Wikipedia:Reference desk/Archives/Mathematics/2009 January 6

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

January 6

Product (not normal) symbol in LaTeX

Hello, I am reading chapter 8 of Royden which is on topological spaces. He uses a big X for product topologies instead of $\prod$. Does any one know the LaTeX symbol for this? StatisticsMan (talk) 01:47, 6 January 2009 (UTC)

The Comprehensive LaTeX Symbol List is the place to look. You seem to have two choices, \bigtimes in the mathabx package or \varprod in txfonts/pxfonts. Base LaTeX has $\bigotimes$. -- BenRG (talk) 01:55, 6 January 2009 (UTC)

See the following macro. 142.103.8.38 (talk) 22:25, 26 November 2009 (UTC)

does 0.999... equal 1 ?

Please bear with me... I'm no mathematician...

If 0.(9) = 1

then 1 = 1.(0)1

and 0.(9) = 1.(0)2

therefore any x = any y

I'm saying that if 0.(9) is accepted to be 1, then any number is equal to any other number. Or from another perspective, the difference between any two numbers is 0.

I guess another tone to put the question would be:

Given any two numbers, one can be equated with the other, by adding or subtracting an (infinite) series of infinitesimal amounts. An infinitesimal amount equates to zero, therefore the difference between the two numbers is an (infinite) sum of zeros.

Yet, in real life, numbers can be conceptualized to be different from each other, and this "discreteness" of them does provide basis for feats of science and engineering.

Meaning, people, where is it ?

Lord KRISHNA (talk) 05:07, 6 January 2009 (UTC)

You went wrong in assuming that any two numbers can be equated with each other by adding or subtracting an infinite number of infinitesimal amounts. That assumption holds in the examples you gave, but how about, say, in the case of 2.1 and 2.2? You can rewrite 2.1 as 2.1(0)9 and 2.2 as 2.1(9), but the difference between those two isn't zero. There is no way to rewrite the two numbers in the way you described. --Bowlhover (talk)

Put in another way, there is not such a real number represented by "0.(9)27" or similar, just because that is not an allowed decimal expansion. The rule is that a decimal representation of the fractional part of a real number has to be a sequence of digits indicized on $\mathbb {N} _{+}$ . If the meaning of "0.(9)27" is that first come countably many 9, then the 2 follows all nines and the last 7 follows 2, we just can't use natural numbers to indexize those digits in that order, so that is not a proper representation of a real number. You are forced to use all natural indices for the 9's, and then you need two further indices like $\omega$, $\omega +1$ for the last two digits. Nevertheless your intuition is meaningful: you can in fact consider systems of numbers represented this way by sequences of digits indexed on ordinal numbers; you will get a non-archimedean field (i.e. containing infinitesimals). Similarly, you can also consider formal power series like $\scriptstyle \sum _{k\leq \alpha }c_{k}x^{k}$, and prove nice algebraic properties of the corresponding constructions, but they are definitely different objects--PMajer (talk) 10:11, 6 January 2009 (UTC)

This:
    'then 1 = 1.(0)1'
makes no sense. The '(0)' part means 'zeros on ALL remaining positions', so there is NO place to put your final '1'. Thus it has no meaning, adds nothing to '1.(0)'. --CiaPan (talk) 10:37, 6 January 2009 (UTC)

We have an article about this: 0.999... -- Aeluwas (talk) 10:45, 6 January 2009 (UTC)

Yea but it's still people who will discuss because they feel it logically incorrect. For example, one thing I've always wondered as it shows in the article why 1/9*9 would be 0.(9) and not 1, in the same way it doesn't go logical that 1/2*2 would equal 0.(9) — chandler — 10:54, 6 January 2009 (UTC)

Indeed. Or, turning the argument around, in ternary, 1/9 (decimal) is 0.01 (ternary) and (1/9)*9 (decimal) is 0.001*100 (ternary). In decimal arithmetic (1/9)*9 evaluates to 0.(9), whereas in ternary arithmetic, 0.001*100 clearly evaluates to 1. So if 0.(9) (decimal) is not equal to 1, then the value of (1/9)*9 depends on which base you are working in. Gandalf61 (talk) 11:20, 6 January 2009 (UTC)

The question "how can one ever get from A to B, if an infinite number of events can be identified that need to precede the arrival at B?" was zenos paradox.Cuddlyable3 (talk) 11:40, 6 January 2009 (UTC)

Anyway I wouldn't claim that "1.(0)1" has no meaning... Sure it has not the current meaning of a real number, but it is at least a representation of an idea of Lord KRISHNA. Nor I would say that there is no place left for the last 1, because in fact he was able to put it :). I would not even say that 1.(0) or 0.(9) is the real number one; it's more properly a representation of it. --PMajer (talk) 12:23, 6 January 2009 (UTC)

Thank you all for replying. So basically I get from your answers that I'd need a (much) better grasp of mathematical language and notation to investigate this matter further and in a rigorous way. But I suppose that 1.(0)1 is at least at conceivable as pi, if not more. I mean, 1.(0)1 is the same as 0.(9) but results from sequences coming from opposite sides of the real axis (real is the most complex my untrained mind can go right now). The notion that all real numbers are joined together "in infinity, by infinitesimals" through additions or subtractions is also not difficult to conceive for me (so much so that this matter has really been bugging me recently).

This is why I gave up on pure logic in high school; clearly to me any logical thought is in itself dry and sterile, and the conclusion is always a reflection of the premise. The only meaningful things are empirical experiences. Said more poetically, the only meaningful thing is communication of reflections. It's like quantum physics or something. Maybe numbers don't actually exist. Maybe I'm just trying to look smart..

Anyway thanks for endulging the cravings of my mind !

---

You are thinking about interesting things. Here's an apparent paradox that seems to be similar to your idea of "infinitely many infinitesimals": Think about the idea of "length". Clearly a single point, by itself, has no length—its length is zero. And if I have five or ten or a million points, the sum of all of their lengths is still zero. Yet somehow, if I have enough points (infinitely many of them—uncountably many, to be technically precise), they can form a line segment, which does have a (nonzero) length. So somehow it must be possible to add together an infinite number of infinitesimals and get something other than zero, which I think is at the heart of your question. This thought experiment leads into a profound and fundamental area of mathematics called measure theory, in which the idea of "length" is formally defined in a logically rigorous and unambiguous way, and leads to some rather surprising results. —Bkell (talk) 00:34, 7 January 2009 (UTC)

For Euclid, points and lines were disparate elements. Imagine his surprise at your thought experiment in alchemistic transmutation. Cuddlyable3 (talk) 10:21, 7 January 2009 (UTC)

Euclid used the word "element" to mean "fundamental"; that is, the theorems he proves in "Elements" are fundamental to an understanding of mathematics at the level that the Greeks had reached in Euclid's time. The title "Elements" in fact comes from the title of a lost textbook by Hippocrates of Chios, "Elements of Geometry". There is no evidence for either Euclid of Hippocrates being atomists; in particular the Greek notion of atomism did not develop until after Hippocrates' time. Thus the title of Euclid's Elements has no connection with the chemical/physical notion of a chemical element.

However, Euclid did think of points and lines are very different "kinds" of things; he would not have regarded putting together an infinite number of points to form a line as a valid geometric operation to be used in proofs. That is not to say that he hadn't thought of it already; the Greeks had come up with the concept of an infinitesmal, but were not able to manipulate it in a rigorous method. Greek geometers, even before Euclid's time, informally calculated areas of two dimensional figures by regarding those figures as being composed of an infinite number of lines; but, after determining the area of a figure they proved their answer to be correct rigorously using Eudoxus' method of exhaustion, without the use of infinitesmals. Eric. 68.18.17.165 (talk) 15:05, 7 January 2009 (UTC)

...and if you want to explore in another direction you may also enjoy this--PMajer (talk) 09:29, 7 January 2009 (UTC)

max size of disconnected graph

Resolved

 – Shahab (talk) 08:49, 6 January 2009 (UTC)

Suppose G is a graph of order n, i.e. has n vertices. What is the maximum number of edges (or order) possible in G so that G is not connected. I have a feeling that the answer is ${\frac {(n-1)(n-2)}{2}}$, (Cnsider the graph obtained by taking a $K_{n-1}$ and adding a vertex but no edges to it.) However I want a mathematical proof that no disconnectd graph with more edges exists. I'll be gateful for any help--Shahab (talk) 06:46, 6 January 2009 (UTC)

First, show that a disconnected n-vertex graph with the maximum possible number of edges must consist of exactly two connected components, each of which is a complete graph (otherwise another edge could be added without connecting the graph). Now, if k is the number of vertices in one of these components, then the number of edges in the graph is $\textstyle {k \choose 2}+{n-k \choose 2}$. Considering n as a fixed constant, this expression is quadratic in k; all that needs to be done is to show that it attains its maximum at $\textstyle k=1$ and $\textstyle k=n-1$ (as k ranges over meaningful values, of course). —Bkell (talk) 07:06, 6 January 2009 (UTC)

Thanks for the quick response and the excellent explanation. (In your last sentence I guess you meant: it attains its maximum at $\textstyle k=1$ or at $\textstyle k=n-1$). The answer is what I expected. Cheers--Shahab (talk) 08:49, 6 January 2009 (UTC)

No, he meant and. It attains its maximum at both ends (since the situation's symmetric). Algebraist 15:29, 6 January 2009 (UTC)

Or, equivalently, consider the complement of your graph of two connected components above. It is a complete bipartite graph between the nodes on one side and the ones on the other. (This is because an edge is not in your graph above iff it crosses between the two components; so the edges in the complement are exactly those that cross between the two components.) The number of edges in this complete bipartite graph is simply (the number of edges on one side)×(the number of edges on the other) = $k(n-k)$. It is easy to see that this quantity (which is a downward pointing parabola centered at n/2) is minimized (and therefore the number of edges in the original graph maximized) at the ends of k = 1 and k = n-1. --Spoon! (talk) 05:15, 7 January 2009 (UTC)

roots of y=0 and definitions

"Find a polynomial with integer coefficients that has ${\sqrt {2}}+{\sqrt[{4}]{2}}$ as a root."

Is "$y=0$" a technically correct answer to this problem? Aren't all numbers roots of "$y=0$"? —Preceding unsigned comment added by Metroman (talk • contribs) 07:25, 6 January 2009 (UTC)

Yes, you are technically correct ("the best kind of correct"), though if this question was on an exam or a homework assignment it should be clear that the answer "$y=0$" trivializes the question and thus would not receive full credit (if any). —Bkell (talk) 08:58, 6 January 2009 (UTC)

Okay, by $y=0$ I assume you mean the zero polynomial i.e. the polynomial with no non-zero terms, which evaluates to zero for all values of y (before someone jumps on me - yes, I know that some definitions of "polynomial" specifically exclude this case). Well, obviously the question should read "Find a non-zero polynomial ..." - it is probably assumed that if you are smart enough to think of the zero polynomial then you are smart enough to realise that it is not the intended answer. Gandalf61 (talk) 11:07, 6 January 2009 (UTC)

${\sqrt {2}}$ is an irrational number and it cannot be defined by any function couched in integer numbers.Cuddlyable3 (talk) 11:33, 6 January 2009 (UTC)

Well, that depends on what you mean by "defined" and "function". ${\sqrt {2}}$ is not equal to the ratio of any two integers. But "square root" itself is a function. If you want to stick to rational functions with integer coefficients, then ${\sqrt {2}}$ can be defined algebraically as the positive root of $x^{2}-2$. Or it can be defined analytically as the limit of

$x_{n+1}={\frac {x_{n}+2}{x_{n}+1}}$

for example. Gandalf61 (talk) 11:50, 6 January 2009 (UTC)

All true. The questioner wants a rational function and not a limit. Cuddlyable3 (talk) 14:50, 6 January 2009 (UTC)

The questioner doesn't want a rational function or a limit, they want a non-zero polynomial over the integers with ${\sqrt {2}}+{\sqrt[{4}]{2}}$ as a root. Such a polynomial exists, but we aren't going to give it, because we don't do homework here. Algebraist 15:26, 6 January 2009 (UTC)

Since the Welcome banner says "The reference desk will not do your homework for you." and there is no evidence that we are asked to do that, WP:AGF must apply. Cuddlyable3 (talk) 17:15, 6 January 2009 (UTC)

AGF does not tell us this, since OP did not ask what the answer was. OP asked if y=0 should be considered an answer.Taemyr (talk) 14:53, 7 January 2009 (UTC)

Aside to Gandalf: who defines the zero polynomial to not be a polynomial? Isn't that just really stupid? Algebraist 15:28, 6 January 2009 (UTC)

Didn't say I thought it was sensible, did I ? I vaguely remembered seeing a big debate somewhere about whether the zero polynomial was really a polynomial, and I just thought I would cover my bases in case someone got all pedantic on me. Gandalf61 (talk) 17:21, 6 January 2009 (UTC)

Gods, I hope no-one's teaching such rubbish to students. Algebraist 17:25, 6 January 2009 (UTC)

jeez... questioner's bones everywhere... is there a piece left, for me...--PMajer (talk) 17:35, 6 January 2009 (UTC)

As the cannibals said to the one who came late to dinner, You're too late, everybody's eaten. Cuddlyable3 (talk) 19:24, 6 January 2009 (UTC)

Eliminate y from the equations x = y²+y, y⁴ = 2 to obtain an equation for x. Bo Jacoby (talk) 00:05, 7 January 2009 (UTC).

I asked this question because it was a problem on one of my math tests. I knew how to find the intended solution but I discovered the shortcut answer. I answered with "y=0" but my teacher did not give me any credit. I just wanted to make sure that I was correct.

For those of you that think I'm just trying to trick you into doing my homework, here is how I could have done the problem. $0=x-({\sqrt {2}}+{\sqrt[{4}]{2}})=x-{\sqrt {2}}-{\sqrt[{4}]{2}}$
${\sqrt[{4}]{2}}=x-{\sqrt {2}}$
${\sqrt {2}}=x^{2}-2{\sqrt {2}}x+2$
${\sqrt {2}}+2{\sqrt {2}}x=x^{2}+2$
${\sqrt {2}}(2x+1)=x^{2}+2$
$2(4x^{2}+4x+1)=x^{4}+4x^{2}+4\,$
$8x^{2}+8x+2=x^{4}+4x^{2}+4\,$
$0=y=x^{4}-4x^{2}-8x+2\,$
Metroman (talk) 03:17, 7 January 2009 (UTC)

To be technical, the question did specify "integer coefficients," so you need at least 2. Where's the second one? You could argue that 0 can be considered a coefficient of x^0 (or any other power of x), but that would be a very unreasonable stretch for which you definitely won't receive any marks. --Bowlhover (talk) 04:56, 7 January 2009 (UTC)

Two coefficients? Hope '7×y+3 = 0' would by okay, so I supose '1×y + 0 = 0' is okay, too... --CiaPan (talk) 07:55, 7 January 2009 (UTC)

however FYI even x is a polynomial in x with integer coefficients, and 0 as well--84.221.209.108 (talk) 09:14, 7 January 2009 (UTC)

In what way is $p(x)=x^{4}-4x^{2}-8x+2$ not a polynomial with integer coefficients? Not only has Metroman found a correct answer, he has the "best" answer, in the sense that $p(x)$ is the unique irreducible monic polynomial with $\alpha ={\sqrt {2}}+{\sqrt[{4}]{2}}$ as a root. The set of all polynomials having $\alpha$ as a root is the ideal generated by $p(x)$; that is, every polynomial $f(x)$ with integer coefficients having $\alpha$ as a root can be written in the form $f(x)=p(x)g(x)$ where $g(x)$ is another polynomial with integer coefficients. (In particular, to answer Metroman's original question, yes $f(x)=0$ is such a polynomial that has $\alpha$ as a root.) Eric. 68.18.17.165 (talk) 14:32, 7 January 2009 (UTC)

Yes, however it seems to me that Bowlhover's objection was about the zero polynomial as a solution. The plural thing is most likely a joke, but in this context it can be misleading, so better to repeat once more that a monomial is a polynomial (and a polynomial is a power series, as well as a 1-Lipschitz function is a 2-Lipschitz function, a linear functional is a nonlinear functional, a positive measure is a signed measure, etc). The language of mathematics follows rules of convenience as any other language, and sometimes the evolution brings the use far away from the etymology. More often than not the number of terms of p(x) is unknown and irrelevant, and we don't want to have to say all times: "let p be a polynomial or a monomial or a constant", which would be another PITA, as it is the she/he form for somebody of unknow and irrelevant sex. We simply don't need a term for a "polynomial which is not a monomial", while we do need a term for "real or complex number which is not rational": that's why "irrational" still means "not rational". --PMajer (talk) 19:52, 7 January 2009 (UTC)

Why not have a look at polynomial ring or I would highly recommend having a look at ring (mathematics). At least it is relevant...--Point-set topologist (talk) 20:28, 7 January 2009 (UTC)

Indeed... the very reason of the current meaning of "polynomial" is that K[x] is something, whereas {polynomials-not-monomials} is devil-knows-what. In some sense the shift in meaning of this and other mathematical terms just follows the shift of interest from "individuals" to "classes" in mathematics. --PMajer (talk) 21:06, 7 January 2009 (UTC)

Determining size of label for cylindrical bottle

My high school geometry days are long behind behind me. If I have a cylinder of 'h' height and 'r' radius, how would I determine the linear dimensions (x & y dimensions) of a paper label that would wrap around the object? --70.167.58.6 (talk) 21:34, 6 January 2009 (UTC)

The height of the paper will be the same as the height of the cylinder, h. To figure out the width, imagine looking down the cylinder lengthwise, so that you just see a circle. Then unwrapping the cylinder is like unrolling this circle to form a straight line; thus the width of the paper is the same of the circumference of this circle. Since the circle has radius r, the circumference is $2\pi r$, and the paper label has dimensions $2\pi r$ by h. Eric. 68.18.17.165 (talk) 21:44, 6 January 2009 (UTC)