Wikipedia:Reference desk/Mathematics

The Wikipedia Reference Desk covering the topic of mathematics.
Mathematics #eee #f5f5f5 #eee #aaa #aaa #aaa #00f #36b #000 #00f mathematics Wikipedia:Reference desk/Mathematics Mathematics WP:RD/MA

Welcome to the mathematics reference desk. Want a faster answer? Main page: Help searching Wikipedia How can I get my question answered? Provide a short header that gives the general topic of the question. Type ~~~~ (four tildes) at the end – this signs and dates your contribution so we know who wrote what and when. Post your question to only one desk. Don't post personal contact information – it will be removed. We'll answer here within a few days. Note: We don't answer (and may remove) questions that require medical diagnosis or legal advice. We don't answer requests for opinions, predictions or debate. We don't do your homework for you, though we’ll help you past the stuck point. How do I answer a question? Main page: Wikipedia:Reference desk/Guidelines The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

skip to bottom Choose a topic: Computing desk Entertainment desk Humanities desk Language desk Mathematics desk Science desk Miscellaneous desk Archives   See also: Help desk Village pump Help manual

December 31

Sinusoidal phase angle

In the article "Phase (waves)", it states that a phase is "the initial angle of a sinusoidal function at its origin". Does this mean gradient at x=0 or something different? — Preceding unsigned comment added by Demiseofanalog (talk • contribs) 22:24, 31 December 2014 (UTC)

I added a title. StuRat (talk) 23:05, 31 December 2014 (UTC)

Something different. In $f(x)=\sin(\omega x+\varphi)$, the phase is $\varphi$ - that is, when $x=0$, the sine function is taken on an angle of $\varphi$. -- Meni Rosenfeld (talk) 00:39, 1 January 2015 (UTC)

"angle" in this context means the argument of the trigonometric function, not the slope of the graph. -- ToE 14:24, 1 January 2015 (UTC)

January 2

Dimension of C(R) as a vector space over R

What is the dimension of the space of all continuous functions from R (the real numbers) to R, as a vector space over R? Is it a named cardinal invariant, or is it just $2^{\aleph_0}$, for some probably embarrassingly trivial reason I haven't noticed? (I'm pretty sure I have a proof that it's at least $\mathfrak{d}$, so in particular it's uncountable). --Trovatore (talk) 07:36, 2 January 2015 (UTC)

It is the same as the dimension of the space of all continuous functions from ℚ (the rational numbers) to ℝ, as a vector space over ℝ. Because ℚ is dense in ℝ. Bo Jacoby (talk) 11:56, 2 January 2015 (UTC).

The answer thus is "interesting" only of CH fails (while a proof may be interesting in either case)? YohanN7 (talk) 13:42, 2 January 2015 (UTC)

It's the continuum. Consider $f_r(x) = \max\{0, 1 - |x - r|\}$, a "spike" with support $[r-1, r+1]$. Then $\{ f_r : r \in \mathbb{R}\}$ is an independent set. For suppose you had some finite dependence relation, and consider the smallest $r$ with $f_r$ having non-zero coefficient in the dependence relation. Then there is some $x$ in the interior of the support of $f_r$ but not in the support of any other function in the dependence relation. This $x$ witnesses that the linear combination is not the 0 function, so it's not really a dependence relation.--173.49.37.187 (talk) 15:54, 2 January 2015 (UTC)

Yes, that does it. Thanks. --Trovatore (talk) 16:19, 2 January 2015 (UTC)

Group theory question

If G is a finite group, and is not Cp^n, n>1, p prime, and has a normal proper nontrivial subgroup N, then is it true that G can always be written as NH, where H is a subgroup of G and N and H have trivial intersection? thanks.2601:7:6580:5E3:69F0:C6A7:1CEB:D852 (talk) 17:36, 2 January 2015 (UTC)

Basically you're looking for a normal subgroup which is not a factor in a semidirect product. I think technically G=C₁₂=<a> meets the criteria, since it has a subgroup H=<a²> of index 2 and only one subgroup <a⁶> of order 2 and this is contained in H. Not really in the spirit of the question though so perhaps the generalized quaternion groups Q_4n are a better example since they have a normal subgroup of order 2n but this contains the only element of order 2. In general, except for groups of small order, the semidirect products are a special case and more general types of group extensions are are more usual. --RDBury (talk) 18:33, 2 January 2015 (UTC)

thanks, good answer.2601:7:6580:5E3:D047:E07F:4765:2B97 (talk) 19:26, 3 January 2015 (UTC)

Intersection of triangles in higher dimensions

The intersection of line segments in 1-D always includes some extreme points. This is not true in the plain. I would like to know if it is possible in d-dimensional space (d>3) to intersect 2 triangles in a way that only interior points are in the intersection. By interior points I mean points of the topological interior by the subspace topology induced via the plane of the respective triangles. Thus, no vertices nor edges would be allowed in the intersection. Is this possible, and if so, starting at which dimension? Is there a generalization to higher dimensional simplexes (tetrahedrons etc.)?

95.115.186.77 (talk) 20:03, 2 January 2015 (UTC)

No. Two intersecting planes always describe a line (unless the planes are identical). Doesn't matter how many dimensions you use. If the intersection is restricted to the portion of a plane described by a triangle, then the resulting line segment must terminate on an edge / vertex of a triangle. In general, two triangle can either A) not intersect, B) intersect at a single point at the boundary of both triangles (e.g. they share a vertex), or C) intersect with a line segment that runs from one boundary point to another boundary point. Dragons flight (talk) 20:21, 2 January 2015 (UTC)

Okay. Apparently I'm just no good at hypergeometry... Dragons flight (talk) 23:17, 2 January 2015 (UTC)

Yah boo, bet even a blind person would do better than you at 3D. ;-) Well actually he'd beat 99.9999...% of the world's population at that! Dmcq (talk) 00:36, 3 January 2015 (UTC)

Sorry if my question was not crystal clear (I kind of suspected incoming misunderstandings). Triangles in higher dimensions (d>3) don't fix a plane, but a hyper-plane, of dimension d minus one. 95.115.186.77 (talk) 20:32, 2 January 2015 (UTC)

Well, maybe you got a point that I can't see in my mind yet. Let's see who's faster, me or you, to pinpoint it out ;-). 95.115.186.77 (talk) 20:41, 2 January 2015 (UTC)

OK, I see my mistake. Those triangles are inside 2-dim planes, not d-1-palnes. Thus, the intersection has dimension 2 at most. Thank you for the ping on my brain. 95.115.186.77 (talk) 20:48, 2 January 2015 (UTC)

Dragons flight: no, two intersecting planes don't always describe a line. See XY and ZT cardinal planes of cartesian coordinates system XYZT in 4D space—they have just one point in common, (0,0,0,0). --CiaPan (talk) 23:11, 2 January 2015 (UTC)

Two triangles may share only an interior point in 4 dimensions. You just need one of the triangles to be in the plane defined by the first two coordinates and the second to be in the one defined by the second two coordinates and for both to have the origin as an interior point. You can always do that in a space which has a dimension which is the sum of the dimensions of the intersecting simplices (except if one of them is a zero dimensional point of course). Dmcq (talk) 22:00, 2 January 2015 (UTC)

As Dmcq said, two planes may have a single-point intersection in a space of four or more dimensions, so you can arrange thing so that the intersection point is an interior point of two planar triangles. However, if two distinct points belong to some plane, the whole line defined by those points belongs to the plane, too. So if you put two triangles in such position that they have at least two points in common, then their respective planes will have the whole line in common. That gives you two collinear and overlapping line segments, which are intersection of the line and the two triangles. An interior of each segment belongs to an interior of the respective triangle, but its endpoints are on the triangle's edge. So the common part of the two segments is a segment, whose endpoints belong to a union of the two triangles' edges. --CiaPan (talk) 06:57, 5 January 2015 (UTC)

January 4

Coupon-collector problem with packages of distinct coupons

Is there a name for, or any research on, the variant of the coupon collector's problem where coupons are acquired in batches of constant size (such as packs of trading cards) and duplication within a batch is impossible? NeonMerlin 04:43, 4 January 2015 (UTC)
Corrected your link Rojomoke (talk) 05:09, 4 January 2015 (UTC)

This paper seems to deal with the problem you described. There are many variants of the CCP; I don't know if this particular variant has been named. --Mark viking (talk) 10:46, 4 January 2015 (UTC)

Is every power series representation of a function a Taylor series?

I'm not convinced. We commonly say things along the lines of "$\sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n}$ is the Taylor series for $e^{-x^2}$" but to me, it is not completely trivial to show that this is in accord with the definition of a Taylor series using derivatives and Taylor's theorem. Likewise the Bessel function of the first kind is definable using a power series, but that series results from using the Frobenius method to solve Bessel's equation and I can't see how to derive the same series using Taylor's theorem (otherwise, it would imply that odd derivatives of the function must be zero at the origin while even ones aren't, which seems rather strange to me). The hypergeometric functions aren't readily identifiable as Taylor series to me either. What gives?--Jasper Deng (talk) 09:50, 4 January 2015 (UTC)

Let $f(x) = \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n}$ and $g(x) = e^{-x^2}$. Let $a_m = \frac{(-1)^{m/2}}{(m/2)!}$ if m is even, otherwise 0. Then $f(x)=\sum_{m=0}^{\infty}a_mx^m$. Since $f(x)=g(x)$ for every x, it follows that $f^{(k)}(x)=g^{(k)}(x)$ for any positive integer k. Assuming we can interchange summation and differentiation (I forgot the exact conditions that guarantee this), we clearly have $f^{(k)}(0)=a_kk!$, so $g^{(k)}(0)=a_kk!$. So the Taylor series of $g(x)$ is $\sum_{m=0}^{\infty}a_mx^m = \sum_{n=0}^\infty \frac{(-1)^n}{n!}x^{2n}$.

I'm sure there are theorems that show that, more generally, arithmetic operations and compositions on well-behaved Taylor series "works". -- Meni Rosenfeld (talk) 11:57, 4 January 2015 (UTC)

Yes. In the interior of the radius of convergence, term-by-term differentiation is allowed (by the ratio test). So every power series representation of a function in an open disc centered at x=a is also the Taylor series of the function centered at a. Sławomir Biały (talk) 12:17, 4 January 2015 (UTC)

January 5

Reverse Polish Notation

Hi all,

I'm trying to understand the following reverse polish notation:

SSP xor DS xor or xor not

If I had the following:

SSP xor DS xor or not

then I know it would be:

NOT((S xor S xor P) or (D xor S))

What would SSP xor DS xor or xor not be?

Letsbefiends (talk) 01:44, 5 January 2015 (UTC)

Ah, I got it! I wasn't pushing the value back onto the stack again. So it is:

NOT(((D xor P) or (D xor S)) xor S)

I had to reread that algorithm on the wiki a few times. - Letsbefiends (talk) 02:25, 5 January 2015 (UTC)

Cool, I'll mark this Q resolved then. StuRat (talk) 03:16, 5 January 2015 (UTC)

Well, RPN is to make it easy for the machine, not for people. :-) Bubba73 ^{You talkin' to me?} 04:24, 5 January 2015 (UTC)

Presumably that first "D" in your answer was a typo, and you meant to state:

not(((S xor P) or (D xor S)) xor S)

-- ToE 06:04, 5 January 2015 (UTC)

Resolved

Here is how a human might resolve this, looking at the expression (and rewriting if necessary) by iterative replacing operators with their respective arguments:

1. SSP xor DS xor or xor not
2. S (S xor P) (D xor S) or xor not
3. S ((S xor P) or (D xor S)) xor not
4. (S xor ((S xor P) or (D xor S)) not
5. not (S xor ((S xor P) or (D xor S))

CiaPan (talk) 05:52, 5 January 2015 (UTC)

There's even a programming language based on this Forth (programming language) which used to be reasonably popular. It has often been described as a Write-only language ;-) Dmcq (talk) 16:32, 5 January 2015 (UTC)

Fermat point of spherical triangles

Where is it? (Defining it analogically as the point where the sides subtend equal angles.) Double sharp (talk) 08:32, 5 January 2015 (UTC)

The article on Fermat point would say something about it if there was anything so it's unlikely there is anything on Wikipedia. There;s a couple of papers on the web if you Google 'Fermat point sphere'. An opportunity to contribute to Wikipedia by adding to that article? Dmcq (talk) 16:43, 5 January 2015 (UTC)

January 6

Area moment of inertia about y axis

If finding the area moment of inertia for a rectangle about the y axis does the equation become depth*breadth^3/12? 94.14.218.233 (talk) 20:27, 6 January 2015 (UTC)

It depends on the orientation of the rectangle. In general it would be given by $\iint_R (x^2+z^2) dS$, a surface integral over the rectangle R. If you are talking about only the xy plane, I still don't have enough information because I still don't know the placement of the rectangle.--Jasper Deng (talk) 20:33, 6 January 2015 (UTC)

Yes if the rectangle is symmetric about the y-axis (the y-axis goes through the middle of the rectangle and is parallel to the depth dimension) - but you need to multiply by the mass-per-unit-area ($\scriptstyle{\rho}$). The calculation is:

$\begin{align}I & = \int _{0}^{\mathrm{depth}}\int _{- \frac{\mathrm{breadth}}{2}}^{\frac{\mathrm{breadth}}{2}}\rho x^{2} \partial x \partial y \\ & = \int _{0}^{\mathrm{depth}}\left[ \frac{1}{3} \rho x^{3}\right]^{x = \frac{\mathrm{breadth}}{2}}_{x = - \frac{\mathrm{breadth}}{2}} \partial y \\ & = \int _{0}^{\mathrm{depth}}\frac{\rho}{12} \mathrm{breadth}^{3} \partial y \\ & = \left[ \frac{\rho}{12} \mathrm{breadth}^{3} y\right]^{y = \mathrm{depth}}_{y = 0} \\ & = \frac{\rho}{12} \mathrm{breadth}^{3} \mathrm{depth} \end{align}$

It is sometimes more convenient to express this in terms of the total mass of the rectangle

$m = \mathrm{breadth} \mathrm{depth} \rho$

So substituting

$\rho = \frac{m}{\mathrm{breadth} \mathrm{depth}}$

gives

$\begin{align}I & = \frac{\rho}{12} \mathrm{breadth}^{3} \mathrm{depth} \\ & = \frac{m}{12} \mathrm{breadth}^{2} \end{align}$

--catslash (talk) 22:55, 6 January 2015 (UTC)

I think by "area moment of inertia" he wants it when the density is equal to 1. Also, the density has to be constant on the rectangle, otherwise the integral has to be done differently.--Jasper Deng (talk) 23:10, 6 January 2015 (UTC)

@Catslash: Use upright or at least normal 'd' in the integral, it's not a partial derivative here:
$I = \int _{0}^{\mathrm{depth}}\int _{- \frac{\mathrm{breadth}}{2}}^{\frac{\mathrm{breadth}}{2}}\rho x^{2} \,\mathrm{d}x \,\mathrm{d}y$
or
$I = \int _{0}^{\mathrm{depth}}\int _{- \frac{\mathrm{breadth}}{2}}^{\frac{\mathrm{breadth}}{2}}\rho x^{2} \,dx \,dy$
instead of
$I = \int _{0}^{\mathrm{depth}}\int _{- \frac{\mathrm{breadth}}{2}}^{\frac{\mathrm{breadth}}{2}}\rho x^{2} \partial x \partial y$
Also separate words denoting values:
with a narrow space: $m = \mathrm{breadth}\, \mathrm{depth}\, \rho$
better with a normal space: $m = \mathrm{breadth}\ \mathrm{depth}\ \rho$
or with an explicit multiplication dot: $m = \mathrm{breadth}\cdot \mathrm{depth}\cdot \rho$
CiaPan (talk) 07:01, 7 January 2015 (UTC)

is this a sound proof?

Is the following a sound proof? I'd like to prove that there is no largest integer.

1. Firstly, suppose there were a largest integer. Let's call this L, for "Largest". Without loss of generality, let's say for example it's one million thirty-six, that this is the largest integer and there is no larger one.
2. Add 1 to L to arrive at a new candidate number, C. In our example, it would be one million thirty-seven.
3. Since L + 1 is greater than L, therefore the candidate C (which was L+1) must be larger than L.
4. Therefore L could not have actually been the largest, since without making any other assumptions except that it is the largest, we have constructed a larger candidate and proved it is larger. We were simply wrong that L was the largest.

Since this applies to any potential largest integer, L, therefore there really cannot be such a thing.

In other words, this doesn't just apply to a million thirty-six, but for any number. For example a million eighty-seven. (Or any other number we suppose might be the largest integer.)

It is a simple contradiction to state that there is a largest integer, L. There is no such integer.

Is this argument basically sound? 212.96.61.236 (talk) 02:34, 7 January 2015 (UTC)

It might be; to do this sort of thing properly, you need to specify what set of axioms you're working from. Step 2 assumes that it's always possible to add 1 to an integer and step 3 assumes that the result is always a larger integer. If you are assuming some version of the Peano axioms including axioms of ordering, then you can indeed make those statements. --65.94.50.4 (talk) 05:45, 7 January 2015 (UTC)

Would it work at the level of formality (rigor) usually adopted by mathematicians? (Albeit it can be a bit too explicit.) Or would it need to be more explicit to be made as above at all? 212.96.61.236 (talk) 07:55, 7 January 2015 (UTC)

That depends on what you mean. Is the above rigorous enough in the sense of people not interested in foundations, yes, its a basic result - if anything, it is overdone, "Assume n is the largest integer, n + 1 > n, contradiction." would work just as well. If you're asking if it is rigorous enough to be considered a proof from axioms, no, not really - however, given standard axioms and your proof, it is obvious how they relate; so, it's not really a problem. If you, however, want a fully formal rigorous proof from basic axioms, then, no, you haven't provided that -- but, unless you have a reason to want one, or just for fun, I don't see the point of bothering (especially for something simple).Phoenixia1177 (talk) 08:40, 7 January 2015 (UTC)

The proof works at the ordinary level of formality (= informality) used in everyday mathematics. Rarely do authors run to the axioms, but it may be instructive to do it in a case like this where the informal proof is easy. YohanN7 (talk) 10:43, 7 January 2015 (UTC)

It certainly wouldn't work for some machine where the largest positive integer is 2³¹−1 = 2147483647 and if you add 1 you get the large negative number −2³¹ = −2147483648. Dmcq (talk) 10:13, 7 January 2015 (UTC)

Yes, that isn't relevant on its own - it is true if the goal is a fully formal proof from axioms, then one needs be real specific. However, mathematicians say, "Let n be an integer", etc. all the time, without any real specification - if the goal is just to provide a proof that there is no greatest integer then, while verbose, the above does the job (it could be rewritten to sound more mathematical, but that's not really the point).Phoenixia1177 (talk) 10:31, 7 January 2015 (UTC)

I would also criticize your use of "Without loss of generality", as you very much are throwing away generality. If you wish to include a specific number to aid in understanding, just say, "Let's say for example it's one million thirty-six, that this is the largest integer and there is no larger one." Invoking WOLOG, as mathematicians use it, would mean that it is sufficient to prove that one million thirty-six is not the largest integer, and that is clearly not what you are asserting. -- ToE 13:28, 7 January 2015 (UTC)

This is symptomatic of people encountering the phrase wlog but never being explicitly taught what exactly it means. So they apply it where it is not appropriate.

In Hebrew, the phrase for wlog translates roughly as "without the limitation of generality". For a while I thought this should be read as "screw generality, we're not going to let it limit us". I think only after I encountered it in English I realized that it is us who are not limiting generality. -- Meni Rosenfeld (talk) 14:10, 7 January 2015 (UTC)

Ha! -- ToE 11:26, 8 January 2015 (UTC)

I agree that it shouldn't be used, but not for the same reasons. There is nothing gained by using a specific number - unless you are unused to using n for some arbitrary number, this is why it makes it look like an example - thus, it just seems out of place. However, I don't think it is an inappropriate use, the proof is easily adapted to the general case, indeed, the exact same logic applies - and this fits with the examples used in the article linked to too - however, as I said, it aids none and looks more like it is attempting to be illustrative than anything else, this makes it seem to sit wrong. It works a lot better, in other cases, because it removes the need for introducing clunky terms or using unpleasant notation (Like with "Wlog assume x <= y", it's more straightforward than two identical proofs or some contrived way to carry around both cases - but, here, it's extraneous, and that makes it awkward).Phoenixia1177 (talk) 17:59, 7 January 2015 (UTC)

I think the criticism of WLOG is wrong. There are two parallel proofs that I presented. One uses L for the number, the other uses one million thirty-six without loss of generality. This makes the proof stronger, not weaker. For example, here is the example the without loss of generality article gives:

“ To prove: "If three objects are each painted either red or blue, then there must be two objects of the same color." Proof: "Assume without loss of generality that the first object is red. If either of the other two objects is red, we are finished; if not, the other two objects must both be blue and we are still finished." ”

This is how it reads in my parallel construction:

1. If there are three objects, they can be enumerated 1, 2, and 3.
2. Their colors are C1, C2, and C3.
3. Let us start with C1, and its opposite color OC1. Without loss of generality, we could say C1 is Red for example. The opposite color (OC1) is blue then.
4. If either C2 = C1 or C3 = C1 then the colors of two objects match. (In our example if either C2 or C3 is red - then two objects, C1 and the one that is red, have matched)
5. If neither C2 = C1, nor C3 = C1 then they both must be a color other than C1. Since there are only two possible colors, they must therefore both be of the color OC1. Since they are both OC1, they both match. (In our example, if neither C2 nor C3 is red, then they must both be a different color. Since there are only two possible colors, the different color must be blue, and therefore they are both blue and thus match.)

You see, we have two parallel constructions here. One is a proof using symbols, another using an example without loss of generality. This aids, does not hinder, the quality of the proof in my opinion. For example, only in the first half of my version is the construction of the 'opposite color' invoked. It's not necessary if you use a WLOG argument, as you do not need to refer to the other color symbolically. So, I disagree with you guys and simply believes that I added to the strength of my proof by including a WLOG argument alongside the symbolic one. 212.96.61.236 (talk) 22:47, 7 January 2015 (UTC)

What ToE complained about is not the use of a "wlog argument". It was the use of the phrase "wlog". If you did the whole thing exactly the same and simply omitted the words "Without loss of generality", it would have been better. What you've given is an example, not a "wlog argument".

Anyway, regardless of whether the use of an example (and the phrase wlog) was valid or not, there is simply absolutely no way that it made the proof stronger. Giving the proof symbolically is sufficient. There is nothing extra to be gained by adding an illustrative example.

In general, in formal logic a proof is either valid or not. There is no sense in providing two proofs and hoping that together they are stronger. -- Meni Rosenfeld (talk) 23:34, 7 January 2015 (UTC)

Meni, I am not a working mathematician, but I simply disagree with you completely. The words "without loss of generality" are a very important clue that the example, in fact, serves as a complete independent proof. (A fact which may not be obvious if not mentioned explicitly - after all it could just be an example - and it may not be a valid statement to add to all examples.) By the way it is not always clear exactly how the reader should apply "without loss of generality". You're asking the reader "to repeat this for anything else I could have picked as an example" and basically introduce their mental symbology without being explicit about it.

Let me show you exactly why I stand by this opinion.

The above proof I presented you guys is actually a trick. It's not rigorous enough. You can tell, because I can use the same exact format to convince someone who isn't paying that much attention:

“

I'd like to prove that there is no largest negative integer.

1. Firstly, suppose there were a largest negative integer. Let's call this L, for "Largest". Without loss of generality, let's say for example it's negative one million thirty-six, that this is the largest negative integer and there is no larger negative integer.
2. Add 1 to L to arrive at a new candidate number, C. In our example, it would be negative one million thirty-six.
3. Since L + 1 is greater than L, therefore the candidate C (which was L+1) must be larger than L.
4. Therefore L could not have actually been the largest, since without making any other assumptions except that it is the largest negative integer, we have constructed a larger negative integer candidate and proved it is larger. We were simply wrong that L was the largest.

Since this applies to any potential largest negative integer, L, therefore there really cannot be such a thing.

In other words, this doesn't just apply to negative a million thirty-six, but for any number. For example negative a million eighty-seven. (Or any other number we suppose might be the largest negative integer.)

It is a simple contradiction to state that there is a largest negative integer, L. There is no such integer. If there were, we could construct a larger one.

”

Now I used the EXACT same reasoning, and the exact same level of rigor. If someone really isn't paying attention, there is a good chance they will believe me.

But this time the proof is wrong. This time there is a largest negative integer: −1. If you add one to it, you get something that isn't a negative integer anymore.

There was a hidden assumption that "any negative integer plus one is still a negative integer". But this is not true at all.

This is blatantly obvious from the proof above. But suppose I had presented only the WLOG. Now someone could have not realized htis fact. Since, in fact, the example works quite well for negative one million thirty-seven. It really isn't the largest negative integer, and negative one million thirty-six is an easy counter-example that's bigger by one.

How is the reader supposed to figure this out? Symbolically, it's easy. With a WLOG argument you're asking the reader to come up with all cases, instead of relying on properties of the symbols. That makes corner cases quite a bit harder to spot.

So, I simply disagree. While a WLOG argument can be perfectly valid, in fact the combination of that with a symbolic proof is strongest of all.

Why prove things one way, when you can prove things and show them as well? One for the intuition and as an exercise, the other to keep from being sneaky. Both are rigorous. Use of both is best. 212.96.61.236 (talk) 00:21, 8 January 2015 (UTC)

You're contradicting yourself. By your own admission, with your so-called "wlog argument" it's difficult to see whether the argument is fallacious or not. If a valid argument is indistinguishable from an invalid one, it means that the method of argument is completely worthless as a way of obtaining truths. Hence the wlog argument is unnecessary.

Whereas, in the symbolic argument, again by your own admission, it's clearer whether the argument is valid or not. Hence, if a symbolic argument supports a claim, and the argument appears valid, we can be convinced of the truth of the claim. And, if the argument is written formally and we've verified its validity exactly, we can be sure of its truth*.

And again - I agree that giving examples is instructive and can make the proof easier to follow, but there's no basis for your claim that the example somehow makes the proof "stronger".

Finally, the term "wlog" is used in mathematics almost exclusively when there is symmetry in naming of objects, so choosing a particular naming is arbitrary and irrelevant and does not cause a loss of generality. In the color example, "red" and "blue" are just names with no semantics attached to them, so they are interchangeable. Whereas in the numbers example, "1000000" and "10" are very different things. It's not an arbitrary choice of names. By choosing 1000000 you are very much losing generality - claims that are true for 1000000, will be wrong for 10. And your own examples strengthen this point - the claim "if n is a positive integer, then n+1 is a bigger positive integer" happens to be true regardless of n, so the "wlog argument" seems to work. But the claim "if n is a negative integer, then n+1 is a bigger negative integer" works for -1000000 but not for -1. So choosing -1000000 "wlog" gives an ostensibly valid argument for a wrong conclusion. So the "wlog argument" is worthless, and it's not wlog at all.

*modulo some weird insights into the distinction between provability and truth, which I'm sure Trovatore will be happy to point out.

-- Meni Rosenfeld (talk) 02:34, 8 January 2015 (UTC)

I do think the use of wlog is acceptable, in that it is not wrong, since it applies with regards to the operations used, etc, and the general argument follows an, obviously, identical course. The problem with using it is that it gains you nothing and, indeed, just makes the whole thing clunkier. Consider: if you are presenting this to people who are comfortable with mathematics, then it looks like an example, clunky, and obfuscating (and pointless) - if you are presenting it to someone who is not comfortable with mathematics, the construct is confusing and it still feels like an example. In short, sure, you can use it, but it's not a great usage and, certainly, adds nothing (and definitely takes something away). Good math is like a good novel, it's not just the story that matters, but also how well it is written - you're telling a good story for the topic, but wlog is pretty bad writing in the way you've used it.Phoenixia1177 (talk) 11:53, 8 January 2015 (UTC)

When using WLOG, then either it has to be obvious that the assertion follows from the restricted case under consideration or it has to be shown explicitly. This is not entirely clear in your proof. It is just about as clear as the original, yet unproven, assertion to begin with, making the WLOG argument, well, murky at best. It doesn't help. There is a false impression that a proof that is easier to read is always easier to understand. This is not the case. The reader may well think he/she understands it because of added gibberish, while, in fact, he/she does not. There is also the risk of logical fallacy as pointed out above. YohanN7 (talk) 12:08, 8 January 2015 (UTC)

Regarding your "trick", that was addressed in the very first response to your question, where 65.94.50.4 said that to be rigorous you need to show that any integer n has a successor, n+1, and that the successor is always a larger integer, that is, that n+1 is an integer, and that n+1 > n. Dmcq gave an example of a set (not ℤ, the set of Integers that mathematicians deal with, but an integral data type from computer science), where n+1 < n for a n = 2147483647, and you now provided an example using a different set, the negative integers, where n+1 is not a member of that set, even though n=-1 is a member. Did that first response not answer your intended question? -- ToE 17:01, 8 January 2015 (UTC)

January 8

Group Theory Question

Does anyone know the status of the following question: Given a finite group $G$ and a subgroup $H$ of $G$, is it true that the automorphism group of $H$, $Aut(H)$ isomorphic to a subgroup of the automorphism group of $G$, $Aut(G)$? My initial instinct was that this is false, but I haven't been able to produce a counterexample yet (admittedly, I haven't really put a lot of effort into that search yet). 76.14.232.104 (talk) 12:47, 8 January 2015 (UTC)

In order for it to be a subgroup it must, to begin with, be a subset. It isn't because elements of Aut H are maps from H to H while elements of Aut G aren't. YohanN7 (talk) 13:10, 8 January 2015 (UTC)

The real question is if you can extend any element of Aut H in such a way that it is an element of Aut G and such that group multiplication in Aut G is compatible with the extension. YohanN7 (talk) 13:25, 8 January 2015 (UTC)

... or one can read the question as "is isomorphic to a subgroup of" instead of "is a subgroup of", which is the way I think it would normally be read when considering questions of this nature (one is rarely interested in more than isomorphism here). I guess the OP's initial reaction is that because the structure of the group H might have automorphisms that do not extend to the full group G. While I cannot answer the question, if hunting for counterexamples, a comment in Automorphism#Examples hints that it might make sense to examine a subgroup H with a nontrivial centre. —Quondum 14:45, 8 January 2015 (UTC)

The isomorphism formulation may lose too much information. Consider a small subgroup of some infinite group, like a two-element group. Ah, finite groups. Well, the objection remains. A subgroup may occur many times in a containing group. YohanN7 (talk) 15:30, 8 January 2015 (UTC)

Consider e.g., G the symmetric group on 6 symbols, H the subgroup generated by the 2-cycles (12), (34)(56). There is an automorphism of H interchanging the two generators, but this is not induced by any automorphism of G. Sławomir Biały (talk) 16:11, 8 January 2015 (UTC)

My original intent was that it only be isomorphic to a subgroup of $Aut(G)$, I've edited the OP to reflect that. 76.14.232.104 (talk) 01:35, 9 January 2015 (UTC)

Let G be C₄×C₂ and H the subgroup isomorphic to C₂×C₂. Then #Aut(G) = 8 and #Aut(H) = 6 and the orders preclude Aut(G) from having an isomorphic copy of Aut(H). --RDBury (talk) 07:43, 9 January 2015 (UTC)

Number of edges in an nxnxn cube

The formula for the number of edges in an nxn square is 2n(n+1). What's the formula for the number of edges in an nxnxn cube?? I know that for n=1, the answer is 12. For n=2, I know that:

• There are 27 vertexes, but I want to know the number of edges.
• The number is bounded below by 36, because there are 36 edges on a group of three 2x2 squares.

Georgia guy (talk) 17:42, 8 January 2015 (UTC)

I'm pretty sure it's $3n(n+1)^2.$ There are 3 ways to choose the direction, $(n+1)^2$ ways to choose the location of the ray, and $n$ ways to choose the segment along the ray.

This generalizes to $dn(n+1)^{d-1}$ for a d-dimensional hypercube. -- Meni Rosenfeld (talk) 17:57, 8 January 2015 (UTC)