Share small, cool mathematical proofs:
Proof that [math]2^{\frac{1}{n}}[/math] is irrational for all [math]n > 2[/math]:
Suppose [math]z^{n} = 2[/math] where [math]z[/math] is a rational number.
[math](\frac{p}{q})^{n} = 2[/math]
[math]\frac{p^{n}}{q^{n}} = 1 + 1[/math]
[math]p^{n} = q^{n} + q^{n}[/math]
This is a contradiction of Fermat's Last Theorem, therefore [math]z[/math] must be irrational. QED.
Prove that [math]i^2 = -1[/math]
[math]\sqrt{i} = -1[/math]
Real number fags BTFO.
>>7889955
unfortunately this proof is not powerful enough to prove sqrt(2) is irrational
>>7889964
Your latex isn't displaying for me but I think you said sqrt(i)=-1.
You're a real number fag.
sqrt(i)=sqrt(2)+sqrt(2)i
>>7890032
pretty sure it is
>>7890047
>2016
>Doesn't know Fermat's last theorem.
I can provide a proof that sqrt(2) is irrational if you'd like, btw.
>>7890063
my imbecility clouded my vision
>>7890047
rofl
Is there some browser plugin for those [math] LaTeX formulas? One for chrome? There was too many LaTeX plugins to search them all.
Proof that raising a number to an irrational power can give you a rational number:
Regard: [math] 2^{sqrt {2}}[\math] Either this is rational then q.e.d. If not than raise this irrational number to the power of [math] sqrt {2}[\math] which will give you 2 which is obviously a rational number. q.e.d
>>7890092
Just use the TeX preview that's in the post window.
Have this for reference:
\int = integral
_ = subscript
^ = superscript
\frac{x}{y} = pretty fucking obvious what it is
\sum = -1/12 meme (summation)
\lim = get off this board if you dont know this
\rightarrow = self-explanatory
>>7890106
That's not really a profound proof. Let k be irrational and n rational. Then 1/k is irrational. n=n^1=n^(k*1/k)=(n^k)^1/k.
Hence, any rational number can be expressed as a real to the power of an irrational.
This stronger case is still kind of trivial tbqhwy.
Proof that for a group [math]G[/math] and [math]a \in G[/math] such that [math]o(a)=n[/math],
[math]o(<a^i>) = \frac{n}{gcd(n,i)}[/math].
First, note that [math](a^i)^\frac{n}{gcd(n,i)} = (a^n)^\frac{i}{gcd(n,i)} = e^\frac{i}{gcd(n,i)} = e[/math].
Now, suppose that there exists some integer [math]m, 1<m<\frac{n}{gcd(n,i)}[/math] such that [math]o(a^i)=m[/math].
Then, clearly, [math](a^i)^{m*gcd(n,i)} = e[/math], but [math]m*gcd(n,i) < n = o(a^i)[/math], so such an [math]m[/math] cannot exist.
Hence, [math]o(<a^i>) = \frac{n}{gcd(n,i)}[/math].
>>7890281
shut up nerd
Does anyone have any requests for theorems to be proven? No troll Riemann hypothesis, please.
I have a pretty nice proof of the Bolzanno Weierstrass theorem if anyone wants that.
>>7890281
Try \langle \rangle
[eqn] o(\langle a^i\rangle) = \frac{n}{gcd(n,i)} [/eqn]
Also does [math] \gcd [/math] work?
>>7890302
What did you start with for BWT?
I really like my proof of the point peak lemma.
>>7890325
[math]\langle[/math]That's much nicer. Thanks, anon.[math]\rangle[/math]
[math]\gcd[/math] versus [math]gcd[/math], so I suppose so.
>>7890336
In normal LaTeX style, gcd would be three variables, whereas \gcd is a math operator.
Many of them are predefined...\sin \cos \max \inf (this is why \inf isn't \infty btw)
I had to define my own for atan2 though
>>7889955
You almost got it
You also need to use the fundamental theory of arithmetic
Show that q and p are prime or the product of primes and go from there
>>7890328
I'll outline the proof that a bounded sequence has a convergent subsequence / accumulation point. An upper and lower bound of the relevant sequence is chosen, say m and M, and then two intervals are constructed: [m, k] and [k, M] where k is the midpoint of m and M. Then, a new interval is chosen such that it contains a non-finite number of points of the sequence. Two new intervals are constructed by splitting this interval, and the process continues recursively. The reals are complete, so a point is chosen in the intersection of all of these intervals, and then it is shown that there exists a subsequence which converges to this point.
What is your proof of the peak point lemma like?
>>7890302
Prove that the gamma function of 1/2 is sqrt(pi)
>>7890365
Let's try that again...
I recently learned the Banach fixed point theorem.
I know that this theorem is so obvious that it is almost trivial, but I found the way you prove it really cool yet simple (probably because I think it's the first time I saw an actual use of a sequence defined by iterating a function).
>>7890219
That assumes that a rational number to an irrational power is irrational, which is simply false. Try again.
>>7890400
I think you're suggesting that I tried to prove the PPL? I was talking about the BWT, not the point peak lemma, friend. Remember how you asked me how I started the BWT? Well I gave a quick overview of the proof.
>>7890431
No, it doesn't assume that at all. I was responding to someone who said a number to the power of an irrational can be rational. You surely accept that a rational raised to an irrational is a number, right?
>>7890396
http://jekyll.math.byuh.edu/courses/m321/handouts/gammahalf.pdf
I tried to do it with Laplace transforms but it didn't work. Here's how someone who isn't retarded did it.
>>7890449
Sorry for the confusion.
No that's my proof of the peak point lemma.
My first try at posting it cut off the conclusion.
Your BWT looks like it uses the nested intervals theorem.
>>7890510
Ah, no worries then.
I haven't really encountered the peak point lemma much before. Is there anything especially interesting that isn't obvious on the surface?
Yup! That's why I mentioned that the reals are complete. I often find that that nested-intervals definition of completeness is more useful than the Cauchy/convergence definition.
Anyone remember that guy who posted his three page proof on a^3+b^3=/=c^3 using the unit circle in november 2014? That was based.
Wonder if he's still around here, cause the name on the paper is John Blutarsky, and it's not turning up when I search for it. I downloaded a copy though, so I know I'm not imagining it.
>>7890519
Peak point lemma is used directly with monotone convergence theorem to prove BWT.
I kinda completely ignored BWT in my proofs, and just use those two.
> I often find that that nested-intervals definition of completeness is more useful than the Cauchy/convergence definition.
Haha. I like the monotone convergence theorem definition.
>>7890531
I don't remember that, but please post it if you find it
← proof by picture
>>7890562
>>7890535
Ah, nice, anon.
Proof that raising a number to an irrational power can give you a rational number:
Regard: [math]2^{\sqrt {2}}[/math] Either this is rational then q.e.d. If not, then raise this irrational number to the power of [math]\sqrt{2}[/math] which will give you 2 which is obviously a rational number. q.e.d
Proof that there are infinitely many primes:
Consider any finite list of prime numbers p1, p2, ..., pn. It will be shown that at least one additional prime number not in this list exists. Let P be the product of all the prime numbers in the list: P = p1p2...pn. Let q = P + 1. Then q is either prime or not:
If q is prime, then there is at least one more prime than is in the list.
If q is not prime, then some prime factor p divides q. If this factor p were on our list, then it would divide P (since P is the product of every number on the list); but p divides P + 1 = q. If p divides P and q, then p would have to divide the difference[2] of the two numbers, which is (P + 1) − P or just 1. Since no prime number divides 1, this would be a contradiction and so p cannot be on the list. This means that at least one more prime number exists beyond those in the list.
>>7890661
Babby challenge: Prove that every integer greater that one has a unique prime factorization.
(Hint, use strong induction and evaluate the case where n+1 in the inductive step is prime or composite.)
Proof that [math]e[/math] is irrational:
Taylor's integral formula gives us: [math]\displaystyle e = \sum_{k=0}^{n+1} \frac{1}{k!} + \int_0^1 \frac{(1-t)^{n+1}}{(n+1)!}e^t dt[/math].
We can then write [math]\displaystyle 0 < n! e - \sum_{k=0}^n \frac{n!}{k!} = \frac{1}{n+1} + \int_0^1 \frac{(1-t)^{n+1}}{n+1}e^t dt \le \frac{1+e}{n+1} < 1[/math] for each [math]n \ge 3[/math].
We finally get [math]\displaystyle \lfloor n! e\rfloor = \sum_{k=0}^n \frac{n!}{k!}[/math] for each [math]n \ge 3[/math]. In particular, [math]n!e[/math] is never an integer and, consequently, e is irrational.
Proof that $\binom{n}{k}=\binom{n}{n-k}$:
Let $T^n$ be the $n$-torus. By the Kunneth theorem, $dim H_k(T^n)=dim H^k(T^n)=\binom{n}{k}$.
By Poincaré duality, we thus have $\binom{n}{k}=\binom{n}{n-k}$.
>>7890281
Someone is clearly in abstract algebra. I think factor groups are much more mind blowing though
>>7890785
What can I say? You got me.
We skipped over quotient groups pretty quickly because retards can't understand equivalence relations.
Do you know anything about sigma algebras and measure theory, by chance?
Proof that [math]\mathrm{GL}_m(\mathbb R) \simeq \mathrm{GL}_n(\mathbb R) \Leftrightarrow m = n[/math].
Assume m < n. We will prove that [math]\{\pm 1\}^{n}[/math] can not be embedded in [math]\mathrm{GL}_m(\mathbb R)[/math].
Let [math]H \subset \mathrm{GL}_m(\mathbb R)[/math] be a subgroup such that [math]\forall h \in H, h^2 = I_m[/math]. Then H is abelian and every element in H is diagonalizable with eigenvalues among 1 and -1.
Furthermore, H is conjugate to a subgroup of [math]\displaystyle \begin{pmatrix} \pm 1 & & \\ & \ddots & \\ & & \pm 1 \end{pmatrix}_{[m]}[/math]. In particular, we have [math] |H| \le 2^m[/math].
Therefore, [math]\{\pm 1\}^n[/math] is not isomorphic to a subgroup of [math]\mathrm{GL}_m(\mathbb R)[/math]. However, it is a subgroup of [math]\mathrm{GL}_n(\mathbb R)[/math], namely [math]\displaystyle \begin{pmatrix} \pm 1 & & \\ & \ddots & \\ & & \pm 1 \end{pmatrix}_{[n]}[/math].
>>7890839
>>7890862
what's with this horrible proof?
the thing on the left is an m^2-dimensional vector space, the other is an n^2-dimensional vector space. vector spaces are isomorphic iff same dimension, qed
>>7890865
GL(n,R) is not a vector space, but only an open subset of a vector space. Subtlety, but I agree that anon's proof is overly complicated.
>>7890871
>but only an open subset of a vector space
elaborate please
>>7890885
GL(n,R) is a group which is a subset of R^(n^2)
>>7890865
But this would only prove that these two things are not isomorphic as topological groups (in the case of GL_n, dimension is not an algebraic notion, rather a topological notion), which is not enough (R and C are not isomorphic as topological groups, but, very surprisingly, they are isomorphic as groups).
This is a very nice (I think) algebraic proof that only uses linear algebra.
>>7890885
The null matrix is not in GL(n,R) ...
>>7890810
Fuck equivalence relations; it's all about that universal property.
>>7890871
It is not *overly* complicated tbqh. Group theory is hard and if you want to prove that two groups are not isomorphic, then there will be some work to do to find something that one can do and the other can't. In this case, it is being able to reflect in n independent directions.
If [math]2^k - 1[/math] is prime then [math]2^{k-1}(2^k - 1)[/math] is perfect.
Proof. Let [math]n = 2^{k-1}(2^k - 1)[/math]. Since [math]2^k - 1[/math] is prime, we know that [math]\sigma(2^k - 1) = 2^k[/math]. Then, noting that [math]\sigma(2^m) = 2^{m+1} - 1[/math] and that [math]2^{k-1}[/math] and [math]2^k - 1[/math] are relatively prime, we have
[math]\sigma(n) = \sigma(2^{k-1}(2^k - 1)) = \sigma(2^{k-1})\sigma(2^k - 1) = 2^k(2^k - 1) = 2n[/math].
Thus, [math]n[/math] is perfect.
>>7890676
Well, that certainly wouldn't fall in the category of small proofs.
>>7891058
It is small?
>>7891058
It's like one step.
>>7891058
First, note that 2 has the prime factorization [math]2^1[/math].
Now, suppose that all integers 2,...,n have unique prime factorizations.
Case 1: Assume n+1 is prime. Then the prime factorization of n+1 is simply [math](n+1)^1[/math].
Case 2: Assume n+1 is not prime. Then, there exist integers n and m, 2<=n, m<= n+1 such that mn=n. Now, by inductive assumption, m and n each have a unique prime factorization. It is left as an exercise for the reader to take the product of m and n.
Q.E.D.
>>7891078
>proof by exercise for reader
>>7889955
Do you know of any alternative proofs for this? I'm interested to see how it would be done without needing to invoke one of the most difficult results in mathematical history (however cool that may be).
>>7892416
Assume [math]\sqrt[n]{2} = \frac{p}{q}[/math] with p and q coprime and n > 1. Then [math]p^n = 2q^n[/math]. Therefore 2 divides [math]p^n[/math] and consequently divides p (this follows from the uniqueness of the prime factorization of p).
Now, we can write [math]p = 2p'[/math] and we have [math]2^n p'^n = 2q^n[/math] and then [math]2^{n-1}p'^n = q^n[/math]. By the same argument as before, 2 divides q. But then p and q are not coprime, contradiction.
A similar argument proves that any algebraic integer that is rational is actually an integer.
