# Wikipedia:Reference desk/Mathematics

Welcome to the mathematics reference desk.

Main page: Help searching Wikipedia

How can I get my question answered?

• Explain what you need to know.
• Provide a short header that gives the general topic of the question.
• Tell us what part of the world your question applies to.
• 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.

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers directly address what the questioner asked (without tangents), are thorough, are easy to read, and back up facts with wikilinks and links to sources. Please assume good faith, especially with users new to Wikipedia. Don't edit others' comments and do not give any medical or legal advice.
Choose a topic:
 Computing desk Entertainment desk Humanities desk Language desk Mathematics desk Science desk Miscellaneous desk Archives

# December 3

## Are There Any R to C or C to R Bijective Functions ?

$f:\R\to\C,\quad f(x)-$ bijective, or $f:\C\to\R,\quad f(x)-$ bijective. I'm asking this with the following in mind: We know that $\N\to\N^2$ and $\N^2\to\N$ bijections do exist, given that they're both countable sets; and we also know that, say, a $\N\to\R$ or $\R\to\N$ bijection can't exist, for prescisely the opposite reason: But can such a bijection between two uncountable sets exist ? — 79.113.223.146 (talk) 09:47, 3 December 2013 (UTC)

The obvious one to me would be even and odd powers of ten being broken out so that in the R to C. 1234.567 maps to 24.6 + 13.57i. I know that only maps to half the complex plane (depending on where a negative value for the Real value goes to. But it is a start.Naraht (talk) 10:04, 3 December 2013 (UTC)
A space filling curve serves this purpose does it not? — Preceding unsigned comment added by 144.82.188.210 (talk) 11:57, 3 December 2013 (UTC)
No. A space-filling curve approaches arbitrarily close to every point, but doesn't actually pass through every point. Naraht has shown how to construct such a bijection, though. Looie496 (talk) 16:59, 3 December 2013 (UTC)
And then you take apart the resultant half of the complex plane on one unit wide stripes and rewallpaper alternating sides of the Complex axis.Naraht (talk) 17:30, 3 December 2013 (UTC)
Space-filling curves do hit every point; what you describe is impossible, since their image is closed. The problem is that they're not injective. It follows from algebraic topology that this is essential; there is no continuous bijection between R and C.
Naraht's approach also has a few details left to work out, since $11.000000\dots$ and $10.090909\dots$ both map to the same place. Fortunately, it's only a countable set of problem points.--80.109.80.78 (talk) 21:43, 3 December 2013 (UTC)
True. Just a first off the cuff idea. I'd be curious to see other possible solutions.Naraht (talk) 22:06, 3 December 2013 (UTC)
Well, you can use arctan to map C into the unit-square, and then use your idea to get an injection from C to R. Since there's an obvious injection from R to C, Schroeder-Bernstein finishes the job. Alternatively, you can note that the only problems with your idea come at decadic rationals, so fix some countable sequence of reals, slide them around to make room, then slide the decadic rationals in amongst them.--80.109.80.78 (talk) 22:30, 3 December 2013 (UTC)

Any infinite set A has a bijection to A x A, see Cardinal_number#Cardinal_multiplication (proofs probably require the axiom of choice). —Kusma (t·c) 14:08, 3 December 2013 (UTC)

In this case at least, the axiom of choice is not required. It's possible to construct explicit surjections from R to C and C to R. The Cantor-Bernstein-Schroeder theorem now does the heavy lifting. Sławomir Biały (talk) 17:49, 3 December 2013 (UTC)
Having surjections both ways, in the absence of choice, doesn't get you a bijection. You need injections both ways. But those are not hard to come by, so your first sentence is true. Actually all three of them are true; just the logical connection isn't quite there :-). (By the way, Schroeder–Bernstein theorem. That's a redirect, but Cantor's argument doesn't help you here, so I think it makes more sense to leave him out; it's not like there's a shortage of stuff named after him.) --Trovatore (talk) 21:52, 3 December 2013 (UTC)
Just a little post-hoc note: My "by the way" parenthetical above was written before another user corrected Sławomir's link — it was a redlink at the time. I wouldn't have bothered to add that, just to complain about Cantor in the name. --Trovatore (talk) 01:26, 4 December 2013 (UTC)

### Two surjectives doesn't equal one bijection?

I guess I'm a little confused why you need the Axiom of Choice to take a Surjective function from R to C and a Surjective function from C to R and get a bijective function between R and C. Is it that you simply need the Axiom of Choice to create a specific bijective function or do you also need the Axiom of Choice to prove that such a bijective function exists?Naraht (talk) 03:04, 4 December 2013 (UTC)

OK, for the specific case of R and C, you can prove the existence of a bijection (even give a rather "explicit" one), with no need to invoke choice.
In general, though, if you know that there are surjections from X to Y and vice versa, you need choice to prove that there's a bijection between X and Y. If you know that there are injections in both directions, you don't need choice.
Here's a specific example: Let X be the reals, and let Y be the disjoint union of the reals with the set of all countable ordinals. Clearly there's a surjection from Y to X (send every real to itself, and every countable ordinal to some fixed real). To get a surjection from X to Y, take some surjection from the negative reals to all the reals, and apply that to every negative element of X. Doesn't matter what you do with 0∈X (for definiteness, send it to the 0 of the copy of the reals in Y). For every positive element of X (positive real), either it codes a countable ordinal or it doesn't. If it doesn't, send it to the ordinal 0; if it does, send it to the ordinal it codes.
But in a model of the axiom of determinacy, there can be no bijection between X and Y, because that would give you an injection of ω1 into R, which contradicts AD. --Trovatore (talk) 03:14, 4 December 2013 (UTC)
I've been trying to see why that contradicts AD, but I'm stuck. Is there a straightforward construction of a game with no winning strategy from such an embedding?--149.148.254.102 (talk) 08:50, 5 December 2013 (UTC)
AD implies that every set of reals is either countable, or has a nonempty perfect subset. This is probably easier to see on 2ω — something along the lines of one player plays 0 or 1 and the other player plays finite sequences (and that's the player trying to get into the set). A winning strategy for the finite-sequence player implies that the payoff set contains the branches of an ever-branching tree, whereas from a winning strategy for the first player, you can read off the countable set. Or something like that; I forget the details but you can probably work them out.
So suppose there's an injection from ω1 into R. Its range is an uncountable set of reals, so it has a nonempty perfect subset, which therefore has the cardinality of the continuum. But that implies a bijection between ω1 and R, which implies R can be wellordered. --Trovatore (talk) 09:12, 5 December 2013 (UTC)
Great, thanks. FYI, here's the details for the perfect set property, as I was able to work them out: suppose the avoiding player has a winning strategy. Fix a point $z$ in our set. There must be some partial run of the game consistent with the avoiding player following this strategy, such that it is the non-avoiding player's turn, the sequence played so far is an initial segment of $z$, and every play by the non-avoider generates a response from the avoider that takes us away from $z$. For if not, we could always extend to stay on $z$ and thus get a win against the strategy. But this means that the strategy computes $z$: have the non-avoiding player play the next $n$ bits of $z$, and the avoider will necessarily respond with $(1 - z(n))$. So every element of our set is computable from the strategy, and thus our set is countable.--149.148.254.102 (talk) 10:24, 5 December 2013 (UTC)

# December 6

## Confused regarding necessary criterion for the sum of two irreducible fractions to be reducible

While doing some research on a recreational mathematical problem I stumbled on an interesting Wolfram Student Support thread from 2005. Interesting, because applying it will significantly reduce the complexity of the recreational problem.

Here the thread author refers to having recently read an (uncited) paper on the reduction of fractions. A result from this paper is apparently that

If $\frac{a}{b}$ and $\frac{c}{d}$ are both irreducible fractions, i.e., $\gcd(a,b) = \gcd(c,d) = 1$, then it is a necessary criterion that in the prime factorizations of the two denominators b and d, at least one prime factor shall occur an equal number of times for the sum
$\frac{a}{b} + \frac{c}{d} = \frac{a \cdot d+b \cdot c}{b \cdot d}$
to be reducible!

It is written with more words in the thread, but that is what I extract out of it.

The thread auhor states that such a prime is called a "balanced" prime between the two denominators, a quite different definition than the usual for a balanced prime.

For instance, if

$b = 12 = 2^2\cdot3$

and

$d = 20 = 2^2 \cdot 5$,

the number 2 is a balanced prime between the two denominators as it occur twice in the prime factorizations of both denominators.

Thus, the sum

$a/12 + c/20 = (20 \cdot a + 12 \cdot c)/(240)$

may be reducible for some a semiprime with 12 and some c semiprime with 20.

For instance, with

$a = c = 1$

the sum is $32 / 420$, which can be reduced to $8 / 105$.

The denominator is actually 240, so reduces to $2 / 15$ 109.151.42.94 (talk) 10:34, 6 December 2013 (UTC)

But here comes my problem, as I appear to be able to find a counter example of the opposite?!?

If

$b = 18 = 2\cdot3^2$

and

$d = 27 = 3^3$,

no "balanced" prime number exist between the two denominators as the only common prime number in their factorizations (3) does not occur an identical number of times (2 vs 3).

Thus, for any a semiprime with 18 and any c semiprime with 27, the sum $a/18 + c/27 = (27 \cdot a + 18 \cdot c)/(486)$ will not be a reducible fraction according to this rule. But, that does not appear to be true (or I am making an embarassing mistake)!? For instance, if $a = c = 1$, the sum is $45/486$. But $\gcd(45,486) = 9$, thus the sum can be reduced to $5/54$, in contradiction with the necessary criterion?!?

So, I am confused. Have I understood the necessary criterion for reducability correctly?

If so, what is wrong with my counterexample?

Is this rule called anything?

Thanks in advance, --Slaunger (talk) 20:10, 3 December 2013 (UTC)

Apparently the thread author is using a modified form of the addition formula. The example 13/36 + 7/45 = 93/180 is given but (ad+bc)/(bd) would give 837/1620. I'm guessing that what the author actually means is the way some of us were taught to add fractions in elementary school, write both fractions with the least common denominator the add the numerators. If so, the formula the author is actually using (though not writing down) is a/b + c/d = ((ad+bc)/gcd(b,d))/(bd/gcd(b,d)). With the example you give the least common denominator is 54, so the result of 5/54 fits. --RDBury (talk) 21:40, 3 December 2013 (UTC)
Ah, yes, of course it has to be the the summed fraction in the form where it is reduced with $\gcd(b,c)$ in the nominator and denomninator as you so clearly state!
Still, I find the result that
$\frac{a}{b} + \frac{c}{d}$
written as the fraction
$\frac{a\cdot[d/\gcd(b,d)] + [b / \gcd(b,d)]\cdot c}{b \cdot d / \gcd(b,d)}$
to be non-reducible if there does not exist a common prime factor in b and d, which occur an equal number of times to be pretty remarkable!
1. Is it correct?
2. Does this rule have a name? (I cannot find anything about it on Wikipedia)
--Slaunger (talk) 15:03, 5 December 2013 (UTC)

──────────────────────────────────────────────────────────────────────────────────────────────────── It's true. To see this, without loss of generality take gcd(b,d) = 1. The only way $ad+bc$ can have a prime factor in common with d is if c has a prime factor in common with d, but this is impossible if the original fraction was written in lowest terms. Likewise, $ad+bc$ cannot have a prime factor in common with b. Sławomir Biały (talk) 16:41, 5 December 2013 (UTC)

Resolved
You are right, thanks! And now I solved the problem, which was finding how many ways 1/2 could be written as a sum of distinct inverse square numbers using terms up to 1/802! --Slaunger (talk) 13:35, 8 December 2013 (UTC)

# December 7

## When is N choose K divisible by P ?

When is $\scriptstyle{N \choose K}$ divisible by P ? — 79.115.135.150 (talk) 23:11, 7 December 2013 (UTC)

Pascal's triangle modulo the prime number 7 for n less than 49.
See Lucas' theorem for one criterion. (This actually implies a pattern in Pascal's triangle modulo p that looks like a Sierpinski's triangle where at each stage you subdivide by p instead of 2.) Sławomir Biały (talk) 14:00, 8 December 2013 (UTC)

# December 8

## where can I find a list of mathematics terms?

I have helped to someone that learn with me at the school to learn math, he knows well English more than my language and I would like sometimes to translate the terms to make the math easier for his, but I don't know the English terms, so maybe you would like to give me a list of mathematics terms on English. It will help us to learn together. Thank you! 213.57.115.126 (talk) 21:46, 8 December 2013 (UTC)

Term by term you could easily look things up on Wikipedia. If you want lists, we have several at Category:Glossaries of mathematics and Category:Mathematics-related lists. Staecker (talk) 23:37, 8 December 2013 (UTC)

## Horizontal transformation factor

This is likely an easy question. I'm a Grade 11 student learning about functions, and this year we learned horizontal transformations (stretch/compression) of various functions (where in grade 10 we were taught only vertical transformations). Now, I had a test on exponential functions, and one of the questions (in the form 2kx) was 20.5x, and I said that it was a horizontal stretch by a factor of 2. My teacher said that it was actually a horizontal stretch by a factor of 0.5, and that the value of the factor was k and not 1/k. Who is right? Thanks. 50.101.203.177 (talk) 22:14, 8 December 2013 (UTC)

It could go either way depending on the terminology, but I'd side with you on this one. When you change from 2x to 20.5x the result is to stretch the graph out in the horizontal direction so that features of the graph appear twice as wide as they were before. See these plots. To me it's reasonable to call that a horizontal stretch by a factor of 2. Maybe your teacher is using some special terminology, or maybe they just got it wrong. If you really want to clarify things with your teacher, draw some pictures so that everybody's clear on what you mean when you say "stretch by a factor of ..." Staecker (talk) 23:34, 8 December 2013 (UTC)

# December 9

## Volume of cone circumscribed about a sphere

A student I'm tutoring had this problem for an exam. Somehow he got the right answer, but the way he did it doesn't seem to make sense. Here's the problem,

A right circular cone is to be circumscribed about a sphere of r=7. Find the height of the cone that has minimum volume.

He solved this problem by setting the volume of a cone equal to the volume of a sphere and plugging in 7 for r, and solving for h (height). But that assumes the volume of a cone and sphere are the same when a cone circumscribes a sphere, but how is that possible? Shouldn't the cone's volume always be greater? He got h=28 which is the correct answer. How else can this problem be solved? ScienceApe (talk) 21:05, 9 December 2013 (UTC)

This site gives a couple of alternative solutions, which result in h=4a (where h is the height of the cone, and a is the radius of the sphere. Equating the volume of the sphere, and the volume a cone where the base has the same radius as the sphere gives 4/3 π a^3=1/3 π a^2 h => 4a=h, the same expression. Note that the radius of the circumscribed cone is sqrt(2)a (easily see in the working of the second solution), so the cones are not identical, just the same height. It's possible your studet knew this identity, and used it, or there's some deeper principle I'm not spotting (or he just took a guess and has the luck of the gods). MChesterMC (talk) 10:19, 10 December 2013 (UTC)
I may be missing something, but isn't there only one way a right circular cone can be circumscribed around a given sphere? Sławomir Biały (talk) 14:41, 10 December 2013 (UTC)
Only if you choose the cone's height to base diameter ratio first. --CiaPan (talk) 08:07, 11 December 2013 (UTC)
"Right" only means that the cone's vertex is directly above the centre of its base (as opposed to a "slant" cone). So a right cone has two degrees of freedom (e.g. height and ratio of height to base diameter) and the condition that it has an inscribed sphere with a given radius only imposes one constraint, leaving one degree of freedom. Intuitively, you can have a tall thin cone with a base diameter only slightly larger than the sphere's diameter, or a short fat cone with height only slightly larger than the sphere's diameter, or anything in between. Gandalf61 (talk) 12:16, 11 December 2013 (UTC)

# December 10

## Reimann's integral from infinity to infinity

The question I am about to ask has been asked on wikipedia before but the answer seemed quite undecided and there was a lot of debate.

In "On the Number of Prime Numbers less than a Given Quantity.", Reimann makes a large jump between this step;

$\Pi(s-1)\zeta(s)=\int\limits_{0}^{\infin}\frac{x^{s-1}dx}{e^x-1}$

and this one;

$2\sin(\pi s)\Pi(s-1)\zeta(s)=i\int\limits_{\infin}^{\infin}\frac{(-x)^{s-1}dx}{e^x-1}$

Is this using a contour integral? (Perhaps a hankel contour would make sense in context? This paper seems to think so; http://www.damtp.cam.ac.uk/user/md327/fcm_3.pdf but seems to arrive at a slightly different result) or is it something else as it appears User:Eric Kvaalen was suggesting?

The original conversation can be seen here;

http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Archives/Mathematics/2012_August_22

Anyone know? — Preceding unsigned comment added by 5.81.8.11 (talk) 20:22, 10 December 2013 (UTC)

# December 11

## Lawson's Klein bottle

I've looked casually for a parametrization of Lawson's Klein bottle in S3, with no luck. Is this it?

w,x,y,z = cosθ cosφ, cosθ sinφ, sinθ cosφ, sinθ sinφ

The circles at φ=0 and φ=π, or at θ=0 and θ=π, coincide but in opposite senses. —Tamfang (talk) 00:21, 11 December 2013 (UTC)

## When is $\int\sqrt[m]{P_n(x)}\ dx$ Expressible in Terms of Elementary Functions ?

When is the antiderivative of the mth root of a polynomial that meets the following criteria expressible in terms of elementary functions ?

$P_n(x)=\sum_{k=0}^{n>2}a_kx^k\quad,\quad\begin{cases}a_n\neq0\\a_0\neq0\\m\in\N\smallsetminus\{0,1\}\end{cases}$86.125.207.160 (talk) 08:05, 11 December 2013 (UTC)

See Elliptic integral which only deals with the square root case and a polynomial of degree 4 or less, for cube roots etc the situation is even less tractable. Dmcq (talk) 10:44, 11 December 2013 (UTC)