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March 22

0/0

why we are not able to divide zero by zero? in artical parity of zero,this tell that zero is even.if zero is even then why it is not divided by itself — Preceding unsigned comment added by Muhammad waqas arif (talk • contribs) 16:14, 22 March 2013 (UTC)[reply]

Please read Division by zero, especially the sections called In elementary arithmetic and Division as the inverse of multiplication. Zero is even, which means it can be divided by 2 with no remainder: 0/2=0. But no number, including zero, can be divided by zero. —Bkell (talk) 16:18, 22 March 2013 (UTC)[reply]

There are various rules for how to handle zero which conflict with each other in the case of 0/0:

"Zero divided by anything equals zero" means the result should be zero.

"Anything divided by itself equals one" means the result should be one.

"Anything divided by zero equals either positive or negative infinity (the sign of the numerator)" means the result should be ±∞.

It can't be all of these things, so the operation is not permitted. StuRat (talk) 16:26, 22 March 2013 (UTC)[reply]

"Anything divided by zero equals either positive or negative infinity (the sign of the numerator)" is a rule in IEEE floating point, but it is not a rule in any mathematical context that I know of. It is true that for a nonzero constant c the limit of c/x as x approaches zero is ±∞, depending on the sign of c, but c/0 itself is not defined. —Bkell (talk) 16:41, 22 March 2013 (UTC)[reply]

See real projective line. --Trovatore (talk) 16:43, 22 March 2013 (UTC)[reply]

Sure, division by zero is allowed there, but there is no distinction between positive and negative infinity. —Bkell (talk) 16:45, 22 March 2013 (UTC)[reply]

I like to think of the tangent graph, which shows how the curve is asymptotic to both positive and negative infinity, at the same points: [1]. So, it's +∞ approaching from one direction and −∞ when approaching from the other. However, at the actual point, it is rather undefined as to whether it's +∞ or −∞, I agree. StuRat (talk) 01:07, 24 March 2013 (UTC)[reply]

Anyway, the contradiction among these "rules" is not the fundamental reason that 0/0 is not defined; 0/0 is undefined because there is not a unique solution to the equation 0x = 0. —Bkell (talk) 16:46, 22 March 2013 (UTC)[reply]

Which is very closely related - StuRat's examples are the solutions 0, 1 and +/- ∞. Rich Farmbrough, 00:53, 23 March 2013 (UTC).[reply]

An interesting subtlety is that, while 0/0 does not usually make sense, it does make sense to say that zero divides zero (because

a|b\iff \exists n(b=na)

). However, unaccountably, one convention is that zero does not divide zero.

That convention, to the extent a convention can be wrong, is just wrong. I can see no advantage whatsoever, and considerable disadvantage, to excluding 0|0. In particular, if you don't say 0|0, then you lose the lattice structure of divisibility on the naturals, with 0 as the greatest element. --Trovatore (talk) 16:48, 22 March 2013 (UTC)[reply]

${\frac {0}{0}}={\frac {\pm 0}{0^{2}}}=\lim _{x\to 0}{\frac {\pm x}{x^{2}}}=\lim _{x\to 0}{\frac {\pm 1}{x}}=\pm \infty$

${\frac {0}{0}}={\frac {a\cdot 0}{0}}=\lim _{x\to 0}{\frac {ax}{x}}=a$ , where a can be any real number whatsoever. — 79.113.218.110 (talk) 20:14, 22 March 2013 (UTC)[reply]

0 being even and 0/0 being undefined don't have any relation to each other. Double sharp (talk) 14:36, 23 March 2013 (UTC)[reply]

Correction:

a

is not 'any' real number, it is 'every' real number, and positive and negative infinity. Plasmic Physics (talk) 01:16, 24 March 2013 (UTC)[reply]

a cannot be every real number, since that contradicts both logic and grammar. Nor is a x 0 defined when a = ±∞. — 79.113.210.71 (talk) 04:21, 24 March 2013 (UTC)[reply]

Care to explain the perceived contradiction? Plasmic Physics (talk) 04:45, 24 March 2013 (UTC)[reply]

My reasoning is that since the product of every element of the set of real numbers; and zero, is equal to zero; and zero divided by poitive and negative infinity, is equal to zero, it follows that zero divided by zero equals every element of the set of real numbers, and positive, and negative infinity. Plasmic Physics (talk) 04:54, 24 March 2013 (UTC)[reply]

As mentioned already, 0/0 is not defined. Nor is 0/∞. The example above shows how you can make a limit of the form 0/0 that actually converges to a for any a. You couldn't make it converge to all a simultaneously, because that would violate the definition of convergence.--80.109.106.49 (talk) 07:45, 24 March 2013 (UTC)[reply]

There must be something wrong with the convergence proof above, because my proof is definitely correct, unless the commutative principle is incorrect. Plasmic Physics (talk) 09:12, 24 March 2013 (UTC)[reply]

The limit of a function approaching a point is not necessarily equal to the value at the point. For the function which is 3 everywhere but 5 at 0 the limit at 0 is 3 but the value at 0 is 5. In the equation above 0/0 = limit x/x as x goes to 0 but that's not necessarily true. If there was a consistent result then that would be a good definition of 0/0 but there's nothing that's anywhere good enough. Dmcq (talk) 12:34, 24 March 2013 (UTC)[reply]

True, but division isn't exactly some random, custom-made, user-defined function. — 79.113.235.6 (talk) 18:57, 24 March 2013 (UTC)[reply]

I'm confused, is that comment for or against my comment? Plasmic Physics (talk) 12:33, 25 March 2013 (UTC)[reply]

Saying 0/0 = limit ax/x as x tends to 0 is wrong. That type equality is only true for functions which are continuous at the limit point. Continuity is a nice property and if there was a value that such limits consistently were equal to then that would be a good value to assign to 0/0. However as you have shown there isn't. It would be possible to assign some standard agreed value to 0/0 but that gives far more problems than it solves so we don't assign a value to 0/0. On the other hand if we had defined 5+x for all values except x=0 we would find that the limit as x tended to 0 for 5+x, 5+ax, 5+x² etc all were 5 and we would feel happy about assigning the value 5 to 5+0 so as to get a continuous function. For 0/0 if we knew we should be getting some definite value we'd assume we had made a small mistake could get rid of the zero's somehow, perhaps t is a limit we re trying to get? perhaps we kept a common factor of 0 in the numerator and denominator we should have cancelled out at an earlier stage? If we had a standard value then some maths software would just churn out that stupid value and we wouldn't have the foggiest idea we were totally wrong. Dmcq (talk) 13:25, 25 March 2013 (UTC)[reply]

Truth is self-coherent and self-contained, devoid of contradictions, and never at odds with itself. The di- in divergence means two. The con- in convergence means together. If one can show that the limit of a function in a point can have at least two different values, this means that the function is divergent. Because of the very meaning and definition of the word divergent. — 79.113.239.66 (talk) 18:47, 24 March 2013 (UTC)[reply]

The first element is actually dis, "apart": dis vergentia, "[to] bend/turn apart". Double sharp (talk) 11:26, 25 March 2013 (UTC)[reply]

dis- itself ultimately comes from the same PIE word for two, just like di-. — 79.113.235.6 (talk) 11:32, 25 March 2013 (UTC)[reply]