work for x also. In a nutshell, ignoring the universal limit of the speed of light, a speed of 0/0 could equate to anything from 0 right through to infinity, in both positive and negative directions. It doesn't matter what concept you apply it to, speed or otherwise. Glooper 07:54, 4 April 2007 (UTC)

my 2 cents

I look at this question as essentially one of general topology, in which case the answer depends on which topology you choose. In this case, "1/0" is just shorthand for "what is the limit of 1/x as x approaches 0?" It depends on the space and topology. If you're in the real numbers R, then 1/0 doesn't "equal" anything, since 1/x has no limit as x approaches zero. If you throw in plus and minus infinity as separate points (the so-called "extended real number system") with the obvious neighbourhoods, then it still doesn't have a limit, so 1/0 still is meaningless. If, however, you identify plus and minus infinity to a single point, then it's perfectly legitimate to say that 1/0 = infinity. The same thing goes for the case of the Riemann sphere. The limit of 1/z as z approaches infinity (z complex) is infinity. Here, infinity is a well-defined "point" that exists just as surely as 0, 1, or pi. My whole point is, it doesn't really make much sense to ask, "what does 1/0 mean??". It's a meaningless question, a red herring. Stop whoever is asking it, and demand that they ask a more well-defined question which has an answer. Revolver 09:11, 15 Oct 2004 (UTC)

(1) I guess I see it as an algebraic problem -- that is, it has mostly to do with sets & operations. It's not necessary to bring limits into play; the motivation to do that is to attempt to guess a value for 1/0, but whether the assigned value was found by a limit or some other means is ultimately irrelevant, since what you want to investigate is whether 1/0 = x implies 1 = 0 x and that sort of algebraic thing. A beneficial side effect is that the article becomes accessible to anyone with some knowledge of algebra, if we can avoid making limits central to the discussion. (2) I'm not sure that assigning 1/0 = infinity on a Riemann sphere solves the problem entirely. Since (presumably) a/0 also = infinity for any nonzero a, one can't conclude a = 0 infinity. But perhaps there is more that can be said here. (3) I'm not convinced that 1/0 = ? is any more "meaningless" than any other question in mathematics. In any event, "stop whoever is asking it" seems a counterproductive tactic for an encyclopedia. Wile E. Heresiarch 14:23, 15 Oct 2004 (UTC)

Looking at the current article, it would be suitable to add a short section on the algebraic geometry case - where rational functions to a complete variety extend to honest functions on a blowing-up. This effectively says the maximum possible, about the case of polynomials which vanish. It also includes the Riemann sphere = complex projective line case, without privileging it unduly.

Charles Matthews 14:49, 15 Oct 2004 (UTC)

Sounds good to me. Have at it! Wile E. Heresiarch 16:17, 15 Oct 2004 (UTC)

Wile, by "stop whoever is asking it", I didn't mean "silence the person", I meant "demand whoever is speaking to ask a better question". I do think that "1/0 = ?" is a meaningless question, simply because, as our responses indicate, it has multiple interpretations. I interpreted it as a topological question, you as algebraic. I don't think there's anyway for either of us to "prove" the other's interpretation is "wrong". This is precisely what I mean by "1/0 = ?" not being a "well-defined" question. You say that "what you want to investigate is whether 1/0 = x implies 1 = 0x, etc." but that is presupposing an algebraic interpretation. Your interpretation comes from viewing the problem algebraically. Mine comes from viewing it from a calculus point of view (i.e. removing discontinuities, etc.) Charles' interpretation is even more broad and fundamental, and further supports my contention that the question must be recast before addressing it. As for the "1 = a for any nonzero a implies contradiction" argument, again that's presuming an algebraic perspective. There is no contradiction from the calculus perspective, in fact in complex analysis classes, the function 1/z is often defined from the Riemann sphere to itself, in the usual way for finite complex numbers, with the stipulation that 1/0 = infinity and 1/infinity = 0. This is not a statement about binary algebraic operations, but a statement about extending the function 1/z on C to be continuous on the Riemann sphere. Revolver 21:22, 15 Oct 2004 (UTC)

Well, I could quibble about various things. Instead I'll just ask how you would like to improve the article. I think that will help us stay focused. Regards, Wile E. Heresiarch 22:36, 16 Oct 2004 (UTC)

OK, time for some mathematics. For X and Y compact spaces, and F: X → Y a function but not everywhere defined, one can take the graph G of F and its closure G* in XxY. The main idea is to project G* back onto X and say it must be a closed set. Which is true if spaces are Hausdorff. For something like 1/x this suggests Y should be some compactification. What we are looking for is that G* should actually be the graph of a function. In this purely topological case one can't really argue that compactifying the real line to the extended real line is better or worse that to a circle (i.e. we could add two points or one to compactify R).

What the algebraic geometers do is a bit more definite, in that projective space acts as a compactification (non-Hausdorff, but everything is covered by the complete variety/proper morphism properties of projective space, plus the fact that the Zariski topology on XxY is not defined as the product topology). The added feature here is that there is more control of G* when it has a 'vertical component' projecting down to X.

Charles Matthews 08:22, 16 Oct 2004 (UTC)

OK, thanks for the information. How would you like to use this to improve the article? Wile E. Heresiarch 22:36, 16 Oct 2004 (UTC)

I would improve it by first noting that there are multiple ways of addressing the problem. Then, I would address it by interpretation, starting with the algebraic (binary operation) way. This includes much of the article (e.g. "incorrect arguments in dividing by zero" is essentially a misuse of algebra) and is easiest to understand. Then, maybe a mention of the other interpretations, but clearly set off in other sections. Again, I don't object to what's there, or object to the usual algebraic interpretation, just to its priveledged status. Revolver 22:50, 16 Oct 2004 (UTC)

Well, it doesn't seem like what you want is all that different from what's already in the article. Why not just go ahead and make the changes you want -- at least it will give us something different to talk about. Wile E. Heresiarch 16:36, 17 Oct 2004 (UTC)

I worry a bit about going back and forth between "concrete" and "abstract" discussions, that it might be something that only a relatively few who read it will be interested in. (Or worse, that it might confuse... "2*4 = 2"?) The only way to do this seems to be to break up the organisation by interpretation. Revolver 06:12, 19 Oct 2004 (UTC)

0/0 and differentiation

I can see a great deal of serious thought went into this article and it is extremely well and formally reasoned. However, it perpetuates the (formally accepted) statement that 0/0 is "just as" undefined as, say, 127/0. I'm quite sure this is formally correct, and I'll thank you not to spank me for saying it is not correct at a lower level, somewhere down among primitive life-forms such as engineers. Here's the text:

Limits of the form

$\lim _{x\to 0}{f(x) \over g(x)}$

in which both f(x) and g(x) approach 0 as x approaches 0, may converge to any value or may not converge at all.

Okay, this is all true, but still, what you are doing, in a sneaky sort of sliding-up fashion, is 0/0. Please don't throw a blizzard of complicated math at my statement; I know what you're after. My point is that 0/0 is what differentiation feels like when I imagine the limit approaching. It is useful -- to me, if to nobody else -- to believe in 0/0 as the limit: the point at which anything becomes possible.

Look at this from the amphibian point of view of an engineer. Think of the tangent to the curve as a long, massive stick; at an initial approximation, it is fixed to the curve at two points, (x,y) and (x1,y1). These points are widely spaced and the tangent does not wiggle. But, as the limit is approached, the distance between the two points decreases and the stick is liable to wiggle more and more; at the limit itself, the tangent is only attached to the curve by a single point and is free to rotate throughout a complete circle. However, we know better than to apply any torque to the tangent-stick as we approach the limit; we very carefully edge (x1,y1) up to (x,y), so when we finally reach the limit, the tangent is left pointing in the correct direction.

All I'm saying here is that it is a useful tool to the student to suspend disbelief and imagine 0/0 = the tangent. It makes differentiation real.

By the same token, when integration is demonstrated by area under a curve, first numerical integration approximates the area as a large number of thin slices. As the limit is approached, the student can think of an infinite number of zero-width slices, which add up to the total area. This is ∞ × 0 = the area. This may not be formally correct, but it is a good and useful way to think about it.

I remember I once listened to my 8th grade math teacher lecture my class on this subject. He said it was "impossible" to divide any number by zero, since no number could be multiplied by zero to yield anything except zero. I raised my hand and said, What about 0/0? He said that was "undefined" and looked pissed off. I accepted that 0/0 was mathematical nonsense. Later, when I got into the calculus, I had a tough time of it, despite being very good at algebra, until I freed my mind and allowed myself to simply imagine the process as a clever way of defining the undefined.

I would like to see if there is anyone bold enough to step away a moment from the purely theoretical and formally correct. Who will join me in wondering where these thoughts belong? I truly believe they are of value to the beginning student of the calculus, who is ill-served by adamant denial of 0/0. --Xiong 06:27, 2005 Mar 11 (UTC)

Difference between 1/0 and 1/infinitesimal and x/0

Can someone explain in simple terms, if 1/0, 1/infinitesimal, or 1000000/0 or 1000000/infinitesimal or any different?

In standard analysis, an infinitesimal dx is just an independent variable, while any fraction symbol with zero in the denominator is undefined. (Exception: 0/0 is defined as a L'Hopital Form, but is not a number -- really it's is just a way of remembering the hypothesis of L'Hopital's Rule.) Thus anything/0 is a meaningless symbol, while anything/infinitesimal is a function of the independent variable infinitesimal (usually denoted dx). But for another way of looking at things, see the article on Non-standard analysis. Rick Norwood 17:42, 23 December 2005 (UTC)

More simply, an infintesimal is some value that is approximately 0. 1/0 is undefined because 1/x has two limits as x->0 . However, 1/infintesimal is simply a positive really really big number (as long as the infintesimal is positive) - such a big number it can be thought of as infinity. One thing to remember though, is that if x = infintesimal, then x/x = 1. but if the infintesimals are different values, than their ratio might be somthing other than 1. This is the basics of derivatives. Imagine that .002 is an infintesimal and .001 is another. .002/.001 obviously = 2. That sort of thing would happen in an equation like Zdy = dx where Z= 10^9999. .002 would most definately be approximately infintesimal, so we could set dx = to .002, in which case dy would = .002/10^9999 which = approximately 0. Maybe that got convoluted.. but its a different explanation Fresheneesz 01:06, 4 February 2006 (UTC)

Simply to clarify, it is merely because an infinitessimal represents the smallest possible number greater than zero. The existence of two limits at zero for 1/x is because the reciprocal of a number og signum x will be of signum x. Since an infinitessimal is greater than zero, yet theoretically equal to it at once, x/dx will equal infinity, and only positive infinity. He Who Is 20:24, 7 June 2006 (UTC)

An infinitesimal is NOT "the smallest possible number greater than zero." One can divide a positive infinitesimal hyperreal number by 2 and get a smaller positive infinitesimal. Michael Hardy 18:40, 9 November 2006 (UTC)

There exists no infinitesimal element other than 0 in real numbers. --Kprateek88(Talk | Contribs) 14:11, 9 November 2006 (UTC)

three impossible things before breakfast

I was going to let my Star Trek reference go, though I tend to favor interesting articles over dull ones, but then came a mathematically incorrect edit, and since I was reverting anyway... : )

Division by 0 is in no way analogous to squaring the circle. Division by zero is a question of definition. Any attempt to define division by 0 forces us to give up more than we are willing to give up -- if we define division by 0, no matter how we define it, then we have to give up the cancellation property, because 0 * 2 = 0 * 3 but 2 does not equal 3. On the other hand, squaring the circle is easy, just not with straightedge and compass. You need other tools. Rick Norwood 17:37, 23 December 2005 (UTC)

Yes. I don't see how anyone would see those as analogous... That's to saying "2 has no square root because its irrational... He Who Is 20:30, 7 June 2006 (UTC)

Let's use the correct definition of division!

Intermediate Algebra by Barnett and Kearns has the actual definition of division and the explanation as to reason for division by zero is not defined:

"We say that a divided by b equals c if and only if there exists a unique value c such that b times c equals a."

Division by zero (b=0) is not defined because if a is nonzero, c doesn't exist, and if a is zero, c is not unique.

B.Wind 21:41, 24 December 2005 (UTC)

There is more than one "correct" definition of division. There is an article in the current issue of Focus that shows just what a great variety of definitions have been proposed. Rick Norwood 01:47, 25 December 2005 (UTC)

So I'm curious - is there a "correct" definition of division that allows for division by zero in the real numbers? The division algorithm is pretty clear, and does not allow division by zero. This is key, isn't it? It IS after all a definition issue. 12/29/2005 DB

Any definition of division that allows division by zero necessarily results in giving up something, usually giving up lots of things, for example, left and right cancellation, closure, well defined, the relationship between multiplication and division, and/or the commutative, associative, and distributive laws. Except for a few very simple systems (example: the only number is 0 and for every binary operation * we define 0*0 = 0, so 0/0=0) defining division by zero looses you more than you gain. However, see non-standard analysis.

All "standard" definitions of division give the same answer for all divisions, and all leave division by zero undefined. Rick Norwood 00:46, 30 December 2005 (UTC)

You can't divide by zero in non-standard analysis either. Michael Hardy 02:11, 30 December 2005 (UTC)

But the simple extension of reals, $\mathbb {R} \cup \{\infty \}$, where $\infty$ is an unsigned infinity, so ${\frac {1}{0}}=\infty$ and ${\frac {1}{\infty }}=0$ allows division by zero (for non-zero numerator). True, that does give up a few things, but in my opinion is very elegant. --Meni Rosenfeld 13:53, 17 January 2006 (UTC)

The big problem with that is that then you lose ${\frac {a}{b}}=c$ implies $a=bc$, which is in my opinion fairly serious...Mrjeff 14:18, 17 January 2006 (UTC)

Well, you win some, you lose some... As I said, I think this is a very elegant extension, although arguably not very useful (though this is a common extension to the complex plane, that neatly unifies some analytic properties of functions). --Meni Rosenfeld 18:39, 17 January 2006 (UTC)

Star Trek Reference

I previously removed a star trek reference from this page. It got re-added, so I thought I'd have a quick discussion about it here.

I don't feel "In some episode of star trek, someone blew up a computer by asking it to divide by zero" is interesting or relevant. Star trek did many stupid things, and also jokes about dividing by zero have appeared in many different areas. Also, I really disagree about this being in the "computer architecture" section. How about a new section called something like "Division by zero in popular culture"? (there is probably a better and/or more standard name for this section).

I certainly think the Star Trek reference should state which episode this happened in -- unfortunately I don't remember. I do remember, in "Return of the Archons", a computer breaking down on being asked to compute pi to the last digit. Rick Norwood 14:31, 4 January 2006 (UTC)

That kind of backs up my point. Do you want to add that to the pi page? I wonder why so many computers on star-trek break down trying to do things that the laptop I'm working on don't crash on at all...Mrjeff 15:37, 4 January 2006 (UTC)

Hows that?Mrjeff 15:31, 10 January 2006 (UTC)

It is for the same reason that the alien computer in Independence Day is the only computer in the universe that will interface with an Apple. Rick Norwood 15:47, 10 January 2006 (UTC)

Another way to think of it is the Macintosh is the computer that can interface with alien systems. :P --127 12:16, 23 January 2006 (UTC)

From my memory, and from searching on the web, it seems that the reference to dividing by zero in Star Trek is incorrect. The incident was actually computing pi, in the episode "Wolf in the Fold" [1], [2]. So I am going to be bold and make the change. 1diot 17:53, 10 January 2006 (UTC)

The evidence that the statement was incorrect notwithstanding, I would have argued to keep it. In the article on Pi for example, there is a section called "Fictional References," containing many similar references.

Rewrite to remove bias

This article currently seems to have a bias against division by zero, something in the sense of "There are those who foolishly define ${\frac {a}{0}}=\infty$, but the ultimate truth is that division by 0 must never be defined." I have a mind to rewrite several sections (some of which have factual errors) to give a more balanced view, that defining division by 0 doesn't make much sense in algebra, but can be useful in analysis if defined properly. That is, of course, as long as no one objects. --Meni Rosenfeld 06:39, 18 January 2006 (UTC)

I'm sorry to say it, but the way you have phrased this does not give me confidence that you can handle this successfully. Why not try it out here, first? Rick Norwood 13:32, 18 January 2006 (UTC)

You're not giving me much credit are you? Well, if you insist, I'll try to compose something and put it up here. But I sure would like to know what made you uncomfortable. --Meni Rosenfeld 13:58, 18 January 2006 (UTC)

What made me uncomfortable? Accusations of bias. Putting words such as "foolish" in other people's mouths. Assertions of factual errors without examples. Rick Norwood 19:07, 18 January 2006 (UTC)

I did not put the word "foolish" in anyone's mouth. I thought it was clear that the sentence inside the quotation marks is an exaggeration, and I apologize if that wasn't clear. All I said is that this is more or less what several phrases in the article sound to me. As for bias, I did not mean it in as strong a sense as you have possibly interpreted it. Since I'm not interested in politics and the like, to me it doesn't carry such a strong negative connotation, as it possibly does to you. As for factual errors, I could of course have given examples such as

For any nonzero a, it is known that
$\lim _{b\to 0{+}}{a \over b}={+}\infty$

Which is actually true only for positive a (a small error, but an error nonetheless). But I thought it would be much better to fix such errors than talking about them.

All in all, it seems that you have been offended by my comment, as I have been from yours. I meant no harm, and I'm sure you didn't either, but I wish you will be more careful when putting potentially offensive comments (as I will when putting potentially controversial comments).

Like I said, I will try to compose an improved version, but this could take a while. --Meni Rosenfeld 10:39, 19 January 2006 (UTC)

Let's agree that both of us want to improve the article. The error you point out certainly needs to be corrected. Rick Norwood 15:29, 19 January 2006 (UTC)

Completed (more or less)

There, I've made the changes I wanted (though not on as large a scale as I originally intended). Any format corrections are welcome, as are any comments regarding the content. -- Meni Rosenfeld (talk) 19:04, 21 January 2006 (UTC)

Most of your changes seem good, though I may want to adjust the wording slightly. Also, in reading over the article again, I see a need to say a few words in the introduction for the non-mathematical reader. Rick Norwood 21:07, 21 January 2006 (UTC)

I've done a little rewriting, caught one capitalization error. I also corrected a few other minor errors that had somehow gone unnoticed for a long time -- for example, the assertion that division by a non-zero number was always defined in the ring of integers.

I'm going to pause here, because I do not like to change too many things at one time. Rick Norwood 21:24, 21 January 2006 (UTC)

I've mad some more changes, mostly slight fixes, but also another reword of the Bhaskara thing, adding a few points that I believe your version missed. -- Meni Rosenfeld (talk) 06:32, 22 January 2006 (UTC)

Fractions

Zack, what you did with the fractions doesn't look very good to me (the shrinking thing). I think they are much less readable this way, and also, using a matrix construct to make them smaller seems not very systematic. Please reconsider this change. Putting the fractions inline and consistently using \frac was great though. -- Meni Rosenfeld (talk) 19:24, 22 January 2006 (UTC)

It is substantially more readable for me this way. I find the enormous gaps between lines produced by the old markup to be a worse readability problem than the tiny print.

\begin{matrix} ... \end{matrix} is the documented method of doing this (see Help:Formula#Fractions, matrices, multilines). It would be better if the distinction between $ ... $ and $$ ... $$ in TeX were directly available in wiki markup, but this is what we have now.

Zack 20:17, 22 January 2006 (UTC)

I agree with Zack. Rick Norwood 20:25, 22 January 2006 (UTC)

Formal calculation

Rick, I don't agree with you that a formal calculation is a bad idea. Sure, since it's not rigorous it can lead you astray. But an experienced mathematician will usually know what is "right" or "wrong" to do (for example with 1/x^2 you know that "the zero is positive" and you get a positive infinity, whereas in 1/x you know "the zero is unsigned" and you can't do anything). And the point in a formal calculation is not to prove something, but rather to get an intuitive result, which you can later prove if you want to be convinced about it. Also, any criticism regarding this concept is of course welcome, but in its proper place, the article formal calculation which I have started. My point in this section was simply to present a useful substitution for a/0, not to discuss all the rights or wrongs of a formal calculation (which belongs, again, in the appropriate article). Please consider rewording the section in a less negative way. -- Meni Rosenfeld (talk) 19:44, 22 January 2006 (UTC)

And I thought I asked for comments regarding my changes, not mocking of them. I never said that the real projective line and the Riemann Sphere are fields, or that you can define $\infty -\infty$ in them. You can't, for the exact reasons you described. Nor did I say that these structures preserve all the properties of their respective original fields. All I said is that those structures are interesting, consistent, potentially useful, and you can define division by 0 in them. I do wish you will reconsider your last edits. -- Meni Rosenfeld (talk) 19:52, 22 January 2006 (UTC)

BTW, the article's name is "Division by zero", not "Division by zero in the real numbers field". In either case, effectively deleting my edits without discussing them first, is just uncivil. -- Meni Rosenfeld (talk) 19:54, 22 January 2006 (UTC)

You underestimated the problems involved in division by zero. I did not revert your changes, but I did remove statements that were misleading. I'm sorry if you were offended, but you should be sure of your facts. Rick Norwood 20:00, 22 January 2006 (UTC)

First, I will have to request that whatever the case, you will be civil in your edits. When you and Zack did changes I didn't agree with, I didn't go about reverting them (and what you did is effectively reverting); I started a discussion about them. I expect you to do the same.

Second, I'm not sure I am the one who needs to do my homework. If you take a look at Riemann sphere and extended real number line, you'll know that the fields of Reals and complexes can be extended, at some cost. In the extended real number line, you can't divide by 0, because positive and negative infinities are separated, so you can't give a sensible definition. However, you will see that the article comprehensively defines arithmetic operations in it, leaving those that cannot be sensibly defined, undefined. For example $a++\infty =+\infty$ where $a\neq -\infty$, but $+\infty -+\infty$ is undefined. The advantage of the extended real number line over the real projective line is that you have the order relation; The advantage of the latter is that you have division by zero. Again, you can't expect everything to be defined; $\infty +\infty$, $\infty *0$ and $0/0$ are all undefined. This may seem like a great sacrifice, but the advantages of having division by zero are many.

It all comes down to what you want to have and what you are willing to sacrifice. If you want an ordered field (which is a big plus, explaining the popularity of reals), you will use the reals numbers. If you want to add in elements representing infinity, while keeping the order, you'll use the extended real number line. If you are willing to sacrifice order, and some of the properties of the field, while adding infinity and the option of division by zero, you'll use the real projective line. And if you dealt with complex numbers in the first place, and want to add infinity and division by 0, you'll use the Riemann sphere (which is very popular).

These stuctures aren't "wrong" because they lack field properties, just as the reals aren't "wrong" because they don't have division by zero. You can't win them all; Each structure has its own features, and you choose which one to use depending on the features you desire. If you want to write an article "The proof that it is impossible to define division by zero in the field of real numbers", go ahead; But the purpose of an article named "Division by zero" is to discuss structures where it is and is not possible to define it.

To conclude, I hope you will have the common sense to restore the content that you have erased. But I'm more than willing to do it myself if necessary. -- Meni Rosenfeld (talk) 07:09, 23 January 2006 (UTC)

And one more thing: Please do check out [3] for additional insight. -- Meni Rosenfeld (talk) 08:29, 23 January 2006 (UTC)

I did not delete any of the sections you added. I'm certainly familiar with the Riemann sphere. In fact, I'm teaching a course in Complex Analysis this semester. I just added warnings that your sections lacked, and also removed contractions and what seemed to me to be unnecessarily informal adjectives. Of the various sections you have written, the one I like least is "formal calculation", which I have seen too often used, especially by students but sometimes by people who should know better, to take limits "the easy way", often resulting in wrong answers. Note that the link you directed me to on Mathworld is very careful to say that if you define division by zero equals (unsigned) infinity, then you must make infintiy minus infinity undefined. Your section lacked that cautionary note.

In any case, I'm not trying to fight with you on this, just to keep the article mathematically accurate. Rick Norwood 14:12, 23 January 2006 (UTC)

Okay, maybe I'm beginning to understand you. A few points though; What I've written is not instructions for a nuclear plant or anything, it doesn't require words of caution at every step in the way. For example the $\infty -\infty$ thing, is just one example that the Riemann sphere is not a field, and that not all aritmetic operations are defined in it. Such notes are great for the article on the Riemann sphere (and indeed, that article in Mathworld is about the real projective line), but this article is about division by zero. So perhaps a few words that these structures aren't fields (which isn't a bad thing) are in order, but specific examples are too much to be put here. And one definitely should not imply that it is wrong to use these structures because they are not fields. The "many problems arise...so $0=\infty$" comment certainly has no place here, it's inaccurate and non-encyclopedic, and in fact added considerably to my irritation. And writing that "$+\infty =-\infty$ is nonsense" is just as aggrevating and inaccurate - The equality is just another way of saying that having division by zero must imply that infinity is equal to its negation, that is, it is an unsigned infinity.

About the formal calculation, it is a tool, and like any tool, what you get from it depends on how you use it. And it is well known that misusing this tool can lead to errors. So perhaps a reminder that care should be taken can be placed, but not your comment that is basically an attempt to discredit it, which can be considered POV (and I'm sorry if that word offends you). And again, any neutral critcism of formal calculations is welcome at that article.

All in all, I really don't like the way the article looks now, and will improve it, taking into consideration your concerns. I can't do this currently, perhaps tomorrow. -- Meni Rosenfeld (talk) 20:31, 23 January 2006 (UTC)

I hope that you do improve the article, but feel strongly that the cautions I provided, or something like them, are necessary. Rick Norwood 21:22, 23 January 2006 (UTC)

I made some changes, now it looks more reasonable to me. I hope you will approve. Too bad about the small fractions though, I still don't like them but will not argue with you. -- Meni Rosenfeld (talk) 08:45, 24 January 2006 (UTC)

I've made a few small changes, of which the most important is an example in which formal calculation fails. Rick Norwood 15:24, 24 January 2006 (UTC)

Looks reasonable. But unsurprisngly, I'm not comfortable with your version of the formal interpretation section. A counter-example is, maybe, appropriate (though I'm inclined to think it does not belong here), but I very much doubt that a definition of the concept is. I think you are making too big a deal out of a small section. I will re-scrutinize this section in a few days, after I've made sure your ideas have sufficient representation in the formal calculation article. And I wish you will re-read my comment on the matter at the very beginning of this thread, to remind you what I think of the issue. -- Meni Rosenfeld (talk) 16:20, 24 January 2006 (UTC)

I reread your comments -- and discovered that I had remembered them correctly. Of course, mathematicians use "formal calculations" all the time. But beginners need an example of how they can go wrong. Rick Norwood 17:00, 24 January 2006 (UTC)

Can I just ask a question, what "formal calculation" is going on here? From my nearby maths book..

lim(x -> a) f(x) = b if and only if: (exists d > 0) such that (forall 0 < e < d) there (exists f > 0) such that |f(e) - b| < f.

Thats the only formal definition of lim I could find. That clearly doesn't apply in this case. What is being used? Mrjeff 12:11, 25 January 2006 (UTC)

See formal calculation. "Formal" as in "formal definition" is not the same as in "formal calculation". And also, you should check the definitions for infinite limits, the one you give is for finite limits. -- Meni Rosenfeld (talk) 12:44, 25 January 2006 (UTC)

Formal calculation, as I understand it, means you go through the forms of doing a calculation without worrying about the meanings of those forms. It is almost the exact opposite of a formal definition, which is a definition where care is taken with both the form of the definition and the meaning of the symbols. Rick Norwood 01:31, 4 February 2006 (UTC)

Signed Zeroes

The article mentions how one can use signed or unsigned infinities to try to make sense of division by 0. It seems that you can also use signed zeroes. As long as $\pm \infty \mp \infty$, ${\frac {\pm \infty }{\pm \infty }}$, ${\frac {\pm 0}{\pm 0}}$, and $\pm 0\times \pm \infty$ are left undefined, then it seems consistent. For example, if $a$ is a positive number, then:

${\frac {a}{+\infty }}=+0$

${\frac {a}{-\infty }}=-0$

${\frac {-a}{+\infty }}=-0$

${\frac {-a}{-\infty }}=+0$

$a\times +\infty =+\infty$

$a\times -\infty =-\infty$

$-a\times +\infty =-\infty$

$-a\times -\infty =+\infty$

and so on. Is there a name for this type of system? --BrainInAVat 02:55, 13 February 2006 (UTC)

One can, of course, invent any system one likes. This would be a system in which only multiplication was defined, not addition (else you would have +0 = +0 + -0 = -0). Also, you loose the betweenness property -- there is no number in between +0 and -0. It will then be isomorphic to the closed interval [-1,1] with one additional point -0. I don't know a name for such a system. Rick Norwood 13:44, 4 March 2006 (UTC)

I also don't know a name, but these definitions are written at −0. I guess you could call them "IEEE 754". Melchoir 17:26, 4 March 2006 (UTC)

Unicode Mathematical Symbols?

I was wondering what everyone's reaction would be to the idea of using Unicode characters in place of images of mathematical symbols in this article (where possible). My connection has been stubborn this evening and it took forever for the equation images to load. 67.142.130.32 05:50, 4 March 2006 (UTC)

The vertical fractions can't be changed, but there is indeed too much math-mode in the article. I'll try to cut it down. Melchoir 06:43, 4 March 2006 (UTC)

Just in from Lewis Carroll

The late reverend C. L. Dodgson just channelled me a message that he loved this phrase: "Nothing can be divided into zero parts." He added: "And doing it requires no effort at all!". Lambiam^Talk 23:39, 6 April 2006 (UTC)

Reversion of Psb777's rewrite of the intro paras

OK. What's wrong with my edit that it needs reverting? Paul Beardsell 23:35, 24 April 2006 (UTC)

The very first sentence. "division by zero is not defined" is directly contradicted by the section "Other number systems". Melchoir 23:41, 24 April 2006 (UTC)

You are right but at what cost? If you define division by zero then it is no longer undefined. But, as is shown again and again in the article, defining it reduces the usefulness of the resulting mathematics. The remainder of the article is simply a set of examples demonstrating why we choose to leave division by zero undefined. To avoid confusion I think we should be plain about this at the beginning of the article. Maybe I didn't do a good job of it. Paul Beardsell 23:55, 24 April 2006 (UTC)

But see, we don't always choose to leave it undefined. That's kind of the point. See Riemann sphere. --Trovatore 23:56, 24 April 2006 (UTC)

I think we taint maths by such construction. What I do get annoyed by are ongoing discussions here and elsewhere (outside of Wikipedia) as to what the value of 1/0 is. And what 0/0 is. It would be very useful to simply have an authoritative reference I can point to that says "division by zero is undefined". I would like to revert to my version and then edit it to say "Division by zero is undefined except in certain rarified, esoteric branches of mathematics. Division by zero is always undefined with real numbers." Would that be acceptable? Paul Beardsell 00:08, 25 April 2006 (UTC)

That's POV; besides, IEEE 754 is hardly rarified mathematics, and limits are high-school material. Now, the article already says "In ordinary (real number) arithmetic, the expression has no meaning." Perhaps you'd like to emphasize that statement somehow? Melchoir 00:18, 25 April 2006 (UTC)

NPOV does not require or consider desirable that all possible interpretations need be given the same weight. IEEE 754 might not be difficult but it is both esoteric and rarified BY DEFINITION. (Strange how definition keeps popping up!) See http://dictionary.reference.com/search?q=esoteric and http://dictionary.reference.com/search?q=rarified Paul Beardsell 00:34, 25 April 2006 (UTC)

The Riemann sphere is to maths what rubber balloons is to architecture. Yes you can construct with balloons but we don't give it much weight in an encyclopedia article on architecture. Paul Beardsell 00:34, 25 April 2006 (UTC)

No one is suggesting that the Riemann sphere be given equal weight with the real numbers. And it is perfectly correct to examine the properties of rubber balloons in an encyclopedia article on rubber balloons. Melchoir 00:49, 25 April 2006 (UTC)

Yes! But this is an article on division by zero not on Riemann spheres. I am not saying do not mention balloons in the architecture article! And I am not saying do not mention Riemann spheres here. Paul Beardsell 00:58, 25 April 2006 (UTC)

Eh, I meant something else; such are the perils of analogies. Melchoir 01:23, 25 April 2006 (UTC)

I agree that we need to say clearly and unambiguously that in the real number system, division by zero is undefined. I know personally one grade school teacher who teachs her students that any number divided by zero is zero, and she smugly tells me where I can take my opinion and shove it when I try to convince her otherwise. It would be nice if her students, who must number in the hundreds by now, could come to wikipedia and get good, clear advice. Starting the article with the statement that division by zero is undefined in the real number system does not prevent other number systems in which division by zero is allowed being discussed further down in the article. Rick Norwood 00:24, 25 April 2006 (UTC)

It does say that unambiguously, in the very second sentence. Let's get our priorities straight here. This is an encyclopedia, not an educational aid for high school students. The first priority is being mathematically correct. --Trovatore 00:37, 25 April 2006 (UTC)

So: Is what I propose above acceptable to you? Paul Beardsell 00:44, 25 April 2006 (UTC)

No, not remotely. The existing article is fine. --Trovatore 00:46, 25 April 2006 (UTC)

Until this very moment I thought we agreed. No Wikipedia article is ever finished. Paul Beardsell 00:48, 25 April 2006 (UTC)

Let me clarify. Of course I am not saying the article should never change. I am saying it should not change in the direction of saying more unequivocally that division by zero is not defined. It is defined, in important contexts. Saying that it is never defined is the sort of laziness that makes it easier to teach a high-school algebra class, and I have some sympathy for algebra teachers (having been one) but it does not need to be any part of our purpose to make life easier for them. --Trovatore 01:18, 25 April 2006 (UTC)

Surely the article has room for improvement... that doesn't involve belittling its own topic or simply being wrong. Any ideas? Melchoir 00:53, 25 April 2006 (UTC)

Is the idea I proposed acceptable to you? I think so: You have suggested a re-emphasis in line with my suggestion. Paul Beardsell 00:58, 25 April 2006 (UTC)

If you're still referring to "certain rarified, esoteric branches of mathematics" then no. Melchoir 01:05, 25 April 2006 (UTC)

Let's get away from the high school. You are appointed mathematical advisor to the prime minister. You are called in urgently. "Tell me about division by zero." What do you do? Talk about Riemann spheres first or last? I think first you say that the operation is not defined on real numbers. "Real numbers? What do you mean?" barks the impatient prime minister. No one is suggesting any material be culled, just re-ordered and re-emphasised. Paul Beardsell 00:58, 25 April 2006 (UTC)

Are you saying we should shorten "In ordinary (real number) arithmetic, the expression has no meaning" to "In ordinary arithmetic, the expression has no meaning" for the benefit of readers who can't deal with parentheses? Melchoir 01:08, 25 April 2006 (UTC)

If the prime minster asked me that, I would say "It is undefined in normal arithmetic". Then I might talk about strange and unusual arithmetics in which it is defined. I would definatly talk about riemann spheres last. First I would talk about arithmetics where A/B = C means A = B*C, which is I feel an important rule of arithmetic, and one reimann spheres throw away. —The preceding unsigned comment was added by Mrjeff (talk • contribs) 08:30, 25 April 2006 (UTC)

I remember a baseball announcer talking about a certain pitcher who had allowed an earned run in the majors, but had not gotten anyone out. He said something along the following lines:

The next time this pitcher gets someone out, his ERA will go down [from what it was when the announcer was speaking]. Because it is infinity.

The announcer wasn't wrong. I don't think he knew anything about the map x |-> 1/x being an anti-order-preserving autohomeomorphism of [0,∞], or at least not in those words. But he was correct that this pitcher's proportion of earned runs to outs was worse than any finite value, and would be better than that as soon as he got someone out. The announcer had intuitively chosen a mathematical structure that was right for the job, rather than one that satisfied some fixed list of algebraic properties. --Trovatore 14:58, 25 April 2006 (UTC)

There are perfectly reasonable algebraic systems on the closed infinite interval

$[0,\infty ]$

for which divison by zero is defined and useful.--CSTAR 15:10, 25 April 2006 (UTC)

I think that's what I just said. But more entertainingly :-). --Trovatore 15:11, 25 April 2006 (UTC)

Hmm, entertaining, assuming you know what an ERA is. That's a bit too culturally specific to th US. Also, in reference to Paul Beardsell's question, there are no prime ministers (yet) in the US where baseball is played. There many be a King, but no prime minister.--CSTAR 15:20, 25 April 2006 (UTC)

I wouldn't be shocked to see the Prime Minister at a Blue Jays game, though I know his sport is really hockey. --Trovatore 15:49, 25 April 2006 (UTC)

Added a section which may help

I have added a section on elementary arithmetic. There are many people who read encyclopedias whose needs are met by simpler explanations; inclusiveness would indicate that "naive" is not a good term for those needs. It is also the case that those who teach elementary concepts require support with concrete examples. 1diot 18:59, 25 April 2006 (UTC)

I agree; the "naive" bit is mostly a language barrier issue though, since mathematicians don't intend the word to be pejorative. (I can certainly see how it can be interpreted that way.) The introduction could use a bit more meat, so I'll try restoring the explanation with a different adjective. Melchoir 19:28, 25 April 2006 (UTC)

Err... on second thought, I'm not sure what to write. The introduction still ought to provide a brief overview of the difficulty, though. Melchoir 19:32, 25 April 2006 (UTC)

I'm not too sure if this section is necessary, but for the time being I've made some work on it. -- Meni Rosenfeld (talk) 14:32, 26 April 2006 (UTC)

Note on the "limits" section

This section makes a lot of hay out of the sign ambiguity. That's irrelevant in the case of the real projective line, and it's also irrelevant if you look at the interval [0,∞], leaving out the negative numbers.

The latter point is important because it has real-world interpretations; there are lots of situations where a quantity that you use as a denominator can meaningfully be zero, but cannot meaningfully be negative. The example of earned run average I've already given (you can retire no batters, but not a negative number). For an example that might resonate more with Europeans—how do they measure automobile fuel efficiency? In Canada it's done in litres per hundred kilometres. Again, a negative denominator is not meaningful (because a car driving in reverse uses positive fuel), but a zero denominator is, and an idling car has infinite fuel consumption per kilometre. --Trovatore 19:43, 25 April 2006 (UTC)

Even better, sometimes it's useful to switch between two different systems. In everyone's favorite example, negative temperature, the inverse temperature resides on the extended number line with signed infinities and an unsigned zero, while the temperature scale has signed zeroes and an unsigned infinity. I'm not sure if it would be original research to mention this stuff, but in a perfect world, it wouldn't be! Melchoir 20:13, 25 April 2006 (UTC)

Temperature has signed zeroes and an unsigned infinity? This seems strange to me. Absolute zero is the lowest possible temperature, and so temperature can only approach positive infinity, not negative infinity. And a temperature of -0 is the same as a temperature of +0. Rick Norwood 21:04, 25 April 2006 (UTC)

I think negative temperature will clear all that up. Melchoir 21:12, 25 April 2006 (UTC)

(Actually, the quote in that article doesn't make the case for unsigned infinity, but it's made elsewhere in the text.) Melchoir 21:13, 25 April 2006 (UTC)

Very interesting. I learned something new today. Can the temperature "wrap around, and pass -0, if it gets "hot" enough? Rick Norwood 22:32, 25 April 2006 (UTC)

Hmm... I have to doubt it. To try to connect back to our topic of dividing by zero, at a temperature of -0, any state of the system with an energy E less than the maximum energy E-star will be suppressed by a Boltzmann factor of

$e^{-(E-E^{*})/-0}=e^{(E^{*}-E)/-0}=e^{-\infty }=0.$

So the system is confined to its highest-energy state(s). Intuitively, this means that it's already as hot as possible. Melchoir 22:55, 25 April 2006 (UTC)

This section, as does most of the article, assumes that the starting point for the discussion is the reals. It shows a possible line of thought for defining division by 0 in it, identifies a few difficulties with this approach, and gives a hint as to what can be modified in order to make something out of it. It demonstrates that negatives are a problem, so we should either identify +∞ with -∞ - leading to the real projective line, or dismiss negatives altogether - leading to [0, ∞], which was not trivial at the starting point (we had to throw away half the objects of our original discussion). The latter should probably also be added to "other number systems".

In short, the section makes a lot of hay out of the sign ambiguity exactly to show why we should restrict ourselves to structures where that's irrelevant. -- Meni Rosenfeld (talk) 14:45, 26 April 2006 (UTC)

I did it through, and...

I have come up with the following: A divided by 0 is infinity. That's right, infinity. There. Does that count as defined? (no, I'm not trying to sound smart) Random the Scrambled 19:12, 19 May 2006 (UTC)

That does count as defined, but the question is how consistent this definition is with the properties we wish division to have. Defining it as infinity is discussed in the article (you should read it again, and perhaps also Real projective line), but there are some subtleties involved (you'll have to let +∞ = -∞, and 0/0 is still a problem). -- Meni Rosenfeld (talk) 09:04, 20 May 2006 (UTC)

Wow, 2=1?

It's amazing how 2 can equal 1, very postmodern. Before you know it, pigs will start flying and I'll be 7 feet tall! 152.163.100.201 19:50, 24 May 2006 (UTC)

The point is of course that if you can divide by 0 the way you divide by other numbers, then 2 = 1; but since 2 ≠ 1, you cannot so divide by 0. (Most people see this particular argument when they're in about 9th grade, don't they?) Michael Hardy 21:22, 24 May 2006 (UTC)

I believe the logic is usually the other way round. You cannot divide by 0, therefore the proof is fallacious. Elle _{vécut heureuse} ^{à jamais} (Be eudaimonic!) 21:50, 24 May 2006 (UTC)

Doesn't this 2 = 1 proof prove that 0/0 is 1/2 :P

No, it does not. --trulyelse01:55, 9 October 2008 (UTC) —Preceding unsigned comment added by 222.155.221.34 (talk)

I don't like how it's worded.

It says:

However, this definition fails for two reasons.

First, positive and negative infinity are not real numbers.

This is as absurd as saying the square root of -1 isn't i because i isn't a number. I think this should be reworded. (I don't know how it should be reworded, though, so I'll leave that to someone else) --Zarel 01:16, 28 June 2006 (UTC)

To say it's not a real number means it's not an element of the set R of real numbers in the same way that i is not a real number. I'm probably missing you're point.--CSTAR 01:40, 28 June 2006 (UTC)

I realize that it's not a real number. However, the article implies that the aforementioned is a reason why the definition fails, which is, in my opinion, absurd. --Zarel 21:21, 2 July 2006 (UTC)

How about this? Melchoir 02:21, 3 July 2006 (UTC)

Much better; thank you. --Zarel 04:30, 5 July 2006 (UTC)

12 / 0 = 0, says school division chart

Your local K-4 school can easily buy a division chart for classroom use (either wall-sized or notebook-sized), which "proves" the opposite:

  Any number divided by zero equals zero

https://www.highsmith.com/webapp/wcs/stores/servlet/Production/Search.jsp?category=88747&A=15015&NSave=0&Au=CategoryId&catalogId=10060&NaoSave=0&An=0&langId=-1&NttSave=division&storeId=10001

Wow, that's both funny and awful. I guess if you want to be generous, you could trace their error back to Brahmagupta... Do you suppose we should mention this in the article? Melchoir 18:41, 2 August 2006 (UTC)

If evidence is found that there actually are schools which have sunk so low as to use this chart, it just might be worth a mention. -- Meni Rosenfeld (talk) 18:46, 2 August 2006 (UTC)

It's actually highly educational. It's a good early lesson that grownups don't always know what they're talking about, and just because you see something on a laminated chart doesn't make it true. --Trovatore 18:50, 2 August 2006 (UTC)

I looked at that web page, and I can't see anything on the chart that says 12/0 = 0. It's too small to read. How do you know it's there? Michael Hardy 20:29, 4 August 2006 (UTC)

Save the image to disk, and blow it up in ImageMagick or Eye-of-gnome or whatever you use. --Trovatore 20:50, 4 August 2006 (UTC)

That is really scary raptor 14:07, 19 August 2006 (UTC)

OK, here it is. BARELY legible:

File:H731660F.jpg

Blatant educational malpractice. If I had a child attending a school that used this, I'd sue. Michael Hardy 21:55, 19 September 2006 (UTC)

(I've now deleted the image for copyright reasons.)

At this web site where the chart saying 12/0 = 0 is sold, the publisher is now soliciting opinions of the product. Everyone: please go there and tell them what you think. Michael Hardy 20:42, 9 November 2006 (UTC)

Update

They have now taken the product off the web and report that they will put up a corrected version. Michael Hardy 03:30, 19 November 2006 (UTC)

I can't believe no one (myself included!) has ever mentioned that the school chart is absolutely correct, provided we interpret division as multiplication with the Moore-Penrose pseudoinverse. -- Meni Rosenfeld (talk) 01:19, 30 December 2007 (UTC)

Which, of course, we normally don't do, in secondary education. FilipeS (talk) 11:04, 31 December 2007 (UTC)

Abstract algebra

From Abstract algebra

Similar statements are true in more general algebraic structures, such as rings and fields. In a field, every nonzero element is invertible under multiplication, so as above, division poses problems only when attempting to divide by zero. However, in other rings, division by nonzero elements may also pose problems. Consider, for example, the ring Z/6Z of integers mod 6. What meaning should we give to the expression 2 / 2? This should be the solution x of the equation 2x = 2. But in the ring Z/6Z, 2 is not invertible under multiplication. This equation has two distinct solutions, x = 1 and x = 4, so the expression 2 / 2 is undefined.

This makes no sense to me, I understood everything in this article till this section. This should be explained better, or atleast cited. . HighInBC 23:34, 12 September 2006 (UTC)

As the name of the section suggests, knowledge of abstract algebra is required to understand it. Links to ring and field are provided for readers without such knowledge. -- Meni Rosenfeld (talk) 05:10, 13 September 2006 (UTC)

Trigonometry in the real projective line

(copied from User Talk:Meni Rosenfeld) Why does the comment you removed not belong there? The range of the tangent and cotangent functions should be viewed as the real projective line with only one point at infinity. Michael Hardy 17:45, 19 September 2006 (UTC)

It just seemed to me that, while this may be perfect for the real projective line article, it is too much detail for this article, where I think that no more than a mention of what it is and its relation to division by zero is required. If, taking this into account, you still think it should be mentioned here, then I have no objections. -- Meni Rosenfeld (talk) 18:00, 19 September 2006 (UTC)

OK, I'm going to try to rephrase it to alter the emphasis. Apparently you though my comment was about the real projective line rather than about how to interpret division by 0 in a certain particular case. Michael Hardy 21:22, 19 September 2006 (UTC)

0/2=0.0.5 ?

I think if you look at it from a quantum perspective (if that's how to word it), zero divided by two, would be zero point zero point five. 0/2=0.0.5 What do you think? - Infurnus 02:40, 23 September 2006 (UTC)

Would you mind clarifying this? For starters, what does a.b.c mean? -- Meni Rosenfeld (talk) 05:59, 23 September 2006 (UTC)

I'm not sure, but let me put it like this: There is no such thing as nothing in our universe, it is all made of SOMETHING. Zero is a placeholder of sorts, so if you're trying to divide "zero" into two, then you would get "half of zero", whatever zero may be. I hope this helps. EDIT: Also: There's no such thing as nothing because everything is made of an infinite amount of things that get infinitely smaller. - Infurnus 11:03, 24 September 2006 (UTC)

Nothing is not a thing; it is the lack of things. How many humans live on the sun? That's right. 0. There are humans, and there is a sun, but there are no humans living on the sun.

The closest mathematically meaningful statement to your idea I can find, is that if ε is a positive infinitesimal (which, of course, cannot be a real number), then ε / 2 is less then ε. -- Meni Rosenfeld (talk) 11:39, 24 September 2006 (UTC)

By the way, I'm not sure there is scientific evidence supporting your claim that "everything is made of an infinite amount of things that get infinitely smaller" (for all we know, the universe could be discrete), although it's a nice idea. -- Meni Rosenfeld (talk) 11:51, 24 September 2006 (UTC)

10/0 = infinity

Another approach to teaching divison to elementary students is to discuss equal parts. To divide 10 by 2, we explain that we can take ten objects and put them in groups of 2. The quotient will tell us how many groups we have. Therefore, to understand, ten divided by zero we can say with ten objects how many sets of zero can you make? You can make an infinite number of sets.

Yes, but this subject is too subtle to be dealt with using purely intuitive notions. For a more serious discussion, take a look at real projective line. -- Meni Rosenfeld (talk) 12:38, 26 September 2006 (UTC)

That example makes me want to answer, "0 with a remainder of 10." Hmmm... --BekiB 17:19, 16 October 2006 (UTC)

Not quite, the remainder must be less than the divisor. That would be the correct answer to "10 divided by infinity", though. -- Meni Rosenfeld (talk) 21:21, 16 October 2006 (UTC)

If 10/infinity can not be "0". Therefore, 10/0 is not infinity.71.114.92.95 16:05, 17 February 2007 (UTC)

Perhaps, what would work would be a new set of infinites, each corresponding to the real numbers, like symbol_x sort of thing to represent the result,which is multiplied by zero to get the value.

eg. x/0 = symbol_x.

symbol_x*0 = x EDIT: sorry, section lower covering this... --trulyelse02:08, 9 October 2008 (UTC)

0/0 = All Real Numbers, n/0 = No Real Numbers

Let us consider two simultanious linear equations, with a, b, c, and d representing constants and x and y representing variables: ${\begin{cases}y=ax+b\\y=cx+d\end{cases}}$ Then, using substitution,

$ax+b=cx+d$

$ax-cx=d-b$

$x={\frac {d-b}{a-c}}$

In the case of these two equations to be lines on the cartesian plane, then a and c represent their respective slopes, b and d represent their respective y-intercepts, and x and y represent the coordinates of the intersection of the two lines. Therefore, when the two lines are parallel, meaning that they have the same slope but different y-intercepts, they will never intersect. And what will the above equation give you as the coordinates of their intersection? Some non-zero number divided by zero. And if the two lines are the exact same, they will have the same slope and the same y-intercepts, and they are constantly intersecting. And what does the equation give you? Zero divided by zero. Therefore, wouldn't zero divided by zero mean "for all x" and a non-zero number divided by zero mean "For no x"? For further proof, consider a logarithm base-1. By the law of changing base, we can dictate that

$\log _{1}x={\frac {\ln x}{\ln 1}}$

And, as the natural log of 1 is equal to zero,

$\log _{1}x={\frac {\ln x}{0}}$

Therefore, the log base 1 of 1 would be equal to 0 divided by 0, and the log base 1 of any other number would be 1/0. This means the same thing as the aforementioned statement of the terms for division by zero. Just wondering if my logic is flawed or anything. --User:ThatOneGuy

Your logic is correct, but uses an argument much too complicated than necessary for the conclusion - 1/0 should be a number x such that 0x = 1, and of course no such real number exists; And 0/0 should be a number x such that 0x = 0, and every real number satisfies it. Since the result of every arithmetic operation is expected to be a unique number, neither is defined in the context of real numbers. -- Meni Rosenfeld (talk) 19:52, 21 October 2006 (UTC)

Forget about all that rubbish!

Can't we just say that division by 0 is undefined, full stop, and there's nothing we can do about it?--67.10.200.101 02:52, 15 November 2006 (UTC)

No. --Trovatore 03:04, 15 November 2006 (UTC)

Yes!--67.10.200.101 03:23, 17 November 2006 (UTC)

Well, literally speaking, of course, we can say it. But it would be incorrect, as the article explains at length. --Trovatore 03:29, 17 November 2006 (UTC)

Why do you say that? I always thought that division by 0 was always undefined, no matter what.--67.10.200.101 03:27, 18 November 2006 (UTC)

Ya thought wrong, friend. As I say, read the fine article. --Trovatore 03:32, 18 November 2006 (UTC)

You're saying that 0 divided by 0 is any number? That's not true!--67.10.200.101

What I said was that you should read the article. If you want more specific direction, concentrate on the bit about the real projective line and the Riemann sphere. --Trovatore 01:40, 23 November 2006 (UTC)

That is, of course, as long as you're satisfied with dividing a / 0 where a ≠ 0. If you want 0/0 to exist, you'll have to resort to wheel theory or something. -- Meni Rosenfeld (talk) 20:22, 25 November 2006 (UTC)

No, it's not a number. That by itself doesn't mean it's undefined in all contexts. Also, even when it's undefined, there's more to say: if y and z both approach 0 as x approaches c, then y/z can approach a specific number, whose value depends on how y and z depend on x. Michael Hardy 02:46, 26 November 2006 (UTC)

Removed paragraph saying 0/0 is well-defined

I removed this :

However, 0 divided by 0 is well-defined. Suppose you had an equation ${\frac {a}{b}}=c$. This statement is always true: ${\frac {a}{b}}=c$ if and only if $bc=a$. So if you set both a and b as zero, the quotient, c, would be all numbers, because 0, which is the value for b in this case, mutliplied by anything, would always equal 0, which is also the value for a in this case.

There seems to be a misconception about the notion of "well-defined". One need a unique real number c for a/b to be well-defined. --Kprateek88(Talk | Contribs) 06:21, 23 November 2006 (UTC)

0/0=1

0/0 = 0¹/0¹= 0⁰

Since any number to the power of 0 = 1 it can be shown that 0/0 = 1—The preceding unsigned comment was added by 86.131.253.224 (talk • contribs) 13:55, November 26, 2006

0/0 = (2*0¹)/0¹ = 2*0¹/0¹ = 2*0⁰

= 2

So 0/0=2. That's why 0/0 is indeterminate - it can be pretty much anything. --Zarel 02:48, 27 November 2006 (UTC)

I was experimenting with that. a^0 = 1 when a =/= 0. If you've ever that qualifier in algebra books ("when x =/= 0"), its usually there in any fraction problem where they must make sure it is defined by not making the bottom zero. The very premise of dividing by zero is undefined. Jaimeastorga2000 05:14, 7 December 2006 (UTC)

Actually, a^0 = 1 even when a = 0 (in other words, 0^0 = 1). Of course, this doesn't make the calculation above correct, since division by zero is not defined in real numbers. -- Meni Rosenfeld (talk) 21:49, 13 December 2006 (UTC)

That is not always so. Different textbooks deals with $0^{0}$ in different ways. Mathematica says $0^{0}$ is indeterminate. It is hard to find a natural definition in general. It is natural to say that $x^{0}=1$ as polynomials, which would suggest (via evaluation $x\mapsto 0$) that $0^{0}=1$. On the other hand $0^{x}=0$ is the natural choice if $0^{x}$ is considered as an exponential function, if we want it to be continuous. In complex analysis it is natural to define in general $a^{b}=\exp(\log(a)b)$ where $\exp(z)=1+z+{\tfrac {z^{2}}{2}}+{\tfrac {z^{3}}{3!}}+\cdots$ and log being the (multivalued) inverse of that function, and with this definition, $0^{0}$ is left undefined. Finally any commutative monoid can be viewed as an $\mathbb {N}$-semimodule, and in that perspective it is natural to say that $a^{0}=1$ holds in general. In summary, the choice depends very much on context. In an encyclopedian article I suggest it would be sound to say $0^{0}$ is undefined. Jesper Carlstrom 10:59, 14 December 2006 (UTC)

IMHO, the existence of textbooks which avoid defining 0^0 is mostly due to narrow-mindedness of their authors (the same way some authors say the vector space {0} has no basis, where it is clear that with any sensible definition, the empty set is its basis). I needn't tell you about the different ways in which 0^0 = 1 makes sense, but I'll comment that as long as we are dealing with 0 as an integer, this is the only possible definition (0^0 is the product of no factors, so it doesn't matter that those factors that aren't there should be 0's). I expect the definition of powers of real (or complex) numbers to be, first and foremost, a generalization of integer powers, so 0^0 should remain 1, even if it doesn't go along with those neat formulae and results in a function which is discontinuous at a point in its domain. Having 0^0=0 won't make 0^x continuous (you're not going to define 0^x=0 for negative x, are you?), only continuous from the right, so it's not a good justification either. -- Meni Rosenfeld (talk) 20:41, 14 December 2006 (UTC)

You are right that if you view $a^{n}$ as a repeated product, the natural definition is $a^{0}=1$ for any $a$ (this was what I said using the language of semimodules). Therefore, if you "expect the definition of powers of real (or complex) numbers to be, first and foremost, a generalization of integer powers", then, of course, you should let $0^{0}=1$. The problem is that this is not the only possible view. For example, the common definition of $a^{x}$ ($a>0$) in real analysis is to say it should be a continuous extension of exponentiation to rational powers. If this should define anything at all in the case $a=b=0$, you must let $0^{0}=0$ (but the problem is unsolvable in this way for negative $a$, which is often taken as an argument for letting $0^{0}$ be undefined as well). I'm not arguing for this way out (in fact, I do agree with you that your view is probably the most natural one), I'm just saying that there is no consensus. Jesper Carlstrom 12:24, 15 December 2006 (UTC)

Indeed, there is no consensus - all I'm saying is that this is very unfortunate. -- Meni Rosenfeld (talk) 12:38, 15 December 2006 (UTC)

Nullity

Take a look at this: http://www.bbc.co.uk/berkshire/content/articles/2006/12/06/divide_zero_feature.shtml It seems someone has solved the problem (at least sort-of). Medevilenemy 19:30, 6 December 2006 (UTC)

Meh. I find that article sensationalistic and naive to the point of plain ignorance. It doesn't explain how "nullity" is anything but a synonym for "indeterminate form" with a fancy symbol, or how a software implementation would be any different from using NaN. I'm willing to keep an open mind and believe it's worthwhile, but only if someone shows me. Or, to put it in Wikipedia terms, if we come up with a decent source. Has anything related to this "solution" been published in a mathematical or educational context? Melchoir 20:34, 6 December 2006 (UTC)

Check out [4], I suppose. I still don't put much into it. He actually mentions one of the Wikipedia examples of division by zero in the 2nd paper, and says it is fallacious because of the assumption that 0/0 = 1. I don't like how he uses infinity as a number - he has defined infinity as 1/0, and negative infinity as -1/0. Not only is infinity not a number, but isn't it trivially demonstrated, since 0 = -0, that this statement is meaningless (+infinity != -infinity)? --RandomPrecision 09:50, 7 December 2006 (UTC)

Is it different from NaN? --MarSch 14:46, 7 December 2006 (UTC)

Aha, I knew I've seen something like this before: Wikipedia:Articles for deletion/Transreal number line. Melchoir 15:47, 7 December 2006 (UTC)

take a look at the howls of derision in the comments box beneath the story. it seems he's passing something trivial and of not much use (rather than something non-mathematical) as something spectacularly original --Mongreilf 15:10, 7 December 2006 (UTC)

I've seen them. The link above to the article[5] is more enlightening though. I don't see anything which is much dfferent from NaN. Not that I checked thoroughly. It is the authors job to do such things and present them clearly. For now I'm going to assume it's just NaN by another name. --MarSch 15:19, 7 December 2006 (UTC)

His biggest fallacy is when he says 1/0 is positive infinity, yet division of a positive number by a negative sign yields a negative answer, and since 0 has no sign, you cannot divide by it and get EITHER positive OR negative infinity. The concept is flawed, and should be promptly deleted from wikipedia.220.253.57.87 13:27, 10 December 2006 (UTC)

He defines 1/0 to be positive infinity and definitions cannot be wrong in the sense of false. --MarSch 14:02, 11 December 2006 (UTC)

Interesting article, but needs references

This is a pretty interesting article. However, I noticed that it currently appears to have no references at all. It really needs citations for its statements and claims to meet Wiki standards for verification and to show that the article isn't mostly original research. Dugwiki 21:30, 7 December 2006 (UTC)

Seconded, I would like to see some textbook references

Done (added Suppes). He offers 5 approaches to the problem. Suppes is very good (is at undergraduate level), cost me $15.00 at Borders a month or so ago.

Also I just found that Tarski 1941 (also in print, Dover) makes a brief 1-page discussion showing how contradictions can appear when what he calls the "identity sign" (equivalence sign) is used (cf page 181-183). I'll add this one too. Bill Wvbailey 16:16, 30 October 2007 (UTC)

It's dividing by zero! It's impossible. What the hell do we need references for? —Preceding unsigned comment added by Larsvolta (talk • contribs) 22:40, 18 May 2008 (UTC)

Did you even take a look at the article? Dividing by zero is not necessarily impossible. -- Meni Rosenfeld (talk) 23:32, 18 May 2008 (UTC)

The Zero-Infinity(zi) Theory

"The Zero-Infinity(zi) Theory" by David Tulga is a complete algebra for dealing with division by zero. It's way more interesting and much more important than the "easy" identity that professor Anderson "discovered", although his identity and solution is also interesting. [6]

ZI and Nullity are much like the complex number system opening new horizons, well, dimensions really.

Please add the external link to David's site to this article, about division by zero, as it is most amazing and people need to find out about it.

- Peter

Sorry, but Anderson's material isn't in this article either. Also, Wikipedia's external links aren't there for promotional purposes, and they have standards: see Wikipedia:External links. Melchoir 00:33, 8 December 2006 (UTC)

I'm not affiliated with David Tulga. I don't know him. I'm not "promoting him". He has made a significant breakthrough in mathematics and I'm trying to have something about this breakthrough included in Wikikpedia on the appropriate page - which is this page. All the material he has published on this amazing theory is on the linked page as far as I know. In what way can I add material and get by the snobish wikipedia attitude that I'm promoting him or am some sort of spamer - these are not the case. This is frustrating, especially so since I've aready been to the web page that you linked to and that page directed me here. Arrrgg.. Would someone please add something to the this page about this breakthrough. Thank you. —The preceding unsigned comment was added by 69.251.244.134 (talk) 00:39, 8 December 2006 (UTC).

This is not a breakthrough, at all. The idea has been brought before several times, and it just doesn't work. We even have a reference desk question about this subject. Also, unless it can be cited through peer-reviewed papers, we can't mention it on Wikipedia. That link in this page would violate the no original research rule. ☢ Ҡi∊ff⌇↯ 01:43, 8 December 2006 (UTC)

First, let me assure you that none of us is against division by zero, and as you can see in the article, there are well-defined mathematical structures in which such a thing exists.

What we are against is inconsistent mathematical inventions, and while I'm sure that David Tulga is a very nice person, unfortunately he doesn't really know what he's talking about. For starters, he doesn't define his structure in a way that makes sense mathematically; and even if you try to put some sense into it, you will end up with multiplication not being associative (0*(2*Zi) = 2 ≠ 1 = 0*Zi = (0*2)*Zi), which is not what we want. -- Meni Rosenfeld (talk) 10:47, 8 December 2006 (UTC)

Problem with Section 3.1 "Proof that 1=2"

Has anybody noticed that although this 'proof' shows how dividing by zero leads to absurdities it's very first step is fatally flawed? It claims that you can factor $x^{2}-x^{2}$ (which is itself zero) in two different ways the first of which being;

$(x^{2}-x^{2})(x^{2}+x^{2})$

Am I the only person who has noticed that if you factor this out it becomes

$2x^{2}-2x^{2}$

so really even the second line claims 2=1!

Perhaps this derivation (based on the same principle) would be better

$a=b$

$a^{2}=ab$

$a^{2}+a^{2}=a^{2}+ab$

$2a^{2}-2ab=a^{2}-ab$

$2(a^{2}-ab)=1(a^{2}-ab)$

$2=1$

Again because a=b you are dividing by zero in the last step. —The preceding unsigned comment was added by 83.216.146.141 (talk) 12:24, 12 December 2006 (UTC).

Well, yes and no. The general formula $(a+b)(a-b)=a^{2}-b^{2}$ is used to deduce that $(x+x)(x-x)=x^{2}-x^{2}$. True, this is also equal to $2x^{2}-2x^{2}$, but this is obvious since all those expressions are 0. If you ask me, any variation of this argument is silly, it's just an obfuscated way to state that 1*0=2*0, so if you allow division by zero then (under certain assumptions) 1=2. -- Meni Rosenfeld (talk) 18:16, 12 December 2006 (UTC)

Clearly the <a href="http://wiki.riteme.site/w/index.php?title=Division_by_zero&oldid=107222503">current example</a> and 83.216.146.141's example are overly complicated. I'm elaborating on Meni Rosenfeld's "1*0=2*0" to cut down on distractions. --Dreddlox 03:42, 14 February 2007 (UTC)

Fraction formatting (again)

There are three ways to format fractions that occur inline within text:

• Fraction ^a/_b within text → ^a/_b.
• Fraction $\textstyle {\frac {a}{b}}$ within text → <math>\textstyle\frac{a}{b}</math>.
• Fraction ${\begin{matrix}{\frac {a}{b}}\end{matrix}}$ within → <math>\begin{matrix}\frac{a}{b}\end{matrix}</math>.

The first form is the standard HTML way, but does not always mix well with <math> markups. And while it's been pointed out that the third form is the documented way to do inline fractions in T_EX, the second form appears to work just as well, and has the benefit of being easier to edit. So that's what I've used. Purists may want to revert my changes, but please choose a consistent format if you do. — Loadmaster 16:42, 12 December 2006 (UTC)

There are other formats, but they are not really fractional forms and look terrible:

• Fraction a/b within text → a/b (no markup at all).
• Fraction $a/b$ within text → <math>a/b</math>.

These kinds of formatting that should be fixed. — Loadmaster 16:47, 12 December 2006 (UTC)

1 / 0 = Q^-1

Here's a theory that can actually solve division by zero errors in computers: comp.theory Division by Zero. It starts by defining x / x = 1 for any x, including zero. This then makes it possible to compute (a / x) * x without risk of division by zero error; by keeping the value of a but setting a binary field that indicates division by zero. By multiplying with zero, the division by zero is cancelled.

Algebraically it introduces a number Q, defined as 1 * 0 = Q. It has an inverse Q^-1 so that Q * Q^-1 = 1, ultimately leading to 1 / 0 = Q^-1. It is the exponent of Q that corresponds with the value of the above mentioned binary field. Q has some similarity with the imaginary i: while i only becomes a real number after squaring it, Q only becomes a real number after multiplying with Q^-1. But it's orthogonal to i. The square root of -4 / 0 would be 2*i*Q^0.5, which immediately illustrates that the exponent of Q can be a real number and i and Q define a three-dimensional space. For computer use, Q^-2, Q^-1, Q⁰ and Q¹ would likely suffice, for which only two bits are required. C0d1f1ed 23:10, 1 January 2007 (UTC)

Looks like a reasonable theory, but of course, without real references we cannot include it in the article. Also, does this theory really help? Seems to me that any result obtained after division by zero has occured is bound to be nonsensical and useless. -- Meni Rosenfeld (talk) 21:55, 1 January 2007 (UTC)

One benefit is that for can go into the Q domain and then recover the original state by performing the inverse operation without hitting NaN. i.e., it will give (5/0) = 5*Q, so (5/0)*0 = 5*Q*0 = 5*(Q*(1/0)^-1) = 5*(Q*Q^-1) = 5*1 = 5. This might be useful. deeptrivia (talk) 22:14, 1 January 2007 (UTC)

Allow me to make a comparison: Is it sensical to take the square root of a negative number? If division by zero is useless then we might question the usefulness of complex numbers as well. And indeed every engineering problem can be solved without them. Computers implement complex numbers as nothing more than a tuple of scalars. Also, like I said before, every complex number where the imaginary component is not zero is meaningless in real life, and cannot be used as an end result. Likewise, 1 / 0 = Q^-1 only becomes meaningful again after multiplying by zero. Anyway, there certainly is some real life use to it. Say we have an external software library that gives us a / x, but we really want a regardless of the value of x. We can multiply by x but when it's zero a classical implementation will return NaN and throw a division by zero exception, potentially crashing the system. Even if we do catch the exception and prevent the NaN from propagating, there is no way to retrieve a. Now, it is arguable that the library should really be able to give us a directly, but for performance and storage reasons it's not unreasonable that it only keeps a / x around. An alternative would be that it returns a / x, except when x = 0, it returns a. The application then has to check whether or not x = 0 before doing the multiplication by x. This is also inefficient, but essentially the software implementation of the above theory. By adding the exponent of Q in a binary field to the floating-point representation, we'd get an efficient hardware implementation. And it doesn't only work for situations where the division by zero is intended. Lots of software has (risk of) unintended divisions by zero, just because of human error. By introducing the Q exponent, division by zero can be harmless (where the programmer intended x / x = 1). On a system with complex numbers the programmer also doesn't have to beware of taking square roots of negative numbers, as long as the imaginary component is zero in the end result... There is no real reference for this theory. I, Nicolas Capens, started thinking about division by zero after reading about Dr. Anderson's nullity number and realizing that it doesn't solve related computer crashes. So I devised a theory that does attempt to fix the problem, and posted it on comp.theory. C0d1f1ed 23:10, 1 January 2007 (UTC)

What is the error in this computation: $2=2\cdot 1=2\cdot (0\cdot Q^{-1})=(2\cdot 0)\cdot Q^{-1}=Q^{1}\cdot Q^{-1}=1$? Jesper Carlstrom 14:11, 3 January 2007 (UTC)

I think it should go:

$2=2\cdot 1=2\cdot (0\cdot Q^{-1})=(2\cdot 0)\cdot Q^{-1}=2\cdot (1\cdot 0)\cdot Q^{-1}=2\cdot Q^{1}\cdot Q^{-1}=2$

In that, 2*0 is not Q¹, but 2Q¹ — Kieff 14:57, 3 January 2007 (UTC)

Are you saying that $2\cdot 0=0$ is not true in this system? Because according to the axioms: $2\cdot 0=2\cdot Q^{1}$. Jesper Carlstrom 16:40, 3 January 2007 (UTC)

I'm not personally saying anything, that's just how I got the theory described above. If we defined Q¹ as 1 * 0 = Q¹, then what is 2 * 0? 2Q seems like the logical option. As far as I could understand, x * 0 is not zero in this system. Either way, it seems that any attempt at validating division by zero would break commutativity in our current multiplication rules. — Kieff 16:47, 3 January 2007 (UTC)

As I see it, this theory doesn't really allow division by 0, it just doesn't have a 0. The number 1 is characterized as being the multiplicative identity, so I would expect that Q = 1 * 0 = 0 (otherwise, we let go of more alegbraic content than I am willing to accept). So your "0" is not really zero, it is the positive number Q. We then see that 0 = Q ≠ 2Q = Q + Q = 0 + 0, so this Q (or 0, however you call it) is not an additive identity. So as I said, this structure does not contain 0, it contains all numbers of the form $aQ^{n}$ where a is a nonzero real, n is an integer and Q is a formal element taken to be positive but less than any real number. There is nothing fundumentally new here (mathematically speaking, anyway), and the lack of a 0 element makes me still skeptic about the applicability.

Also, I reiterate my point that if a division by zero was caused by some error, then the results are unreliable, and it might actually be better for the program to crash than to continue to run and give misleading results. I understood that you also intend to add a flag indicating that a division by zero has occured, presumably for the purpose of controlled termination of the program - but if we agree that termination is desired in such a case, I don't see how adding these extra elements is better than just giving some arbitrary result (say 0) and turning on the flag. -- Meni Rosenfeld (talk) 16:58, 3 January 2007 (UTC)

It certainly doesn't just add a flag to terminate the application, it actually neutralizes certain division by zero errors to allow the application to continue normal execution. I'll start with a practical example to clarify the potential usefulness. In 3D computer graphics, coordinates (x, y, z) are projected to the screen by dividing x and y by z to create the perspective effect. For all of these projected points (X, Y) we can draw a straight line from the middle of the screen through the point. But when z is zero, the best we get with real numbers is ($\infty$, $\infty$), with absolutely no information about the original x and y (needed for drawing the line). This is clearly bad since z should have absolutely no influence on the angle of the line, even if it's zero.

By introducing Q^-1 for division by zero, the projection when z = 0 gives us (x Q^-1, y Q^-1). Now we can compute the angle of the line with atan2, a function in the C programming language, which is defined as: atan2(x, y) = 2 atan(y / (sqrt(x² + y²) + x)). Evaluating this for (1Q^-1, 2Q^-1) gives us:

= 2 atan(2Q^-1 / (sqrt((1Q^-1)² + (2Q^-1)²) + 1Q^-1))

= 2 atan(2Q^-1 / (sqrt(1Q^-2 + 4Q^-2) + 1Q^-1))

= 2 atan(2Q^-1 / (sqrt(5Q^-2) + 1Q^-1))

= 2 atan(2Q^-1 / (2.236Q^-1 + 1Q^-1))

= 2 atan(2Q^-1 / 3.236Q^-1)

= 2 atan(0.618)

= 2 * 0.554

= 1.107

No division by zero exception, no infinities or NaNs to deal with, just the angle we expected.

Indeed it's arguable that 1 would no longer be the multiplicative identity if 1 * 0 = Q, unless Q = 0. I didn't choose the symbol by accident, the visual resemblance to 0 is intended. I could have just as well written (5 / 0) * 0 = 5*0^-1 * 0 = 5, but that would have been very confusing (watch the spaces, 5*0^-1 is one element). It's a lot more comprehensible to introduce an element Q that has an exponent and that we can multiply and add. This also allows to make a difference between 0*Q⁰, 0*Q¹ and 1*Q¹. Just to illustrate: 1 - 1 = 1*Q⁰ - 1*Q⁰ = 0*Q⁰, 0 * 0 = 0*Q¹ = 1*Q² and 1 * 0 = 1*Q¹. I haven't had the time yet to make a complete set of sound axioms but I hope this shows we don't necessarily have to make big algebraic sacrifices to make this theory work. It's all based on preservation of information, just like complex numbers allow sqrt(-x) to return sqrt(x)*i instead of NaN... C0d1f1ed 21:26, 3 January 2007 (UTC)

Some questions to Nicolas: it seems from your proposed definition $Q=1\cdot 0$ that you do not suggest that $1\cdot 0=0$, because if you did, you would not need to introduce Q. Is this so? Then I find it natural that also $2\cdot 0$ be different from zero. (I think in terms of a free abelian group generated by the real numbers and divided by some subgroup of elements that should be equal to 1, if that helps.) But I am puzzled by your axioms. What about the following calculation: $0=(1^{-1}\cdot 1)\cdot 0=1^{-1}\cdot (1\cdot 0)=1^{-1}\cdot (1\cdot Q)=(1^{-1}\cdot 1)\cdot Q=Q$? I am aware that I use a bit more than the axioms in your list, but it seems from your motivation that you accept implicitly rules like associativity and inverses of each element. I would be interested in knowing exactly at which step of my calculation I violate your intuition. Best, Jesper Carlstrom 21:14, 3 January 2007 (UTC)

Indeed, I introduce Q to preserve more information. 1 * 0 is not the same as 0 * 0 or 2 * 0. These would be Q, Q² and 2Q respectively. In your calculation you assume that 0 = 1 * 0 = (1^-1 * 1¹) * 0 for the first step. This is only true if you actually started from Q instead of 0. You could have instead written 0 = 1 - 1 = 1*Q⁰ - 1*Q⁰ = 0*Q⁰. C0d1f1ed 22:34, 3 January 2007 (UTC)

[Edit conflict]Well, if 1 is not an identity, I assume there is no identity, so there is no sense in discussing inverses. So I think that's the problem with this particular calculation. Anyway, I'm still skeptic that such a structure can be defined with enough algebraic properties to be worthwhile, but good luck trying to find such a construction! -- Meni Rosenfeld (talk) 22:48, 3 January 2007 (UTC)

Nicolas explicitly states that $Q^{0}$ is an identity, so I used $1^{-1}$ as notation for the solution to $x\cdot 1=Q^{0}$, which Nicolas seems to accept as existent. If this is not in the lines with your notation, Nicolas, please let me restate my calculation as follows. Consider $(Q^{0}/1)\cdot (1\cdot 0)$. You suggest the formula $(a/b)(bc)=ac$. With this we can simplify $(Q^{0}/1)\cdot (1\cdot 0)=Q^{0}\cdot 0=0$. But you also say $1\cdot 0=1\cdot Q$ so we could also simplify $(Q^{0}/1)\cdot (1\cdot 0)=(Q^{0}/1)\cdot (1\cdot Q)=Q^{0}\cdot Q=Q$. Both seem to be according to the rules you suggest. Jesper Carlstrom 07:58, 4 January 2007 (UTC)

Thanks all for the useful critique. Based on the remaining conflicts in the theory I've started the following subsection. C0d1f1ed 12:16, 4 January 2007 (UTC)

No zero gets left behind

Q really equals 0. But the problem with just writing 0 is that we'd be tempted to think $0\cdot 0=0$, which would be a loss of information. The real reason why division by zero is a problem in real number theory is that multiplying by zero annihilates any number. When we write $x\cdot 0=0$ there's no way to retrieve x again. So to have any hope of recovering from a division by zero we need to make sure that the right-hand side still has information about x. In a first attempt I could write $x\cdot 0=x0$, where $x0$ really represents one element of the new algebra. But it's far too easy to mistakenly evaluate $x0$ to 0. To prevent that I'd rather write $x\cdot 0=xQ$. I'll get to the problem of keeping 1 the multiplicative identity right after the following analogy.

i really equals ${\sqrt {-}}1$. However if we didn't use a new symbol we'd be easily tempted to write ${\sqrt {-1}}{\sqrt {-1}}={\sqrt {-1\cdot -1}}={\sqrt {1}}=1$. Also writing $a+b{\sqrt {-1}}$ would confuse people because they can't evaluate this, while it's actually just one complex number. Hence writing i is really useful. For the same reason I believe it makes sense to introduce Q even though at first it appears to bring nothing new. It really adds a new dimension. Every number $aQ^{x}$ can also be written as a 2-tuple $(a,x)$.

So let's go head-on with $1\cdot 0=Q$ and $1\cdot 0=0$. This already dictates that Q equals 0. To better grasp this let's first define multiplication more precisely. The basic axiom is $aQ^{x}\cdot bQ^{y}=a\cdot bQ^{x+y}$. But since we want to keep all information about multiplications (and divisions) by zero it's necessary to add the exceptions $0Q^{x}\cdot bQ^{y}=bQ^{x+y+1}$, $aQ^{x}\cdot 0Q^{y}=aQ^{x+y+1}$ and $0Q^{x}\cdot 0Q^{y}=1Q^{x+y+2}$. Armed with this we can see that both $1\cdot 0=0$ and $1\cdot 0=Q$ hold if 0 is defined to be equal to $1Q^{1}$. The only problem is that this doesn't comply with the first axiom I wrote on comp.theory: $a=aQ^{0}$. But if Q is really 0 then we can see that $0Q^{0}$ equals $1Q^{1}$. The zero in front of Q just gets 'promoted' to the exponent of Q. This preserves information. Both representations are equivalent. Note that we can project the numbers of this algebra onto the real number axis by replacing Q with 0. Every element $aQ^{x}$ with either $a=0$ or $x>1$ projects onto 0.

The question now arises what's the additive identity. If real numbers are $aQ^{0}$ and we can only add numbers with the same Q exponent then $0Q^{0}$ is the only logical answer. So clearly we do have two representations of the real number 0. We can extend every axiom to ensure consistent behavior for $1Q^{1}$ and $0Q^{0}$, but maybe we can just state $0=1Q^{1}=0Q^{0}$.

So what do we have now: 1 is still the multiplicative identity, also for both representations of 0, so I hope this is a satisfying solution for Meni. And both of Jesper's calculations actually result in $1Q^{1}$, with 0 just being the real number representation. C0d1f1ed 12:16, 4 January 2007 (UTC)

This makes me even more confused. You said that my calculation was erroneous, but that it would be correct if I replaced 0 by Q. But now you say that 0 and Q are the same, only that you prefer the notation Q. I take it that you have changed your mind. If you admit that 0=Q and $1=Q^{0}$ it seems that you have a quite standard algebraic creature in mind: the group G/H, where G is the free abelian group generated by the real numbers and H is the subgroup consisting of the finite products of 1 and $1^{-1}$. In this group $0\cdot 0\neq 0$ holds, and all your axioms hold except $0/b=0\cdot 0$. In fact, I think this axiom of yours is problematic. What about the following calculation? $0\cdot 0=(0/1)\cdot (1\cdot 0)=(0\cdot 0)\cdot (1\cdot 0)=0^{3}$ Jesper Carlstrom 14:07, 4 January 2007 (UTC)

Your calculation wasn't incorrect, but it was confusing to me where the conversions happened and what the correct representations of 0 are. Just writing $0=Q$ is misleading because that's like stating $0=(1,1)$ if we use the 2-tuble notation. They are equal but not identical (just like $0=0+i\cdot 0$ but this does not imply that $i=0$ or that complex numbers are useless). I stand corrected about that axiom, it should be $0/b=1Q^{1}/bQ^{0}=1/bQ^{1}$. Anyway, if this is really a G/H group, has it been used before to define division by zero? Thanks. C0d1f1ed 15:21, 4 January 2007 (UTC)

OK, then it seems we have come to a point where we understand eachother. I don't know if this has been used to deal with division by zero, but the methods are standard in group theory. I wrote down a note very quickly, you can find it here: [7]. I'd be interested in hearing your opinion. The choice of H in the note is a bit different, in order to make the model better. Jesper Carlstrom 16:14, 4 January 2007 (UTC)

Looks pretty nice! I'm not fully familiar with all the notation and algebraic constructs though. But there appears to be a small typo in the commutative axiom. I think it's also worth noting that Q exponents can probably be reals, not just integers. But maybe it's best to only look at integers first... C0d1f1ed 16:50, 8 January 2007 (UTC)

Here's what bothers me. If I understand correctly, you said that 0Q⁰ = 1Q¹ and that this is the additive identity. But that means 1Q¹ + 1Q¹ = 1Q¹ where I would expect 1Q¹ + 1Q¹ = 2Q¹. There's also the thing about addition not cancelling, that is, 1Q³ + 1Q¹ = 1Q¹ = 1Q² + 1Q¹ but 1Q³ ≠ 1Q². Also, if 1Q¹ is what we would call "zero", then 1Q² is in a sense "more zero than 0", which is counterintuitive.

All in all, I think this is a nice idea, but ultimately any way to manifest it turns out to be inferior to standard constructions, such as the real projective line and wheel theory for division by zero, hyperreal numbers and surreal numbers for nonzero infinitesimals, and homogeneous coordinates for applications like computer graphics. -- Meni Rosenfeld (talk) 22:57, 6 January 2007 (UTC)

0Q⁰ = 1Q¹ but only when comparing as a real number. Both project to the real 0, but the real 0 does not map to both representations simultaneously. Like I said before, they are equal but not identical. Ok maybe that's too misleading. They have equal real values.

Basically, the identity element for addition is not the same as the cancelling element for multiplication. There really is no cancelling element for multiplication. This is unavoidable, because what we want is that zero (the multiplicative of the two) doesn't cause cancellation (and hence division by this zero is well defined). Anyway, 0Q⁰ is clearly the additive zero and 1Q¹ the multiplicative zero (that doesn't cancel). I still have to take the time to write down all the axioms... C0d1f1ed 16:50, 8 January 2007 (UTC)

It is impossible to tell if what you say makes sense because you have not defined the terms and obviously you do not use them in the ordinary way. Before you admitted that 0=Q, now you say they are equal but not identical. You have not defined what "equal but not identical" means. The claim that "0Q⁰ is clearly the additive zero" is impossible to verify, because we do not have any axioms to compute with. I think you have to lay down the axioms before we can continue the discussion. You seem to admit that your structure is a group under multiplication and that the distributive law holds. Then it cannot be a group under addition. Hence the question is only which one of the group axioms you do not want to have. For instance, does the equation 2+x=0Q⁰ have a solution? Jesper Carlstrom 09:41, 12 January 2007 (UTC)

0/0 = all real numbers

Simple algebra can solve this. We're solving for x. x = 0/0. Multiply each side by zero. You get: 0x=0. Since 0 times any real number = 0, x = all real numbers. In essence, you're asking how many times does zero go into itself. Zero goes into itself any number of times.76.21.45.13 07:34, 17 January 2007 (UTC)

First, multiplying both sides of an equation by 0 is not a valid method of solution, because it creates new solutions.

Second, we would want the result of an arithmetic operation to be a number, not a vague statement like "all real numbers". While it is of course true that for any $x\in \mathbb {R}$, 0x = 0, it is not true that for any $x\in \mathbb {R}$, 0/0 = x, since that would imply 1 = 0/0 = 2, but the real number 2 is different from 1. -- Meni Rosenfeld (talk) 14:20, 17 January 2007 (UTC)

Well, remember that an equation can have multiple solutions. If the square of x=4, then x=both 2 and negative 2. That doesn't mean that 2=(-2). It is still correct to say that 0/0 can be interpreted as any real number. 76.21.45.13 07:36, 1 February 2007 (UTC)

Certainly every real number is a solution to 0x = 0. But what is sought is a unique solution. That doesn't exist. Michael Hardy 21:33, 1 February 2007 (UTC)

We agree on that. 76.21.45.13 12:51, 2 February 2007 (UTC)

I think the problem may lie in the way you have phrased your last observation. If $x^{2}=4$ then x = 2 or x = -2, not "x = 2 and x = -2". Another correct phrasing is that both 2 and -2 are solutions to the equation $x^{2}=4$. However, it is not true that both ${\sqrt {4}}=2$ ${\sqrt {4}}=-2$. ${\sqrt {}}$ is an operator which for every nonnegative real number y returns the unique nonnegative solution of $x^{2}=y$, so ${\sqrt {4}}=2$, period. Likewise, / is an operator which, for every pair of real numbers a and b for which there is a unique solution to the equation $bx=a$, returns this unique solution. A unique solution exists iff $b\neq 0$. Maybe reading up on function (mathematics) will help. Note that it is possible to discuss multi-valued functions, but this concept is not as useful as ordinary functions, and in any case it doesn't allow you to have something equal to two different things. -- Meni Rosenfeld (talk) 12:56, 3 February 2007 (UTC)

Would you agree though that 0/0 "can" be any real number? Just as if x squared = 4, x = -2 or 2 (as opposed to "and"), if x = 0/0, x can be 1 or 2 or any other real number. 76.21.45.13 06:15, 4 February 2007 (UTC)

No. 0/0 first has to be defined in order for it to be anything. As there is no sensible way to define it in the context of real numbers, it is left undefined, and you can see that the definition I gave above for division does not define a value for 0/0. Don't confuse "x = 0/0" with "0x = 0". The former is a meaningless string of symbols, the latter is an equation for which any real x is a solution.

If you really wanted, you could redefine division so that a/b will be the set of all real solutions to the equation bx=a, and then 0/0 would be $\mathbb {R}$. But such a definition seems rather useless (since we can discuss this set directly), and just as importantly, it is not conventional. -- Meni Rosenfeld (talk) 16:31, 4 February 2007 (UTC)

Meni, I'm pretty sure that I understand what you say now. You can't multiply both sides of an equation by zero to get a solution. For instance, if we had x=2+1 as an equation and we multiplied each side by zero, it would change the number of solutions. Even though one is tempted to say that 0/0 means "0 times what number = 0?," it's not actually the case. I have another question though. Is 1/0 "infinity," or is it also undefined? 76.21.45.13 13:30, 13 February 2007 (UTC)

The article actually discusses the 1/0 issue. As you probably know, the most commonly used number system is the set of real numbers, in which there is no element called "infinity". So as long as we discuss real numbers, the answer is that 1/0 is undefined. However, it is possible to discuss extensions of the set of real numbers, such as the real projective line, in which there is an element ∞ and division is defined in a way that 1/0 = ∞. This is not without cost - in this system, $\infty =-\infty$ which is somewhat counterintuitive, and division does not have the usual algebraic meaning - even though 1/0 = ∞, it is not true that $\infty \cdot 0=1$. -- Meni Rosenfeld (talk) 20:22, 13 February 2007 (UTC)

0/0 = nothing, everything, and all inbetween

0/0 equals anything and everything, including zero, any finite number, and infinity.

You can add zero to zero as many finite times as you want, and you still get zero. So x = 0/0 works for all finite positive x. Similarly, you can subtract zero from zero as many finite times as you want, and you still get zero. So x = 0/0 works for all finite negative x.

Also you may choose to add/subtract zero zeros to/from zero, and you would still get zero. Hence, x = 0/0 works for x = 0.

Furthermore, you can add/subtract zero to/from zero infinity times, and you still get zero.

In a nutshell, ignoring the universal limit of the speed of light, a speed of 0km/0hr could equate to anything from 0 right through to infinity, in both positive and negative directions.

Hence x = 0/0 is the same as defining x as having a domain of -infinity right through to +infinity.

Thanks, Glooper 08:30, 4 April 2007 (UTC)

Scientific trivia

Is the example about impedance really correct? It seems to me that it is about low impedance (V=0, C>>0) rather than undefined impedance. An example of undefined impedance would be when there is no connection between the terminals at all (physically very rare). Or am I confusing things? Jesper Carlstrom 08:11, 10 April 2007 (UTC)

You're right, but the section is also irrelevant. And while I was at it, I also removed the section on the pseudoinverse because I couldn't see it relevance. -- Jitse Niesen (talk) 13:56, 10 April 2007 (UTC)

I don't agree that the sections you removed are irrelevant. It is true that pseudoinverses are "irrelevant for scalars", as you say in the history, but who said this article is about scalars? People that visit this article are probably interested in views of division by zero in different contexts. It is also relevant to have examples from physics that give rise to division by zero and suggests solutions. I think the sections should be put back, but the Scientific trivia section should be corrected, and I would prefer a different name, like Examples from physics or whatever. Jesper Carlstrom 11:23, 11 April 2007 (UTC)

Thank you, I added this example because it happened to me once when I was connecting a fused wire from my car battery to the subwoofer amplifier in my car. At the subwoofer end, while reconnecting the wires I accidentally let the positive lead touch the ground terminal, completing the circuit. There was a spark, and the fused connection (near the battery) broke immediately. When I bought the replacement fuse, I found out it was rated for 50 Amps, and it prevented the smaller fuse on the amplifier itself from breaking. In my case, $Impedance=12.5V/0Amps(currentdraw)$, demonstrating the high energy flow that burnt the fuse. But as I rethink this issue, this only applies in physics, since the real idea of divide by zero is the ambiguity created by the non existent mulplicative inverse of the operation. [User:Petershen1984|Petershen1984]] 18:59, 20 April 2007 (UTC) aka User:220.133.92.72

Why anything/0= infinity

Imagine an object traveling at infinite speed. Such an object would instantly cover any distance. Since speed=distance/time, infinity=anything/0. --68.40.85.31 16:14, 10 April 2007 (UTC)

Let 1/x, where x is very very close to zero but non-zero. If this value is applied to the slope of a linear equation, the line would be near (but NOT!) vertical. If x takes on a value that is infinitely close to but non-equal to zero, the resulting value would be infinitely large. However, in linear equations, a slope of y rise over 0 run (y / 0) is undefined simply because the idea of a vertical line (x = A) is that, y could take on any value between negative and positive infinity and does not change the value of x. In other words, it could be 1, could be 2. The idea that divide by zero is undefined is simply because it results in ambiguity as there is not one single mulplicative inverse of x / 0.--Petershen1984 18:57, 20 April 2007 (UTC)

Division by an unknown that may be zero

Would it be appropriate for the article to briefly mention an issue that students of algebra often trip up on, namely that you should not divide an expression by an unknown since the unknown might take the value 0. (If I recall correctly...). Thanks. Itsmejudith 16:08, 1 May 2007 (UTC)

Sounds like it might be appropriate, although you probably should have a reference for it to verify that indeed this is a common problem among algebra students. Dugwiki 16:38, 1 May 2007 (UTC)

Something like this: An equation says

$x^{2}-6x=0.\,$

A student rashly divides both sides by x, getting

$x-6=0,\,$

and concludes, incorrectly, that the solution of the original equation is x = 6. In fact, if

$x^{2}-6x=0\,$

then

$x(x-6)=0,\,$

whence

${\mbox{either }}x=0{\mbox{ or }}x-6=0,\,$

and finally

${\mbox{either }}x=0{\mbox{ or }}x=6,\,$

so the original equation has two solutions. Michael Hardy 19:12, 1 May 2007 (UTC)

Good idea. We hardly need a reference for the claim that this is a common problem. I think that noone that has ever been a maths teacher disagrees. The example above can be simplified as follows:

$x^{2}=6x$

student divides by x,

$x=6$.

In my experience, it is an even more common mistake to divide by x when it is an obvious factor at both sides. Jesper Carlstrom 06:32, 2 May 2007 (UTC)

I wasn't particularly proposing that the article would say that it is a common problem, only that dividing by an unknown can introduce error. The example above could be included, as it is a good reminder to those whose memory is getting hazy. Itsmejudith 11:52, 3 May 2007 (UTC)

Here is a nice example why it is actually good that division by zero cannot be performed. Suppose we know that y is defined in [0,2] and that it is continuous with $y(0)=-1$. We also know that it is differentiable in the interior and that $y'\cos ^{2}x=y^{2}$. We want to compute $y(2)$.

When x is close to zero, we know that $y^{2}\cos ^{2}x\neq 0$ so we can divide by it (assuming that this could be done everywhere yields a problem below). Hence, fore these x we have

${\frac {y'}{y^{2}}}={\frac {1}{\cos ^{2}x}}$

hence $D(-y^{-1})=D\tan x$ so that $-y^{-1}=\tan x+C$, where C is a constant. Using $y(0)=-1$ we get $1=\tan 0+C$, so that $C=1$. Hence

$y={\frac {-1}{\tan x+1}}={\frac {-\cos x}{\sin x+\cos x}}$

One is now tempted to compute y(2) using this formula.

However, the division above cannot be performed for x=π/2, so the solution we got is correct only for smaller x. In fact, the general solution is

$y={\frac {-\cos x}{\sin x+C(x)\cos x}}$

where C(x)=1 for x<π/2, but C(x) is unknown for x>π/2, although it must be constant there and C(x)<-tan 2 (approx 2.18). Hence, it is in fact impossible to compute y(2), except that we can say that it must be positive. Note the potential disaster! If someone believes he can compute as if you could always divide, he would convince himself that the status of some process could be computed at time 2, although the process could actually develop in many possible directions for x>π/2. I think this shows that it is good that we cannot divide by zero. This helps us find unexpected solutions to problems, as above. Jesper Carlstrom 08:49, 4 May 2007 (UTC)

0/0 = God?

Since about fourth grade I began to question the mathematical rule that something divided by zero is undefined. It only made sense to me that something divided by nothing yields infinity. I used to think I was original with this idea until I recently looked on here and found that Bhaskara II made the same conclusion about a little less then 900 years ago. So here comes the namesake of this thread:

What happens if you divide zero by zero? Various mathematical rules come into play.

Rule #1. First things first, officially you can't divide by zero, so it is undefined. Rule #2. Secondly, any fraction with the numerator of zero is zero. Rule #3. Thirdly, any fraction with an identical numerator and denominator equals one. Rule #4. Last, but not least, the disregarded theory of Bhaskara II. Rule #5. As someone else pointed out on the post, it is impossible to use in mathematics because it can cause 1 to equal 2.

Now when I thought about this... there is only one thing that came to mind that can be described this way. One being (Rule #3) with infinite power (Rule #4 and #5) that both doesn't physically exist (Rule #2) nor can be proven to exist (Rule #1), also known as God. Therefore I put it to you that God can be mathematically represented by the fraction Zero over Zero and on that note, if you look at the entire world as a system of numbers, the very center of the Universe, the thing that all things revolve around is that very fraction.

Matthew Biebel 04:50, 5 May 2007 (UTC)

Okay, let's take this one step at a time. First, I suggest you read the article again. You will see that division by zero is undefined in the structure called "The Real numbers", but there are other algebraic structures in which it is defined perfectly. An example is the real projective line where ${\tfrac {1}{0}}=\infty$.

Many people, myself included, have realized in childhood that 1/0 "should" be infinity, but generally do not have at that point the mathematical tools to understand what it really means. You are certainly not alone in this respect.

About 0/0, there is a somewhat esoteric mathematical theory which "sort of" deals with the issue, and it is mentioned in the article. It is true that defining it is tricky, because we would want it to satisfy too many things. However, this is common in mathematics, and deducing that it should in some way represent "god" is in my opinion extremely over-reaching. -- Meni Rosenfeld (talk) 21:43, 6 May 2007 (UTC)

(#4) ⁰/₀ has "infinite power" only in the sense that it might be represented by the arithmetic value ∞. (#1) says it is undefined within the structure of normal arithmetic, and says nothing about physical reality. (#2) says it might be equal to zero, and zero certainly exists as an arithmetic number. So combining all of these to say that ⁰/₀ represents God is simply carrying the metaphor too far. — Loadmaster 20:37, 14 August 2007 (UTC)

1 - I am not a horse

2 - God is not a horse

Therefore, I am God. 147.197.230.174 11:39, 17 August 2007 (UTC)

1. Caterpillars eat cabbage.
2. You eat cabbage.
3. Therefore, you are a caterpillar.
Ain't logic wonderful? — Loadmaster (talk) 18:56, 28 December 2007 (UTC)

sudden thought, just as not all infinties are equal, perhaps not all 0's are either? --trulyelse02:28, 9 October 2008 (UTC) —Preceding unsigned comment added by 222.155.221.34 (talk)

My 1's greater than yours! Mr. Richelieu (talk) 01:13, 17 December 2008 (UTC)

2=1

Here is another method to show that 2 = 1.

Assume x = y.

Multiply both sides by x, leaving x² = xy.

Subtract y², leaving x² - y² = xy - y².

Factorise, leaving (x + y)(x - y) = y(x - y).

Divide both sides by (x - y), leaving x + y = y.

Now assume that x = y = 1, then: 1 + 1 = 1, 2 = 1

Someone can add this to the page, as I am on my final warning for vandalism, even though 3 of my edits were in good faith. I just don't want to risk it. Efansay---T/C/Sign Here Please 11:34, 29 August 2007 (UTC)

It looks like you have already added this. However, the article really doesn't need another obfuscation of the 1=2 proof, so I will remove it. -- Meni Rosenfeld (talk) 12:06, 29 August 2007 (UTC)

You can only divide by (x-y) if x and y are not equal. In grade 10 we had to specify such conditions when we divided by x-y or anything which can result in zero :P Piepants 08:19, 14 October 2007 (UTC)Piepants

That's the point, this is supposed to illustrate the fallacious argument that will result if we do allow division by zero with the usual properties. -- Meni Rosenfeld (talk) 08:27, 14 October 2007 (UTC)

The inverse of Zero

${\frac {1}{{\frac {1}{9}}+{\frac {1}{9^{2}}}+{\frac {1}{9^{3}}}+{\frac {1}{9^{4}}}+{\frac {1}{9^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{9^{n}}}}}=8$

${\frac {1}{{\frac {1}{8}}+{\frac {1}{8^{2}}}+{\frac {1}{8^{3}}}+{\frac {1}{8^{4}}}+{\frac {1}{8^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{8^{n}}}}}=7$

${\frac {1}{{\frac {1}{7}}+{\frac {1}{7^{2}}}+{\frac {1}{7^{3}}}+{\frac {1}{7^{4}}}+{\frac {1}{7^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{7^{n}}}}}=6$

${\frac {1}{{\frac {1}{6}}+{\frac {1}{6^{2}}}+{\frac {1}{6^{3}}}+{\frac {1}{6^{4}}}+{\frac {1}{6^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{6^{n}}}}}=5$

${\frac {1}{{\frac {1}{5}}+{\frac {1}{5^{2}}}+{\frac {1}{5^{3}}}+{\frac {1}{5^{4}}}+{\frac {1}{5^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{5^{n}}}}}=4$

${\frac {1}{{\frac {1}{4}}+{\frac {1}{4^{2}}}+{\frac {1}{4^{3}}}+{\frac {1}{4^{4}}}+{\frac {1}{4^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{4^{n}}}}}=3$

${\frac {1}{{\frac {1}{3}}+{\frac {1}{3^{2}}}+{\frac {1}{3^{3}}}+{\frac {1}{3^{4}}}+{\frac {1}{3^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{3^{n}}}}}=2$

${\frac {1}{{\frac {1}{2}}+{\frac {1}{2^{2}}}+{\frac {1}{2^{3}}}+{\frac {1}{2^{4}}}+{\frac {1}{2^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{2^{n}}}}}=1$

${\frac {1}{{\frac {1}{1}}+{\frac {1}{1^{2}}}+{\frac {1}{1^{3}}}+{\frac {1}{1^{4}}}+{\frac {1}{1^{5}}}\ldots }}={\frac {1}{\sum _{n=1}^{\infty }{\frac {1}{1^{n}}}}}=0$

The inverse of infinity approaches zero and the inverse of zero leads to infinity user:Twentythreethousand 12:40, 18 September 2007

I take that to mean that $\lim _{x\to +\infty }{\frac {1}{x}}=0$ and $\lim _{x\to 0^{+}}{\frac {1}{x}}=+\infty$. That is rather well known. -- Meni Rosenfeld (talk) 14:13, 18 September 2007 (UTC)

Another approach using counter machines to illustrate 6/0 = either "any number" or "undefined": the problem of termination

The following is actually an informal proof, "informal" in the sense of starting with (quasi-)informal premises. It is constructive in the intuitionist, constructivist mathematical sense, meaning that it produces an "object" (the machine and its behavior) to be observed/displayed. We will be working only in the upper right quadrant including the X and Y axes (negative numbers have not been defined). When we plot the output of our machine as it divides by 0 we would see a ascending spire of positive integers, ascending the Y-axis ad infinitum (i.e. going toward any "etc" we care to choose <0, etc > ). (There is a way to approach this for the "negative" numbers, not shown here, and these would be an descending spire going down the Y-axis all the way to -∞. Thus at infinity the two spires meet and thereby, in an intuitive sense, close the circle).

This example (proof?) shows that, depending how the algorithm is defined, one can view N/0 as either (i) a forever- increasing or accumulating of markers in/on a place called a "register" , or as (ii) "undefined".

Start with the notions of "0" = "empty place", | = mark, and invoke the Shepherdson-Sturgis counter machine model (an abstract computational machine that is easy to use because it has a couple extra instructions). In 10 lines, we can write an algorithm that calculates Q = INTEGER(N/X). "Q" is the place called "quotient", "N" is the place called "numerator" (will contain "6" = |||||| in our example), "X" is the input place where the machine finds a "number" (tally-count i.e. "5" = |||||) to copy into the place "D" called the "Denominator". For this demo we can dispense with the remainder (aka residue) and just compute Q(X) = INTEGER(N/X).

The machine is equipped with a TABLE of instructions that proceed one after another unless either (i) a "test-register-for-zero yields "true" and this fact sends the "program" off to "exit E1", or (ii) an unconditional jump occurs and simply forces the "program" to non-sequential "exit E1". Of these 8 instructions, only 3 are mandatory (DCR(r), INC(r), JZ(r)[E1], HALT ); clear, copy, and the unconditional jump can be built from the first three.

The Shepherdson-Sturgis instruction set :

"a. INCrement(r): add 1 to the number in register r

"b. DeCrement(r): subtact 1 from the number in register r

"c. clear(r): "clear" register n, i.e. place 0 in it

"d. copy(r_s,r_d): copy from source-register r_s into destination register r_d, leaving contents of r_s intact

"e. J[E1]: jump to exit 1.

"f. JZ(r)[E1]: jump to exit 1 if register r is empty

(1963:218, Journal of the Association of Computing Machinery (JACM) 10:217-255, 1963, with written-out mnemonics rather than P, D, O, C for increment, decrement, clear and copy, and "JZ" instead of "J" for (f.))

An actual program: Our machine achieves its result by successive subractions of divisor D from numerator N, keeping input X for restoring D. This continues until [N]=0: decrement numerator N and divisor D until [D]=0, when [D]=0 increment quotient Q, then restore divisor D from the copy held in X:

Instruction #:	1	2	3	4	5	6	7	8	9	10
Instruction:	clear	copy	jz	jz	dcr	dcr	j	inc	j	H
r_source	Q	X	D	N	D	N		Q
r_destination		D
jump_to addr			8	10			3		2

Division by 0 -- a partial function:

A simple example using X = { 1, 2, 3, 4, 5, 6, 7, 8, 9, etc } and N = 6, i.e. Q(X) = INT(6/X) will suffice.

What happens when X = D = 0? The machine never reaches HALT, but it does increment the contents of place Q, i.e. [Q], and produce every "number" (count of objects) from 0 to ∞, ad infinitum.

If a person or a machine were to look into place "Q" they would see a different number every time they looked, and the number would always increase. Thus we have produced the following:

[Q] = q = N/0 = steadily increasing number, ad infinitum, Q.E.D. (part Ia)

If they were to take any one of these numbers -- call it Q_any, and multiply it by 0, they would get 0.

Thus N/0*0 = 0. Q.E.D. (part Ib)

The division by 0 sends the values in output-put place "Q" up the Y-axis ad infinitum, generating every positive integer.

Argument			OUTPUT value
INPUT variable	parameter		Q=INT(N/X)
Denominator	Numerator		Quotient
X	N		Q
1	6		6
2	6		3
3	6		2
4	6		1
5	6		1
6	6		1
7	6		0
8	6		0
9	6		0
etc	6		0

An "extension" to output "undefined" in the case the program fails to halt:

Because the first algorithm's division by 0 never progresses to HALT, i.e. it never progrresses beyond instruction #9 in the above example, we can tack on a few instructions right before HALT so that the machine always generates Y = 6/X+1, unless it is dividing by 0. If it is dividing by 0 then output-place Y remains "empty", a condition we might call "null" or "undefined".

If at the outset (as programmers) we define an extension of the program such that the output is too be the contents of an additional place Y (rather than Q) if and only if the program has ended its computation, then the program will never put any thing (mark, pebble, abacus bead) into/onto Y. What we do in this modified, extended program is require the program to clear (empty) Y at the outset, plus we consider this cleared output (empty of marks or counters) to represent undefined. Thereafter, the symbol "0" will be considered a single mark or counter |, the symbol "1" to be two marks ||, etc. Thus if the program terminates its computation for Q, then it will copy this incremented value into Y. So now we watch place Y and see if it puts at least one thing into Y, and thereby confirm the algorithm has terminated. Thus we have produced the following:

Unless [X]=0, the output at Y will be Y(X)=N/X + 1, otherwise Y will be (and remain ad infinitum) "empty".

[Q] = q = N/0 with termination required = "undefined" output. Q.E.D. (part IIa)

6/0*0 = "undefined" output. Q.E.D. (part IIb)

The program modified with an "extension" that includes an additional register "Y". Now, because division by zero results in a never-ending loop that stops the progress of the computation, the machine never reaches instructions 11, 12 and 13. In this case output-place "Y" will remain empty and we choose to rename this condition "Ø" or "null" or "undefined", etc:

Instruction #:

1

2

3

4

5

6

7

8

9

10

11

12

13

Instruction:

clear

clear

copy

jz

jz

dcr

dcr

j

inc

j

copy

inc

H

r_source

Q

Y

X

D

N

D

N

Q

Q

Y

r_destination

D

Y

jump_to addr

9

11

4

3

If we had put a "test for X=0" at the beginning of our program, our machine could terminate with, e.g. a special number, e.g. | in a place called "V" for "valid input". wvbaileyWvbailey 21:20, 7 October 2007 (UTC)

x=x+x

x=x+x would not prove that 1=2 it would merely prove that x=0. —Preceding unsigned comment added by 75.162.139.10 (talk) 04:38, 22 November 2007 (UTC)

Correct. The proof in question is deliberately fallacious, and is presented to demonstrate why division by zero is impossible. -- Meni Rosenfeld (talk) 09:39, 22 November 2007 (UTC)

Division by Zero

I have written a subpage on my user page about my idea of division by zero. I would like some feedback and information on where I've gone wrong if I have. If you could leave any comments on my talk page, that would be great. The page is at User:Freenaulij/Division by Zero —Preceding comment was added at 04:00, 4 December 2007 (UTC)

Before I go into depth in reading that page, I suggest you take a look at real projective line. In particular, there is nothing inherently new in the ideas I've seen you suggest - if everyone you talked to blowed them off, it's because you haven't talked to the right people (though there are of course some debatable details). -- Meni Rosenfeld (talk) 08:34, 4 December 2007 (UTC)

Better yet, read this article.

However, do note the following. While there are many ways to define division by zero, none I've seen maintain the original concept that division is the inverse operation of multiplication - that is, "${\tfrac {a}{b}}$ is the unique c such that $b\cdot c=a$". In particular, we can't both have that and keep multiplication associative - this would lead to 1 = 0 * (1/0) = (0 * 2) * (1/0) = 0 * (2 * (1/0)) = 0 * (2/0) = 2 (where the equality 2 * (1/0) = 2/0 is justified by (2 * (1/0)) * 0 = 2 * (1/0 * 0) = 2 * 1 = 2), which is something we don't want. So this means that if division by zero is defined, then either:

1. Division is not the inverse operation of multiplication,
2. Multiplication is not associative, or
3. 1=2.

Any of those is a much more fundamental change than those that follow from introducting negative or complex numbers, and this is why division by zero is usually frowned upon.

As for applications of defining 1/0 = ∞, worry not, there are plenty. -- Meni Rosenfeld (talk) 08:46, 4 December 2007 (UTC)

Another point. You can't "prove" that 1/0 = ∞. You can only prove anything about something that has already been defined. At the basic level, 1/0 has not yet been defined. What you can do is show that it is plausible to extend the definition by stating that 1/0 = ∞ etc. I am undecided whether your arguments do a good job demonstrating this.

And mathematical books aren't "wrong". In the real numbers system, you really cannot divide by zero. What you can do is discuss an alternative, richer system, but that's not what those books are aiming at. -- Meni Rosenfeld (talk) 08:52, 4 December 2007 (UTC)

Meni is correct about 1/0 not "being defined" but rather always in the process of being defined. For an example, the example above with the counter machine doing division by 0 as repeated subtractions of 0 from (in this case) 6. It bumps its output one count every time it subtracts 0. Because the machine is an abstract machine its "memory registers" are of unbounded capacity, so it can always add 1 one more count, again subtract 0 from 6, add 1 more to the output, again subtract 0 from 6, add 1 more to the output ... (Think about what would happen if you (a robot who can't say NO) were ordered to do this: "Until the hole before you is filled (i) repeatedly throw, once every second, no rocks into the hole, and (ii) tally up the number of times you throw no rocks into the hole." No matter how large or small the hole, you will be there a long, long time (until the end of time), and your tally will be ever-increasing). Bill Wvbailey (talk) 16:10, 5 December 2007 (UTC)

We should probably avoid confusing the OP with these vague OR claims. -- Meni Rosenfeld (talk) 17:16, 5 December 2007 (UTC)

I wouldn't call a demonstration such as the above either vague nor OR, it's just an example from computability theory. It's using the Shepherdson-Sturgis machine model from the literature and just does a simple algorithm which is hardly OR. Don't go overboard here: the mission is to educate and inform, not obfuscate. I believe the article even mentions repeated subtractions of 0. Bill Wvbailey (talk) 17:26, 5 December 2007 (UTC)

"1/0 is always in the process of being defined"? What does that even mean (from a mathematical viewpoint)? No, don't tell me, provide a reference discussing this (you said it's not OR, right?). -- Meni Rosenfeld (talk) 17:30, 5 December 2007 (UTC)

Get a cc of Kleene 1952 in particular the section where he talks about intuitionism versus formalism:

• "The intuitionist refrains from accepting such an existence proof [reductio ad absurdum] because its conclusion there exists an n such that P(n) can have no meaning for him other than as a reference to an example of a number n such that P(n), is not available to him, since he does not conceive the natural numbers as a completed totality." (Kleene 1952:50)

Also, read up on induction and inductive proofs e.g. Kleene 1952:21. While you're at it, this comes from the book about zero that I added to the references of this page (the only three book references on this page are the ones I've added, by the way):

• "When x goes to zero, 1/x gets bigger and bigger and bigger and finally just blows up and goes off to infinity. ... The curve 1/x has a singularity at the point x = 0 -- a very simple sort of singularity that mathematicians dubbed a pole. There are other types of singularities as well; for instance, the curve sin(1/x) has an essential singularity at x = 0. Essential singularities are weird beasts; near a singularity of this sort, a curve goes absolutely berserk. It oscillates up and down faster and faster as it approaches the singularity, whipping from posity to negative and baqck again. In even the tiniest neighborhood around the singularity, the curve takes on almost every conceivable value over and over and over again." (Seife 2000:147)

You are probably bothered about the notion of how 'defined' is defined. The above and the example is my definition of "defined" as synonymous with "algorithm", as in "demonstration" in the intuitionist sense. There is nothing here having to do with novelty. As you yourself wrote (you must be an intuitionist): "You can only prove anything about something that has already been defined. At the basic level, 1/0 has not yet been defined."

In the book by Suppes the chapter is titled Theory of Definition and the section is titled The Problem of Division by Zero. A xubsequent chapter by Suppes discusses 5 approaches to defining 0. One of this is similar to the reductio mentioned above:

• "The second approach is to let x/0 be a real number, but to define division by the conditional definition stated in the last section: IF y≠0 Then x/y = z if and only if x = y*z. In this case for every number x, x/0 is a real number, but we are not able to prove what number it is."(Suppes 1957:167)

This second approach, the computability result of an algorithm to "divide by zero" by repeated subtraction and addition to produce the product, is the example of a definitiont that I posted above. Bill Wvbailey (talk) 18:28, 5 December 2007 (UTC)

No, I'm nowhere near an intuitionist. My statement you quote is as classical as it gets. Yes, I am bothered by your notion of "definition", it looks like you have made quite a leap from the sources you quote. I'll try to take a look at Suppes' book to see if it supports your position. -- Meni Rosenfeld (talk) 18:49, 5 December 2007 (UTC)

Wait, so Meni Rosenfeld, how would you define "definition"? freenaulij 02:59, 6 December 2007 (UTC) —Preceding unsigned comment added by Freenaulij (talk • contribs)

Exerpt from Suppes 1957

Hey Bill, I don't have access to this book, just curious what are the other three [four] approaches. Freenaulij (talk) 03:27, 6 December 2007 (UTC)

Well, let's see. The first subchapter is §8.5 The Problem of Division by Zero. This he follows up with §8.6 Conditional Definitions (where he discusses "eliminating all the 'interesting' cases) and §8.7 Five approaches to Division by Zero, which I will repeat in its entirety. This will take more than one sitting to type in. The following is from Suppes 1957:163-169:

§8.5 The Problem of Division by Zero.

"That everything is not for the best in this best of all possible words, even in mathematics, is well illustrated by the vexing problem of defining the operation of division in the elementary theory of arithmetic. If we introduce the definition:

(1) x/y=z if and only if x = y*z

"we realize immediately that (1) does not satisfy the fourth restriction for equivalences definition operation symbols. For we cannot prove that given any two numbers x and y there is a unique z such that x = y*z. For instance, there is no z such that 1 = 0*z; and any number z has the property that 0 = 0*z.)...

"In spite of the difficulties besetting us there is a formally satsifactory way of defining division by zero. It is a so-called axiom-free definition since it requires no previous theorem to justify it. It does require that when y = 0, then x/y=0. Because of the several quantifiers occurring in the definiens we use logical notation to state the definition:

x/y = z ←→ (z')[z' = z ←→ x = y*z'] V [-(∃w)(z')(z' = w ←→ x = y*z') & z = 0]

[NOTE: This is a rather difficult definition in part because of Suppes's symbolism. Suppes is using the symbols as follows: ←→ means "if and only if" i.e. "is logically equivalent to", = means " identical to ", (z') means "for all (nowadays written ∀) numbers with the symbol z' ", " V " is logical inclusive OR, " - " is logical NOT, ∃ means "there exists at least one example of". The identity sign " = " is actually a predicate, a function that outputs { true, false }. So in words

[ x/y is logically equivalent to EITHER OR BOTH of the following statements:

• (a) For all numbers z', the notion " z' is identical to z " is logically equivalent to the notion " x is identical to y*x' "
• OR (b) It's not the case that a number w exists such that, for all numbers z', the notion " z' is identical to w " is logically equivalent to the notion " x is identical to y*z' " AND at the same time the notion " z is identical to 0 ".

The complicated character of (5) argues strongly for some other solution, which we turn to in the next section. ...

"Here it will suffice to say that there seems to be no method of handling division by zero which is uniformly satisfactory." (Suppes 1957:163-164)

§8.7 Five approaches to Division by Zero:

"We have already mentioned several aspects of the problem of division by zero, and have remarked that there is no uniformly satisfactory solution. In this section we want to examine five approaches to the problem. The next to last of the five yields the soultion which is probably most consonant with ordinary mathematical practice.

• 1 "The first approach differs from the others in that it recommends a change in the baseic logic to deny meaning to expressions like:

1/0 = 1/0

Without attempting to characterize the basic changes necessary, we may still offer some general objections to this approach. The first objection is that it is undersirable to complicate the basic rules of logic unless it is absolutely necessary. In other words, change the foundations of infereence only if all other approaches have failed. Second, if usch a change were adopted, the very meaninfulness of expressions would sometimes be difficult if not impossible to decide. For example, assume that we have added oto our axioms of Chapter 7 sufficienint axioms to obtain the expected theorems on the natural numbers (i.e., the positive integers). Consider now the expression:

(1) For every natural number n, 1/(n*) = 1/(n*)

"where n* is the unary operation defined as follows: n* = 1 if n is an odd integer or n is an eveninteger which is the sum of two prime numbers; n*=0 if n is an eveninteger which is not the sum of two primes. The problem of the existence of even integers which are not the sum of two primes is a famous problem of mathematics which is still unsloved (Goldbach's hypothesis). Thus on the basis of the first approach the meaningfulness (not the truth or falsity) of (1) is an open question.

• 2 "The second approach is to let x/0 be a real number, but to define division by the conditional definition stated in the last section: IF y≠0 Then x/y = z if and only if x = y*z. In this case for every number x, x/0 is a real number, but we are not able to prove what number it is.

(2) 1/0 = 2/0

"The inability to prove or disprove (2) is an argument against the second approach, since we want our axioms to be as complete as possible."

• 3 The third approach agrees with the second in making x/0 a real number, but it differes in making x/0 = 0 for all x. This eliminates the undecidability of statements like (2). For the third approach the appropriate definition of division is a proper definition similar to the axiom-free definition discussed at the end of §8.5:

x/y = z ←→ [y ≠ 0 → x = y*z) & (y = 0 → z = 0)

"An advantage of the third approach is that it permits the definition of zero by a straight forward proper definition fully satisfying the criteria of elminability and non-creativity. The main disadvante of this appraoch is the one mentioned in §8.5: many mathematicians feel uneasy with the identity:

x/0 = 0.

• 4 The forth approach agrees with the second and third in requiring no basic change of logic; it differes in not making it possible to prove that x/0 is a real number. The basic idea is to introduce a predicate 'R' which means 'is a real number.' Hypotheses using this predicate must be added to all the axioms stated in Chapter 7. Furthermore, to guarantee that addition and multiplication of numbers yield numbers, we must add the two closure axioms:

R(x) & R(y) → R(x+y)

R(x) & R(y) → R(x*y).

"The introduction of the predicate 'R' widens our domain of individuals, for now it does not follow that everything in the domain is a real number (in particular the 1/0 is a real number). The indroduction of 'R' has the further consequence of making it natural to make all our definitions of arithmetical relations and operations conditional. There is no point in defining the realition of equal to or less than, for instance, for things which are not numbers. Thus we would ahve:

[R(x) & R(y)] → [x ≤ y ←→ (x = y V x < y)].

"The definition of division would be:

"(3) [R(x) & R(y)& R(z) & y ≠ 0 ] → [x/y = z ←→ x = y*z].

"If x is a real number, with (3) at hand we cannot prove:

"x/0 is a real number

"And we cannot prove:

"x/0 is not a real number,

"But we are not faced with the counterintuitive situation of being forced to call x/0 a real number. The situation can be improved by introducing into our system a primitive symbol for some object which is not a real number. Without saying what the object is, let us designate it by 'v'. They we have the axiom:

"-R(v) [here - is ~ or bent bar, i.e. NOT]

"that is, the assertion that v is not a real number. We may now define division by:

"(4) [R(x) & R(y)] → [x/y = z ←→ [y ≠ 0 → (R(z) & x = y*z)) & (y = 0 → z = v]].

"The virtue of (4) is that it definitely places x/0 outside the domain of real numbers for any number x. Such a consequence would seem to be in closest accord with ordinary mathematical usage. Definition (4) also has the virtue of making

x/0 = y/0 ,

"Where x and y are real numbers, thus eliminating a vast proliferation of odd mathematical entities. On the other hand, to see once for all that all is not for the best in this best of all possible worlds, notice that if we adopt (4) we cannot decide whether or not v/v is a real number. If we use conditioncal dfinitions and insist on not tampering with the basic law of identity for therms, then we must be prepared for the undecidability of the statue so fentities like v/v. We make thake the attitude that everything is in good order in the domain where we intend to use the division operation, and we really do not care what is going on elsewhere if not inconsistencies can creap in.

• 5 "A fifth approach should be mentioned which is of considerable theoretical importance but does not correspond at all to ordinary mathematical practice. The idea is simple: banish operation symbols and individual constants, and use only relation symbols. Thus '0' is replaced by the primitive one-place predicate 'Z', where it is intended that Z(x) means that x is an identity element with respect to addition. the trnary relation symbol 'A' is used to replace the addition operation symbol:

A(x, y, z) ←→ x+y = z

"Similarly the ternary relation symbol 'M' is used for multiplication:

M(x, y, z) ←→ x*y = z

"With this apparatus we may easily give a proper definition of the division relation symbol:

D(x, y, z) ←→ -Z(y) & M(y, z, x). [Here - again is used as NOT]

"In this approach there is no need for unuusual mathematical entities, but it is extraordinarily awkward to work continually with relation symbols rather than operation symbols. For example, the associativity of addition has to be expressed in some manner like the following:

A(x, y, w) & A(w, z, s₁) & A(y, z, v) & A(x, v, s₂) → s₁ = s₂ "(Suppes 1957:1666-169)

---

In the examples I gave above using computability execute some of these axioms/definitions. In particular, the second example defines the number '0' as a single count |, the number '1' as the successor i.e. ||, etc. Thus 'empty counter' becomes a new 'symbol' i.e. (ε or "u") that stands for 'division by 0'. Or a person could simply do as every system on the planet does, check for division by 0 and print 'undefined operation' or something to that regard. Bill Wvbailey (talk) 16:01, 6 December 2007 (UTC)

"Less than" and "greater than" in definitions of division by zero and infinity

• It's better to use the terminology less than and greater than when working with infinity and zero(they are both unknown).

infinity * 0=1

0*1>0 but lesser than one(or unknown) which makes zero times numbers less than infinity less than one or maybe equals to zero.

2>1 or 1<2 Twentythreethousand (talk) 04:26, 5 December 2007 (UTC)

What on earth are you talking about? 0*1 = 0 which is not greater than 0. ∞ * 0 is undefined in our context. -- Meni Rosenfeld (talk) 11:23, 5 December 2007 (UTC)

• Fallacies based on division by zero

0*1<0*2;1<2 There is no fallacies if you used a smaller sign.

When you use division, and one(the dividend) is lesser than the divisor(let's say 17) you multiply by zero in the

quotient and then multiply the remainder by ten. When you multiply 17 by zero does it give you one or is it

undefined?Twentythreethousand (talk) 18:24, 5 December 2007 (UTC)

Please clarify. -- Meni Rosenfeld (talk) 18:49, 5 December 2007 (UTC)

17*0>0 or equal to zero. Zero is unknown if infinity is unknown.Twentythreethousand (talk) 14:54, 6 December 2007 (UTC)

• File:ZerosInInfinity.gif

one divided by infinity equals 0,and 2/infinity equals 0 until infinity divided by infinity equals 1. So the

difference between 1 and zero equals one because we're counting on digits based on one.

Same thing with zero until zero times infinity leads us to one.Twentythreethousand (talk) 06:37, 7 December 2007 (UTC)

Is math broken?

Without even using calculus it is not difficult to prove that x/0 is undefined in the real number system. Let's say x=4. At first glance, some would say 4/0=0 but this is not true because 0*0=/=4. This applies to any real number. 4/0=/=22 because 22*0=/=4. That is not to say that under no circumstances does x/0 equal anything, but simply that if one is using real numbers, x/0 cannot be proven algebraically. However, in one circumstance it can. If 0/0=4 then 4*0=0. This is true, and it applies to any REAL number. Not to mention that 0/0=(0*4)/0. Why is 0/0 different? I understand that in f(x)=1/x that as x approaches 0 y approaches infinity, but because the equation never reaches 0 but instead gets infinitely smaller it does not prove that in terms regarding real numbers that x/0=infinity. Nor am I saying that 0/0=infinity. I am simply asking why, using very basic mathematics one is able to prove that 0/0=all reals. —Preceding unsigned comment added by 70.160.200.187 (talk) 03:14, 18 January 2008 (UTC)

The "problem" (if you can call it that) lies in the definition of division. We know from school that in usual circumstances, $b\cdot c=a$ if and only if ${\frac {a}{b}}=c$. However, just because we learn it at school doesn't make it true. To investigate the issue, we must go back to the basics. The standard definition of division is:

Let a and b be real numbers such that $b\neq 0$. Then ${\frac {a}{b}}$ is the unique real number c such that $b\cdot c=a$.

You can see that this definition doesn't address the case that b is zero, thus it leaves expressions such as 0/0 still undefined. You may think this is too restricting, and look to extend the definition so that it gives a value to 0/0. But what would that value be? We would want it to be "the" number that, when multiplied by 0, gives 0. But there's more than one number satisfying that, so there is no obvious candidate to fit the role of being the value of 0/0. Thus, the best course of action is to realize that "0/0" is problematic and leave it undefined. The side effect is that even though $4\cdot 0=0$, it is not true that ${\frac {0}{0}}=4$. -- Meni Rosenfeld (talk) 12:57, 18 January 2008 (UTC)

Categorization

Why's this classed in Internet Memes? Is it because Chuck Norris can divide by zero? If so, is that worth mentioning? —Preceding unsigned comment added by 69.107.249.31 (talk) 00:57, 26 January 2008 (UTC)

It is an internet meme because of the belief on the internet that a successful division by zero will destroy the universe. --trulyelse02:33, 9 October 2008 (UTC) —Preceding unsigned comment added by 222.155.221.34 (talk)

Infinity/0?

If a / 0 = infinity, then infinity * 0 = a. Therefore, since anything * 0 = 0, a can only = 0? 69.221.124.196 (talk) 17:15, 30 March 2008 (UTC)

If division by zero made sense within the real number system, and if you didn't have infinity on the right hand side of that equation, then 'a' would indeed be unique. However, since neither of these conditions is met, the dividend need not be unique. FilipeS (talk) 19:03, 30 March 2008 (UTC)

The first sentence is false: infinity * 0 = "anything". Infinity is not a particular object, i.e. a fixed "thing"; "infinity" is that which is in the "act of becoming" (e.g. the expanding universe). Bill Wvbailey (talk) 23:08, 30 March 2008 (UTC)

A tomato is a vegetable and a fruit. "Infinity" both is a general name for a certain kind of concept, and can be the name of a particular object in a given setting. For example, in a certain construction of the real projective plane, ∞ is just the set $\{(a,0)|0\neq a\in \mathbb {R} \}$. -- Meni Rosenfeld (talk) 00:38, 31 March 2008 (UTC)

Internet Meme?

Allright, wikipedians. The Meme has to be mentioned in this article or a new one. —Preceding unsigned comment added by 209.7.111.45 (talk) 12:42, 19 September 2008 (UTC)

I would agree. We must remember however, that finding sources for memes is almost impossible, since by the time someone publishes something about it, its so old its no longer funny. On a side note, please in future put new sections at the bottom of the talk page. Zell65 (talk) 02:44, 2 October 2008 (UTC)

On the positioning — I went ahead and moved it.

On the substance — which meme? The Chuck Norris thing? No, that really doesn't need to be mentioned, either here or elsewhere. But I'm guessing that Chuck Norris facts will come up blue, and I suppose you can put it there, if you really have to, and if it isn't there already. --Trovatore (talk) 03:07, 2 October 2008 (UTC)

No the divide by zero meme, Google it :) 138.253.201.111 (talk) 02:32, 12 January 2009 (UTC)

Meme mentioned, and in a way that assumes nothing and makes no spurious mentions of martial artists. —Preceding unsigned comment added by 209.147.139.164 (talk) 19:52, 3 February 2009 (UTC)

1/0 can be a whole number

Another legitimate mathematical use of 1/0 is to indicate a whole number, a possible number of points (of length zero) in any interval of a line of points (of arbitrary unit length). The paper on this is at http://www.metaphysica.de/texte/mp2005_2-Cooke.pdf and an abridged version is at http://www.geocities.com/potential_continuity/tocontinuewithcontinuity.html Username12321 (talk) 09:54, 18 October 2008 (UTC)

Which whole number? There's an awful lot of them around... FilipeS (talk) 16:42, 18 October 2008 (UTC)

My little counter-machine demo further up this talk page (with the graphs) shows that 1/0 can be any old integer you want -- given that you output the value of the counter (the mu-operator) that's trying, and failing, to terminate the computation. Bill Wvbailey (talk) 14:28, 21 October 2008 (UTC)

There are indeed; it's a cardinal number (the number of points in continua) of a realistic (rather than formalistic or constructivistic) kind. Incidentally I've added the link to the abridged version (since it contains a link to the paper) under "Further Reading" (there being no "External Links" section on this page), which I hope was OK? Username12321 (talk) 08:46, 21 October 2008 (UTC)

I have removed it on the basis of WP:ELNO criterion #2:

Any site that misleads the reader by use of factually inaccurate material or unverifiable research. See Reliable sources for explanations of the terms "factually inaccurate material" or "unverifiable research".

Moreover, it was likely to be removed by a bot anyway, since the domain geocities is permanently blacklisted. siℓℓy rabbit (talk) 12:06, 21 October 2008 (UTC)

I guess we can add a reference to the published paper itself under Further reading if there is consensus to do that. siℓℓy rabbit (talk) 12:11, 21 October 2008 (UTC)

How about we say this, we have one cake, we want no slice's from the cake (1/0) how many cakes are still there? One, thus 1/0=1. Anonymous x —Preceding unsigned comment added by 121.72.179.127 (talk) 10:07, 24 October 2008 (UTC)

There is no mathematically legitimate way of assigning a value to 1/0 as a whole number. This remark is at least as old as Bishop Berkeley's criticism of infinitesimal calculus in The Analyst, see Ghosts of departed quantities. In fact, this interesting episode in the history of mathematics could be mentioned in this page. Katzmik (talk) 10:28, 24 October 2008 (UTC)

P.S. On the other hand, it is possible to obtain hyperintegers by dividing 1 by a suitable infinitesimal in non-standard calculus. Katzmik (talk) 10:30, 24 October 2008 (UTC)

I put in a reference to the published paper; hope that's OK. The paper deals with the usual objections to division by zero near its beginning. The idea of mathematical validity is a vague one, of course. There is formal validity and informal, for example, and in the latter constructivistic as well as the more usual, realistic sort. The paper argues for validity in the latter sense, i.e. realistic, since that is the familiar sort to most people. So this whole number is an epistemically possible number of metaphysically possible things. Whereas anything can be made formally valid, by choosing the right (formal) logic.Username12321 (talk) 14:36, 5 November 2008 (UTC)

Berkeley's objections were to magnitudes that are sometimes zero and sometimes nonzero (and so might apply to standard lines of points), but we can obtain something like that quite legitimately in the realm of approximations (lots of negligible grains making heaps). The use of nonstandard infinitesimals is a good example (taking the standard part merely ignores the infinitesimal) since they also show one of the strengths of the formal approach since they are based upon infinite natural numbers, which clearly do not really exist, but can be shown to be consistent if standard arithmetic is. Still, they are more imaginary than imaginary numbers, since at least the latter are instantiated by quantum mechanical wave functions. Whereas nonstandard infinitesimals might suffer from sliding (see Moore, M. E. (2002) ‘A Cantorian Argument Against Infinitesimals’ Synthese 133, pp. 305-330.), were they real (not just formal), the (so-called 'irreal') infinitesimals of my (2005) paper could not, since they are metaphysical posits, in the grander tradition of Cantor et al.Username12321 (talk) 13:25, 6 November 2008 (UTC)

Incidentally, books like Seife's but by Robert Kaplan and John D. Barrow also came out around the turn of the millennium...Username12321 (talk) 13:25, 6 November 2008 (UTC)

Why is this right?

The article currently states:

and for any negative a,

<!-- NOTE : Do not change the following! a is negative here, so the limit from ABOVE is negative. -->

$\lim _{b\to 0^{-}}{a \over b}={-}\infty .$

Why? $b\to 0^{-}$ refers to the limit from the negative side, i.e. the limit from below, which is positive infinity. --Zarel (talk) 07:05, 18 December 2008 (UTC)

Fixed. –Pomte 07:53, 18 December 2008 (UTC)

about Fallacies based on division by zero

in Fallacies based on division by zero

With the following assumptions:

${\begin{aligned}0\times 1&=0\\0\times 2&=0.\end{aligned}}$

The following must be true:

$0\times 1=0\times 2.\,$

Dividing by zero gives:

$\textstyle {\frac {0}{0}}\times 1={\frac {0}{0}}\times 2.$

Simplified, yields:

$1=2.\,$

The fallacy is the implicit assumption that dividing by 0 is a legitimate operation with 0/0 = 1.

However the assumption has no fundamentation, 0/0 can be "probed" to be 0, 1 or any other number.

Also, if 0 changed to 0/0 then it should have change back to 0 not 1 (to keep the equations equal)... making a loop.

...this is a nonsensical argument. --Bodinagamin (talk) 21:25, 5 March 2009 (UTC)

Why is

Why is gravitational singularity linked from this page? There is no mention of gravity on this page, and no mention of division by zero on that page. Ethan Mitchell (talk) 23:29, 26 April 2009 (UTC)

Good question. It's probably a reference to the joke that division by zero causes an implosion of the universe. It seems that the "In Popular Culture" section has been taken out (probably for not satisfying WP:N) so I've removed the link. --Zarel (talk) 07:58, 27 April 2009 (UTC)

0/0=X

Yes that’s right, X, as in any number. How many nothings go into nothing?.........well, if you where to say 1 for example, then you would be correct, since if you have got one load of nothing then you would still have nothing. If you however say 2 you would also be correct, because if you have two loads of nothing you still ultimately have nothing, and the same applies to 3, 4, 5 ect. To put it differently, 1 zero is in zero, 2 zeros are in zero, 3 zeros are in zero, so, X zeros are in zero. I can prove this by asking the following question; If your travelling at 0mph, how long does it take you to travel no distance? The answer is obvious. If you wait for say an hour then you would have travelled no distance, you can't argue with me there. If you wait for 2 hours instead then you still would have travelled no distance, you can't argue with me there either. Say if you waited for an infinite about of time, would that mean that you would be able to travel any distance at your 0mph speed? Of cause not! It doesn't mater how long you wait, you still will never be able to cover any distance if your not moving, so there is no problem in saying that the answer is X. Now we know that in order to find out time we have to divide the speed by the distance, and since we know that both the speed and distance is 0, then 0/0=X. Robo37 (talk) 19:12, 14 May 2009 (UTC)

Yes, if you allow a/b to take on multiple values, then 0/0 can easily be defined as all (complex) numbers, assuming we take the care to interpret laws and rules as we do for other multivalued functions. For the first fallacy in the article, for instance, going from 0 × 1 = 0 × 2 to 0/0 × 1 = 0/0 × 2 would be quite all right under that interpretation, as there are values of 0/0 such that 0/0 × 1 = 0/0 × 2 (the fallacy would then come in the next step). But as the article points out, a/b is not normally defined as the solutions x to the equation bx = a, but rather as the unique solution x to that equation, if one exists. And there is clearly no unique solution to 0x = 0. —JAO • T • C 19:35, 14 May 2009 (UTC)

The most important fallacy here is that this "X" concept you're proposing here is not a number. In fact, it's not anything close to a number. What's $X^{2}$? X+3? Conceptually, yes, 0/0 represents any number, but mathematically, this means that 0/0 is indeterminate and undefined in the real and complex planes. --Zarel (talk) 05:42, 11 June 2009 (UTC)

1 apple + 2 apples = 3 apples. Are apples numbers? No. Does it matter? No. Most of the stuff you do in maths class circulates around letters anyway. I personally can't see why X can’t come up as the answer on every calculator or computer when 0/0, or indeed every answer that is currently described as indeterminate (except for ∞/∞), is entered. Or if not then it should at least be mentioned on this article that 0/0 is any number. Robo37 (talk) 19:19, 11 June 2009 (UTC)

But the point is that 0/0 is not any number. 0/0 could be defined to be any number, but that definition of division is very unusual. The article Division (mathematics) starts off with the normal definition:

"Specifically, if c times b equals a, written:

$c\times b=a\,$

where b is not zero, then a divided by b equals c, written:

${\frac {a}{b}}=c$"

Note the "is not zero" part. The fact that an altered definition of division would yield other results is of course trivial and needs not be mentioned. —JAO • T • C 20:36, 11 June 2009 (UTC)

A number divided by a number should be a number. 1 apple + 2 apple = 3 apples, because "1 apple", "2 apples", and "3 apples" are all numbers associated with units. Similarly, x + 2x = 3x works because x, 2x, and 3x are all expressions.

X doesn't come up when someone enters 0/0, because they're just going to ask "What's X?" and then you explain "X is indeterminate and undefined" and then they ask "So why didn't you just say 'indeterminate and undefined' in the first place?" That's what most calculators currently say: "undefined" (Well, the cheap ones just say "ERROR: DIVISION BY ZERO").

Seriously. So all you're proposing is that we call "0/0" "X"? We might as well just call it "0/0", that doesn't take much longer to write. And if we're going to explain it, we're going to call it undefined and indeterminate, because just saying "X" explains nothing at all.

I mean, practically, if someone asks you "Where's the electron around that hydrogen atom right now?" and you say "X", that doesn't mean anything. "X" is pointless, it doesn't mean anything or explain anything. We have words to do that already. Use them. --Zarel (talk) 02:58, 16 June 2009 (UTC)

0/0 = 0, obviously...

Please someone hit me with the clue stick. The answer seems so obvious to me :

a / b = a * 1/b, right?

0 / 0 = 0 * (at this point who cares, '0 *' is enough to know the result's always 0) 1/0 = 0

Really, a division is a multiplication. And multiplying anything by 0 equals 0. 1/0 may be impossible, but 0 times something impossible equals nothing. It just seems so obvious, I really fail to see how come people don't bring it up more often, even if some genius out there has a reason for saying it's wrong. And I really fail to see how it could be wrong, it's just obvious. --89.127.179.216 (talk) 21:13, 8 June 2009 (UTC)

There's no magical rule that says 0*a = 0 for all a. It's certainly true for all natural a, all real a, and even all complex a—but 1/0, when at all defined, is none of that. —JAO • T • C 21:25, 8 June 2009 (UTC)

Please refrain from using the term "obvious" too much - I've noticed it seems to increase one's likelihood of being wrong. ;)

Anyway, Jao is correct. 0 times any real or complex number is 0, but since 1/0 is not a real or complex number, you can't multiply it by 0 and expect to get 0. A good analogy is infinity, especially since 1/0 has many of the same properties infinity does: 0 times infinity is not 0, but an indeterminate form. --Zarel (talk) 05:38, 11 June 2009 (UTC)

That's pretty stupid, but here, have 0 boxes containing an infinity of cookies for this explanation. ;-) --89.127.179.216 (talk) 23:16, 15 June 2009 (UTC)

On style, please review WP:CIVIL.

On substance, you're confusing two sorts of multiplication here. In cardinal arithmetic it's quite true that κ×0=0×κ=0 for any κ (even if κ is infinite), which is your point about the cookies and is well taken.

But we're not doing cardinal arithmetic here; we're talking about extending real-number multiplication to the extended real number line. Real-number multiplication is not about adding x copies of y — you can't add π copies of anything. There are a couple of conceptual steps in between — first an algebraic one to understand rational multiplication, then a topological one to see what multiplication should be on the reals. When you apply the last step to the extended reals, there is no well-defined result for 0×∞ --Trovatore (talk) 23:51, 15 June 2009 (UTC)

I don't mind that tone of post; "That's pretty stupid" clearly refers to how 0*infy isn't always 0 - it's not a personal attack.

Here's my explanation: Normally, with real numbers, you can abstract it away and ask things like "A * B = the number of cookies you have when you have A many boxes of B cookies each", but since infinity isn't a real number, you can't do it that way. In fact, whenever infinity is involved, you have to do it like this:

"A * B = the unique limit of every sequence A(x) * B(x) where A(x) tends towards A and B(x) tends towards B."

So let's say A = 0, and B = infinity. Then let's say A(x) = 1/x (which tends to 0) and B(x) = x (which tends towards infinity). Then A(x) * B(x) = 1/x * x = 1. So the limit of A(x) * B(x) is 1. So that's an example of a sequence where 0 * infinity is 1 and not 0. You can find sequences where 0 * infinity is any real number, which is why we say it's indeterminate. --Zarel (talk) 02:24, 16 June 2009 (UTC)

Well, my point is that you actually can't do that with real numbers. You can do it with natural numbers, but not with real numbers. It doesn't make sense to say "A many boxes" when A is a real number. (In my opinion it doesn't make sense even when A is a real number that corresponds to a natural number under the canonical embedding, like 3.00000..., but I don't care to argue about that — what's completely clear is that it never makes sense to talk about 3.218742... boxes of cookies.)

So multiplying A by B where A and B are real numbers, really does not mean "add up A copies of B. It means something subtler. --Trovatore (talk) 02:59, 16 June 2009 (UTC)

x/0 = x(infinity)?

Could this be a way to solve the fabled x/0? I mean x(infinity) should always equal a form of infinity, unless it's 0/0, in which 0(infinity) = 0. Many people disagree with some of my statements however. What do you guys think? ExoKiller4 (talk) 22:26, 11 June 2009 (UTC)

x/0 and (x)(infinity) are roughly analogous concepts, but they're nowhere near identical. 1/0 is an indeterminate form with a magnitude of infinity - Mathematica calls it "complex infinity". Not quite the same thing as "infinity", which is generally refers to the one that's positive and collinear with the real line (although not in the real line). --Zarel (talk) 02:45, 16 June 2009 (UTC)

Misunderstanding of Bishop Berkeley's argument

This sentence in the introduction:

"Historically, one of the earliest recorded references to the mathematical impossibility of assigning a value to a/0 is contained in Bishop Berkeley's criticism of infinitesimal calculus in The Analyst; see Ghosts of departed quantities."

shows a massive lack of understanding of what Bishop Berkeley's argument was.

His argument was that taking the limit of something as a quantity approaches 0 was not well-defined. In fact, at the time he wrote this, the taking of limits was indeed not well-defined. (Much later, Cauchy put the process of taking limits on a rigorous foundation.)

The mere fact that there is some similarity between dividing by zero on the one hand, and letting a quantity approach zero on the other hand, is quite different from the two things being the same: THEY ARE NOT.

Oh, and incidentally the "Ghosts of departed quantities" article has been hijacked by one user with no interest in historical accuracy, but only in advancing his agenda to promote non-standard analysis. It says nothing about what Berkeley's objections to calculus were, nor how the were resolved almost 200 years ago by Cauchy.Daqu (talk) 04:00, 15 July 2009 (UTC)