/sci/ - Science & Math

Proof that any nth root of 2 is irrational when n>2:

Assume rationality and derive a contradiction.
If nth root of 2 is rational then
2^(1/n)=a/b for some real a,b
Put both sides to the power n
2=(a^n)/(b^n)
2(b^n)=a^n
b^n+b^n=a^n
But this is clearly impossible because x^n+y^n=z^n has no solutions for n>2.

>>6513235
>But this is clearly impossible because x^n+y^n=z^n has no solutions for n>2.
prove it

Have you read Proofs from The Book?

Proof that log base anything of anything is irrational unless both the base and the number under the log are either even or odd (in this case it still may be irrational, however).
I can't LaTeX so I'll post like this: log(base)(number to evaluate the log of)
log(n)(x)=a/b
n^(a/b)=x
n^a=x^b
This is clearly possible with integer a,b iff n and x are either both even or both odd.

>>6513238
I have a truly marvellous proof, which the character limit of this post is too small to contain.

I just googled it and the book looks excellent. Looking at the PDF now. Thanks, anon!

>>6513240
*impossible

>>6513238
Just reading it now and a lot of the math is too advanced for me. I'll try printing some off and working through it with a pencil but I suspect I won't be able to get much out of it. :/

Probably don't need to post the proof here because everyone knows it, but I really like Euclid's proof of infinite primes.

Seems practical to me

That's what Fermat said.

>>6513235
I fucking love it.

>>6513235
>using FLT
nice job anon, just specify integer n and we're good

>>6513235
This makes me rock hard

>>6513240
I don't get what you are trying to say/show

this one isn't famous but

Theorem: Between any two rationals there is a third rational.

Say we have two rational numbers,

N = p/q and A = a/b

p/q < a/b

There exists a rational number c such that p/q < c < a/b.

To prove that such a number exists, we will take the average of p/q and a/b.

(p/q + a/b)/2

= p/2q + a/2b
= pb/2qb + aq/2qb
= (pb + aq)/2qb

Since pb + aq is an integer and 2qb also an integer, we have shown that a third rational exists between any two rational numbers.

QED

>>6513323
I like that a lot, anon.
>>6513321
i was trying to show that any log base of any number is irrational unless the base and the number are both either even or odd.

>>6513323

Same anon here.

Did you guys know that the sum of any five consecutive integers is divisible by five?

The sum of five consecutive integers can be expressed as

n+(n+1)+(n+2)+(n+3)+(n+4)

which simplifies to

5n+10

Both components of this expression are divisible by five since 5n is clearly divisible by 5 and 10/5 = 2, so the sum of any five consecutive integers is divisible by 5!

QED

>>6513329
Blah. I see what you were saying and I like it!
Here is something I did last semester for my topology class.

>>6513344
I'm not great at proofs, so I'm proud of this even if it does suck.

>>6513338
It may have been clearer if you factored 5n+10 to 5(n+2) but I still like that proof.
>>6513344
I honestly am just getting into math and I wish I could understand your proof but I can't.

I'm not actually sure my log thing is valid, so if anyone out there finds a flaw, please let me know.

>>6513338
Another way:
n-3+n-2+n-1+n+n+1
5n-5
5(n-1)=0 (mod 5)
Or, actually a much clearer way:
n-2+n-1+n+n+1+n+2
5n
5n=0 (mod 5)

>>6513233
I never liked that proof. Mainly because there's a stronger more intuitive proof about identities on groups. I'll write it up and post it in a while.

>>6513344
counterexamples are always good.

>>6513348
Something else I proved in the class
Theorem: Let <span class="math"> A [/spoiler] be a compact set in a metric space <span class="math">X[/spoiler]. <span class="math">A [/spoiler] must be bounded.
Proof:
Let <span class="math">B[/spoiler] be a collection of open balls so that <span class="math"> B=[/spoiler]{<span class="math">{D(c,r):r\in{R^+}}[/spoiler]} for some constant <span class="math"> c \in X [/spoiler].
<span class="math">B[/spoiler] is an open cover of <span class="math"> A[/spoiler] because <span class="math">B[/spoiler] covers all of <span class="math"> X[/spoiler] and <span class="math"> A \subset X[/spoiler].
By the statement of the problem, <span class="math"> A[/spoiler] is compact.
By definition of compact, <span class="math"> A[/spoiler] has a finite subcover. Let the finite subcover equal {<span class="math">D(c,r_{1}),D(c,r_{2},...D(c,r_{3})[/spoiler]}.
The definition of boundedness states: <span class="math"> A~squence~ is~called~bounded~provided~that~there~exists~an~open~disc~D~which~contains~every~term~of~the~sequence.[/spoiler]
Let <span class="math">r_{max}[/spoiler] equal the largest <span class="math">r[/spoiler] value for the discs in the finite subcover.
<span class="math">A\in D(c,r_{max})[/spoiler].
Thus <span class="math"> A[/spoiler] is bounded.
QED

Also, how do you do a black box in TeX on here?

>>6513348
Something else I proved in the class
Theorem: Let <span class="math"> A [/spoiler] be a compact set in a metric space <span class="math">X[/spoiler]. <span class="math">A [/spoiler] must be bounded.
Proof:
Let <span class="math">B[/spoiler] be a collection of open balls so that <span class="math"> B=[/spoiler]{<span class="math">{D(c,r):r\in{R^+}}[/spoiler]} for some constant <span class="math"> c \in X [/spoiler].
<span class="math">B[/spoiler] is an open cover of <span class="math"> A[/spoiler] because <span class="math">B[/spoiler] covers all of <span class="math"> X[/spoiler] and <span class="math"> A \subset X[/spoiler].
By the statement of the problem, <span class="math"> A[/spoiler] is compact.
By definition of compact, <span class="math"> A[/spoiler] has a finite subcover. Let the finite subcover equal {<span class="math">D(c,r_{1}),D(c,r_{2}),...D(c,r_{n})[/spoiler]}.
The definition of boundedness states: <span class="math"> A~squence~ is~called~bounded~provided~that~there~exists~an~open~disc~D~which~contains~every~term~of~the~sequence.[/spoiler]
Let <span class="math">r_{max}[/spoiler] equal the largest <span class="math">r[/spoiler] value for the discs in the finite subcover.
<span class="math">A\in D(c,r_{max})[/spoiler].
Thus <span class="math"> A[/spoiler] is bounded.
QED

Also, how do you do a black box in TeX on here?

>>6513380
Also, I proved this as a lemma trying to work my way to the Heine Borel theorem, which, for whatever reason, I'm still not able to prove. I've resisted looking up proofs of it in the hope that if I ever go back to it I might get it.
I'm able to prove special cases, but not the theorem in general. It doesn't seem like it should be that hard, but I have issues trying to prove things are compact.

>>6513233
Using commutativity was completely unnecessary. Identity is <span class="math">always[/spoiler] defined as a left and right identity regardless of the type of group.

15/100, F--.

>>6513384
When I proved the theorem in the original post I just proved cancellation and OP's proof fell out of that. All you need after you have cancellation is the definition of set equality and associativity (as well as identity, but that is what the theorem deals with).
The theorem for unique inverses is also super similar.

>>6513323
Now prove there's also a irrational between them and that between 2 irrationals there's both a rational and irrational to get the whole theorem.

I like Euclid's proof that there are infinitely many primes: if you multiply all of the known primes, none of them divide that number plus one. Therefore either this new number is prime itself, or has a prime factor not listed. Regardless, we conclude that there is always another prime to be found.

It's nice because it only requires a little bit of number theory to prove a lemma, and then the proof follows easily from there.

>>6513384
some older treatments start by defining a left identity element and a right identity element then demonstrating that they are in fact, one in the same. utter fucking tedium lol.

>>6513392
Hint: prove order density of irrationals and rationals.

>>6513394
AHHHH
In my group theory class a couple years ago, one of the first questions the professor asked was if there was a difference between the left and right identity and asked us to prove that there wasn't. Without commutativity, how do you do that?

Topology guy here: I have a proof that any closed interval in <span class="math"> R [/spoiler] is compact, if anyone wants to see it.
Idk if it's that elegant, but none of my proofs really are.

>>6513323

I like this one.

>>6513394
Only time I've ever seen that is with semigroups

Generally this is given as a definition, but it is totally provable and I love the proof, for whatever reason. (Topology guy, btw)
Proof that <span class="math"> \emptyset \not\subseteq A[/spoiler] for any set <span class="math">A[/spoiler].
By definition of <span class="math"> \not\subseteq [/spoiler], <span class="math"> A \not\subseteq B[/spoiler] iff <span class="math"> \exists x \in A : x \notin B [/spoiler].
Assume <span class="math">\exists A: \emptyset \not\subseteq A [/spoiler].
For this to be true there must <span class="math"> \exists x \in \emptyset : x \notin A [/spoiler].
By definition of <span class="math">\emptyset[/spoiler], <span class="math">\not\exists x : x \in \emptyset [/spoiler].
Therefore <span class="math"> \emptyset \subseteq A [/spoiler].
QED

Generally this is given as a definition, but it is totally provable and I love the proof, for whatever reason. (Topology guy, btw)
Proof that <span class="math"> \emptyset \subseteq A[/spoiler] for any set <span class="math">A[/spoiler].
By definition of <span class="math"> \not\subseteq [/spoiler], <span class="math"> A \not\subseteq B[/spoiler] iff <span class="math"> \exists x \in A : x \notin B [/spoiler].
Assume <span class="math">\exists A: \emptyset \not\subseteq A [/spoiler].
For this to be true there must <span class="math"> \exists x \in \emptyset : x \notin A [/spoiler].
By definition of <span class="math">\emptyset[/spoiler], <span class="math">\not\exists x : x \in \emptyset [/spoiler].
Therefore <span class="math"> \emptyset \subseteq A [/spoiler].
QED

>>6513401
I would like to see it. I love topology! Did you have to prove that the finite complement topology on R is not induced by a metric?

>>6513321
Change the "possible" to "impossible" in the final line and it makes a lot more sense.

>>6513344
Not very rigorous, but it does show the intuition of such a proof. Great job!

>>6513398
Let x be the left identity and let y be the right identity:

x=xy=y

Proof is complete by transitivity.

>>6513394
Yea, I've always found that tedious as well. It is not the method I mentioned in this post >>6513371

>>6513398
I will write down some proofs and definitions if you give me a while.

>>6513235
This fails if anywhere in the ridiculously long proof of Fermat's last theorem, this fact is used. Still pretty funny though.

>>6513416
Is the nullset contained within the nullset?

>>6513233
I like the similar proof of showing that 0*a=0 in a ring. If only because when I did it in my first algebra class we used the notation 0 subscript R for the zero element of a ring R and so some lines read sort of like "oraoraora".

>>6513233
What is 0/?

Sorry for my stupidness

>>6513521
another identity. the point being to prove that 0 and 0' are the same and thus that the identity element is unique.

An easy one but I like it:
<span class="math"> \lim_{n \to \infty} \, x^{\frac{1}{n}} = 1 [/spoiler] <span class="math"> \log (\lim_{n \to \infty} \, x^{\frac{1}{n}}) = \lim_{n \to \infty}\,\frac{1}{n}log(x) = 0log(x) = 0 [/spoiler]
Since the log is 0, <span class="math"> \lim_{n \to \infty} \, x^{\frac{1}{n}} = 1 [/spoiler]

<span class="math">latex\\test[/spoiler]

I know /sci/ hates economics, and with good reason -- the field has been hijacked by politicians and people who blindly believe anything they read.

Personally, I find it disgraceful how this mathematically elegant result gets crudely thrown around as >muh free market

Stolen from proofwiki, but probably my favourite circle area proof

A theorem proved in a variety of ways can be interesting. The Fundamental Theorem of Algebra comes to mind.

I like this one: >>6513471
(http://en.wikipedia.org/wiki/Interesting_number_paradox))

>>6513233
> nobody posts the h-cobordism theorem
;_;

Here is a proof that the definition of a group only needs left identity and left inverse. I quite like it. It also works if you start with right identity and right inverse, but fails if you have one of each (e.g. right identity and left inverse).

I always liked how simple some proofs in maths are, and have some kind of flow when reading them. I'm studying CS and most proofs are just muh induction on terms/formulas/derivations

I like this proof of the cardinality restriction on finite fields that doesn't rely on linear algebra:

Let F be a finite field. Take any two elements a,b of F*.
There exists k in F* such that kb = a(k = a*b^-1).
By distribution k*(n-sum of b) = (n-sum of a) for all n.
Since k is in F*, for any n, (n-sum of a) = 0 iff (n-sum of b) = 0..
Thus, a and b have the same additive order.

As a and b were arbitrary, all elements of F* have the same order. To respect Cauchy's theorem, the order of F must have only one prime factor, thus |F| must be a prime power

>>6513583
anon look at some of the results here lol. what were you seriously expecting?

also, you could base a course on the higher-dimensional proof of the h-cobordism theorem.

>>6513398

This is the definition for groups I carry around in my head along with four extra theorems. It's kind of an older treatment that this anon referred to >>6513394, but personally I prefer it because it's somewhat more minimal (in order to prove something is a group I only have to prove associativity, left inverse, and left identity). I really hope I didn't fuck up the LaTeX, I kind of avoid posting it on /sci/ since there's no way to know till it's been posted.

Let <span class="math">S[/spoiler] be a set. Define a (total) function <span class="math">\circ \colon S \times S \to S[/spoiler]. The tuple <span class="math">M = \left (S, \circ \right )[/spoiler] is called a Magma.
Recall that a function by definition is both well defined and only maps to items in the codomain (closure). Also, recall that a total function is defined for every element in the domain (in other words the pre-image is the domain). As a convention use the notation <span class="math">s_1 \circ s_2 := \circ(s_1, s_2)[/spoiler].

A magma is a semigroup iff <span class="math">(\forall s_1,s_2,s_3 \in S)[(s_1 \circ s_2) \circ s_3 = s_1 \circ (s_2 \circ s_3)][/spoiler]. In other words, an associative magma is a semigroup.

A semigroup is a group iff the following two conditions are met.
1) There exists a left identity in <span class="math">S[/spoiler]. <span class="math">(\exists e \in S)(\forall s \in S)[e \circ s = s][/spoiler]
2) Condition 1 is met and there exists a left inverse in <span class="math">S[/spoiler] for every element in <span class="math">S[/spoiler]. <span class="math">(\forall s \in S)(\exists s^{-1} \in S)[s^{-1} \circ s = e][/spoiler]

Now we'll prove the rest of the requirements that one would typically see in a definition of a group (in the process showing that this definition is sufficient).

(cont.)

>>6513597

We'll start with what is probably the most striking thing about this definition. The inverse and identity are only one sided. We will prove that they must be two sided.

First we'll prove that any left inverse must also be a right inverse.
Let <span class="math">(e \in S)[/spoiler] be a left identity, let <span class="math">(s_1 \in S)[/spoiler] be any element.
By our definition this implies that the left inverse of <span class="math">s_1[/spoiler] must also exist in <span class="math">S[/spoiler]. In other words.
<span class="math">\Rightarrow(\exists s_1^{-1} \in S)[s_1^{-1} \circ s_1 = e][/spoiler]
Since <span class="math">s_1[/spoiler] and <span class="math">s_1^{-1}[/spoiler] are both in <span class="math">S[/spoiler] then so must <span class="math">s_2 = s_1 \circ s_1^{-1}[/spoiler].
Our goal is to show that <span class="math">(s_2 = e)[/spoiler] (and therefore proving that <span class="math">s_1^{-1}[/spoiler] is also a right inverse).
First we prove that <span class="math">s_2 \circ s_2 = s_2/<span class="math">, it makes the proof clearer.
s_2 \circ s_2 = (s_1 \circ s_1^{-1}) \circ (s_1 \circ s_1^{-1})
= s_1 \circ (s_1^{-1} \circ s_1) \circ s_1^{-1}
= s_1 \circ (e \circ s_1^{-1}) \\= s_1 \circ s_1^{-1}
= s_2

Next recall that (s_2 \in S) \Rightarrow (\exists s_2^{-1} \in S)[s_2^{-1} \circ s_2 = e] and proceed to prove that e = s_2
e = s_2^{-1} \circ s_2
= s_2^{-1} \circ (s_2 \circ s_2)
= (s_2^{-1} \circ s_2) \circ s_2
= e \circ s_2
= s_2

Therefore any left inverse must also be a right inverse. Note that we accomplished this by only using left identity, left inverse, and associativity.

(cont.)[/spoiler][/spoiler]

>>6513594
I thought we had some grad students though?

Yeah, Milnor's book on it is really good, although a bit dated. Differential topology in general though, can be very beautiful and intuitive.

Have you taken a look at any banach manifold morse theory? I'm thinking of looking into it a bit.

>>6513597
We'll start with what is probably the most striking thing about this definition. The inverse and identity are only one sided. We will prove that they must be two sided.

First we'll prove that any left inverse must also be a right inverse.
Let <span class="math">(e \in S)[/spoiler] be a left identity, let <span class="math">(s_1 \in S)[/spoiler] be any element.
By our definition this implies that the left inverse of <span class="math">s_1[/spoiler] must also exist in <span class="math">S[/spoiler]. In other words.
<span class="math">\Rightarrow(\exists s_1^{-1} \in S)[s_1^{-1} \circ s_1 = e][/spoiler]
Since <span class="math">s_1[/spoiler] and <span class="math">s_1^{-1}[/spoiler] are both in <span class="math">S[/spoiler] then so must <span class="math">s_2 = s_1 \circ s_1^{-1}[/spoiler].
Our goal is to show that <span class="math">(s_2 = e)[/spoiler] (and therefore proving that <span class="math">s_1^{-1}[/spoiler] is also a right inverse).
First we prove that <span class="math">s_2 \circ s_2 = s_2/<span class="math">, it makes the proof clearer.
s_2 \circ s_2 = (s_1 \circ s_1^{-1}) \circ (s_1 \circ s_1^{-1})
= s_1 \circ (s_1^{-1} \circ s_1) \circ s_1^{-1}
= s_1 \circ (e \circ s_1^{-1})
= s_1 \circ s_1^{-1}
= s_2

Next recall that (s_2 \in S) \Rightarrow (\exists s_2^{-1} \in S)[s_2^{-1} \circ s_2 = e] and proceed to prove that e = s_2
e = s_2^{-1} \circ s_2
= s_2^{-1} \circ (s_2 \circ s_2)
= (s_2^{-1} \circ s_2) \circ s_2
= e \circ s_2
= s_2

Therefore any left inverse must also be a right inverse. Note that we accomplished this by only using left identity, left inverse, and associativity.[/spoiler][/spoiler]

>>6513597
We'll start with what is probably the most striking thing about this definition. The inverse and identity are only one sided. We will prove that they must be two sided.

First we'll prove that any left inverse must also be a right inverse.
Let <span class="math">(e \in S)[/spoiler] be a left identity, let <span class="math">(s_1 \in S)[/spoiler] be any element.
By our definition this implies that the left inverse of <span class="math">s_1[/spoiler] must also exist in <span class="math">S[/spoiler]. In other words.
<span class="math">\Rightarrow(\exists s_1^{-1} \in S)[s_1^{-1} \circ s_1 = e][/spoiler]
Since <span class="math">s_1[/spoiler] and <span class="math">s_1^{-1}[/spoiler] are both in <span class="math">S[/spoiler] then so must <span class="math">s_2 = s_1 \circ s_1^{-1}[/spoiler].
Our goal is to show that <span class="math">(s_2 = e)[/spoiler] (and therefore proving that <span class="math">s_1^{-1}[/spoiler] is also a right inverse).
First we prove that <span class="math">s_2 \circ s_2 = s_2[/spoiler], it makes the proof clearer.
<span class="math">s_2 \circ s_2 = (s_1 \circ s_1^{-1}) \circ (s_1 \circ s_1^{-1})[/spoiler]
<span class="math">= s_1 \circ (s_1^{-1} \circ s_1) \circ s_1^{-1}[/spoiler]
<span class="math">= s_1 \circ (e \circ s_1^{-1})[/spoiler]
<span class="math">= s_1 \circ s_1^{-1}[/spoiler]
<span class="math">= s_2[/spoiler]

Next recall that <span class="math">(s_2 \in S) \Rightarrow (\exists s_2^{-1} \in S)[s_2^{-1} \circ s_2 = e][/spoiler] and proceed to prove that <span class="math">e = s_2[/spoiler]
<span class="math">e = s_2^{-1} \circ s_2[/spoiler]
<span class="math">= s_2^{-1} \circ (s_2 \circ s_2)[/spoiler]
<span class="math">= (s_2^{-1} \circ s_2) \circ s_2[/spoiler]
<span class="math">= e \circ s_2[/spoiler]
<span class="math">= s_2[/spoiler]

Therefore any left inverse must also be a right inverse. Note that we accomplished this by only using left identity, left inverse, and associativity.

>>6513235
>^n+b^n=a^n
>But this is clearly impossible because x^n+y^n=z^n has no solutions for n>2.
The proof of this stage is left as an exercise.

>>6513603
Next we prove that a left identity is also a right identity.
Let <span class="math">(e \in S)[/spoiler] be a left identity. Let <span class="math">s \in S[/spoiler] be some element in <span class="math">S[/spoiler].
<span class="math">(s \in S) \Rightarrow (\exists s^{-1} \in S)[s^{-1} \circ s = e][/spoiler]
Note that <span class="math">(s \circ e)[/spoiler] must be in <span class="math">S[/spoiler] and proceed.
<span class="math">s \circ e = s \circ (s^{-1} \circ s)[/spoiler]
<span class="math">= (s \circ s^{-1}) \circ s[/spoiler]
A left inverse is also a right inverse
<span class="math">= e \circ s[/spoiler]
<span class="math">= s[/spoiler]

Therefore any left identity must also be a right identity. Here we only used left identity, two-sided inverse, and associativity.

(cont.)

>>6513606
The last two proofs

Then we prove that there can only be one left identity.
Suppose you have a second left identity <span class="math">f[/spoiler], in other words.
<span class="math">(\exists f \in S)(\forall s \in S)[(f \circ s = s) \wedge (e \neq f)][/spoiler]
Then
<span class="math">e = f \circ e<span class="math">
but we proved that e is also a right identity, so
= f
This contradicts our hypothesis so there must only be one left identity, similarly there must only be one right identity, and one identity in general.

The last proof is the uniquness of the left inverses.
Suppose that given (s \in S)[\math] you have a second left inverse \widetilde{s^{-1}}.
(\forall s \in S)(\exists \widetilde{s^{-1}} \in S)[(\widetilde{s^{-1}} \circ s = e) \wedge (\widetilde{s^{-1}} \neq s^{-1})]
Then, because e is also a right identity.
\widetilde{s^{-1}} = \widetilde{s^{-1}} \circ e
and because s^{-1} is also a right inverse.
= \widetilde{s^{-1}} \circ (s \circ s^{-1})
=(\widetilde{s^{-1}} \circ s) \circ s^{-1}
=e \circ s^{-1}
=s^{-1}
This contradicts our hypothesis so there must only be one left inverse, right inverse, and inverse in general.[/spoiler][/spoiler]

>>6513606
Lets try this again. It won't let me delete the fucked up post. Apparently you aren't allowed to delete posts that often.

>>6513612

Forgot to post the actual post.

Then we prove that there can only be one left identity.
Suppose you have a second left identity <span class="math">f[/spoiler], in other words.
<span class="math">(\exists f \in S)(\forall s \in S)[(f \circ s = s) \wedge (e \neq f)][/spoiler]
Then
<span class="math">e = f \circ e[/spoiler]
but we proved that e is also a right identity, so
<span class="math">= f[/spoiler]
This contradicts our hypothesis so there must only be one left identity, similarly there must only be one right identity, and one identity in general.

The last proof is the uniquness of the left inverses.
Suppose that given <span class="math">(s \in S)[/spoiler] you have a second left inverse <span class="math">\widetilde{s^{-1}}[/spoiler].
<span class="math">(\forall s \in S)(\exists \widetilde{s^{-1}} \in S)[(\widetilde{s^{-1}} \circ s = e) \wedge (\widetilde{s^{-1}} \neq s^{-1})][/spoiler]
Then, because <span class="math">e[/spoiler] is also a right identity.
<span class="math">\widetilde{s^{-1}} = \widetilde{s^{-1}} \circ e[/spoiler]
and because <span class="math">s^{-1}[/spoiler] is also a right inverse.
<span class="math">= \widetilde{s^{-1}} \circ (s \circ s^{-1})[/spoiler]
<span class="math">=(\widetilde{s^{-1}} \circ s) \circ s^{-1}[/spoiler]
<span class="math">=e \circ s^{-1}[/spoiler]
<span class="math">=s^{-1}[/spoiler]
This contradicts our hypothesis so there must only be one left inverse, right inverse, and inverse in general.

>>6513599
>banach manifold morse theory
I know each word but I've only the vaguest idea how one might stitch it all together lol.

My pursuits are more homotopic/categoric nowadays. I agree that many of the results and proofs from those days in topology are amazing.

>>6513597
>>6513603
>>6513606
>>6513607
>>6513614
> mfw there are people out there right now, right fucking now, that enjoy this type of thing
fuck algebraists, fuck categoryboos, and fuck logicians especially.

>>6513614
Wait, why the fuck is \neq rendering as a normal equals sign!?

>>6513615
>I've only the vaguest idea how one might stitch it all together lol.
its done for example in this book (a horrible book, but its a good reference)
http://www.amazon.com/Fundamentals-Differential-Geometry-Graduate-Mathematics/dp/038798593X/ref=pd_sim_b_1?ie=UTF8&refRID=0N7DC603Q18NMWRV8YRX
basically, instead of manifolds being locally euclidean, make them locally banach spaces!

to motivate it, note that most PDEs can be written as variational problems - finding the critical point of a functional (not necessarily linear) from some banach space (a sobolev space) to the reals. So knowing the topology of level sets of functionals on banach spaces is important. These level sets are banach manifolds. So basically you take a PDE and turn it into a problem in morse theory :)

I'm a mathematical physicist, but I do a lot of work in algebraic topology. Is category theory having any developments these days?

>>6513616
wut

>>6513614
this is pretty cool! thank you.

>>6513615
>homotopic

Speaking of which:
Frank Adams' solution of
http://en.wikipedia.org/wiki/Vector_fields_on_spheres
via the Hopf invariant, his Spectral Sequence, etc. etc. is really beautiful mathematics.

>>6513630
You're welcome anon.
>>6513586
I didn't see this till after I'd finished typing up and posting mine. I like the shorter proof that a left inverse is also a right inverse, I've always thought the proof I posted was more convoluted than it needed to be.

>>6513631
> hopf invariant
boner.jpg
i shouldn't be surprised, but i bet this is nicer than the usual index counting stuff

>>6513605
Made me smile.

>>6513401

urgh fuck how my lecturer teaches topology, just finished metric spaces and had an exam on it. learning 56 proofs wasn't fun.

I'm not an algebraist but I love this proof to bits; reading the last few lines always makes me euphoric.

(The proof for right cosets is similar, by considering <span class="math">ba^{-1}\in H[/spoiler] instead.)

>>6513233
H=W
anybody else a fan of this one? Its my favorite work of mathematics, and its barely over a page. Its terse, but the underlying geometry is beautiful.

>>6513699
...and I just realized that I messed up the definition of the group operator: <span class="math">\#[/spoiler] should be an arbitrary subset of <span class="math">((G\times G), G)[/spoiler] with <span class="math">((a,b),c) \in \#[/spoiler] standing for the proposition: "c=a#b is well defined, and we write c=ab for short". The closure axiom becomes: "# is a well-defined functional relation: for every (a,b) <span class="math">\in G\times G[/spoiler] there is exactly 1 <span class="math">c\in G[/spoiler] such that c = ab".

Hopefully the main result doesn't suffer from my carelessness also.

>>6513338
>Both components of this expression are divisible by five since 5n is clearly divisible by 5 and 10/5 = 2, so the sum of any five consecutive integers is divisible by 5!

What?

(5n+10)/(5*4*3*2*1) = (5n+10)/(120) = (n/24) + (1/12)

It's obviously NOT divisible by 5!, which isn't surprising at all, so I don't understand what's so amazing about your proof?

>>6513735
You're retarded, aren't you?

>>6513616
only analysis ;)

>>6513233
Baire's Lemma but mostly the fundamental theorem of algebra via algebraic topology.

I can't into latex but I found this pdf containing the proof of Euler's partition identitity (that the number of ways to add distinct numbers to form a particular number is equal to the number of ways to add odd numbers to form that number). There's a really pleasant moment of realisation when thing about generating functions clicks, atleast for me. http://www.alexhealy.net/papers/math192.pdf

Very simple proof that there exists <span class="math">p[/spoiler] and <span class="math">q[/spoiler] irrational so that <span class="math">p^q[/spoiler] is rational.

We start from the fact that <span class="math">\sqrt{2}[/spoiler] is irrational (inb4 people read the troll post above in the thread and throw a lame-ass trolling attempt here).

Consider <span class="math">x= \left(\sqrt{2}^{\sqrt{2}} \right)^{\sqrt{2}}[/spoiler].

Clearly, <span class="math">x=\sqrt{2}^2 = 2[/spoiler], so <span class="math">x[/spoiler] is rational.

Now, what about <span class="math">y=\sqrt{2}^\sqrt{2}[/spoiler]?

Well:

1) Either <span class="math">y[/spoiler] is rational. In that case, since <span class="math">y=\sqrt{2}^sqrt{2}[/spoiler], we take <span class="math">p=q=\sqrt{2}[/spoiler] and <span class="math">p^q=y[/spoiler], which concludes the proof.

2) Or <span class="math">y[/spoiler] is irrational. Then we can take <span class="math">p=y[/spoiler] and <span class="math">q=\sqrt{2}[/spoiler], and <span class="math">p^q=2[/spoiler] is rational, which concludes the proof.

Proof is cool because in the end, we don't know whether <span class="math">y[/spoiler] is rational, and we don't even care: it works either way.

>>6513235
>clearly
>use fermat last theorem

yeah it's no big deal, it's not like it took over 350 years to prove that shit...


assume that <span class="math">\displaystyle \sum_{n=1}^{+\infty} \frac{1}{n} < +\infty [/spoiler], so we have :
<span class="math">\displaystyle \sum_{n=1}^{+\infty} \frac{1}{n} = \sum_{n=1}^{+\infty} \frac{1}{2n}+\frac{1}{2n} [/spoiler]
and since that <span class="math">\frac{1}{2n}<\frac{1}{2n-1}[/spoiler] for every integer n and assuming the first hypothesis we have
<span class="math">\displaystyle \sum_{n=1}^{+\infty} \frac{1}{2n}+\frac{1}{2n} < \sum_{n=1}^{+\infty} \frac{1}{2n}+\frac{1}{2n-1} [/spoiler]
but since <span class="math">\displaystyle \sum_{n=1}^{+\infty} \frac{1}{2n}+\frac{1}{2n-1} = \sum_{n=1}^{+\infty} \frac{1}{n} [/spoiler] it implie that <span class="math">\displaystyle \sum_{n=1}^{+\infty} \frac{1}{n} < \sum_{n=1}^{+\infty} \frac{1}{n} [/spoiler], which is absurd.

So the harmonic series is divergent.

i hope i didn't messed up the tex ...

>>6513781
> \displaystyle \sum_{n=1}^{+\infty} \frac{1}{2n}+\frac{1}{2n-1} = \sum_{n=1}^{+\infty} \frac{1}{n}

hm? I don't see how that's true at all.

>>6513233
what do you guys think of godels ontological proof?
>http://en.wikipedia.org/wiki/G%C3%B6del%27s_ontological_proof
has it been debunked?
Any other interesting things that stemmed from it?

>>6513783
fuck latex, I mean

<span class="math">\sum_{n=1}^{+\infty} \frac{1}{2n}+\frac{1}{2n-1} = \sum_{n=1}^{+\infty} \frac{1}{n}[/spoiler]

>>6513784
>Definition 1: x is God-like if and only if x has as essential properties those and only those properties which are positive

yes, that's not slippery and underdefined at all ...

>>6513784
I could even assume the correctness of the proof based on the great logician that wrote it, but that doesn't imply the axioms he started from are acceptable.

>>6513790
>>6513786
thanks

>>6513785
have you tried to express the first terms of the two series ?

left one is (1/1+1/2)+(1/3+1/4)+(1/5+1/6)+... and the right one is (1/1)+(1/2)+(1/3)+(1/4)+...

2n will give you every even integers and 2n-1 every odd integer so...

>>6513785
Not that guy, but you can just write out the terms manually: 1/2 + 1/1 + 1/4 + 1/3 + 1/6 + 1/5 ...
By assuming that the series is absolutely convergent you can rearrange this to get the harmonic sum. Of course this yields an absurdity, so the assumption was false.

>>6513801
Ah yes, I see it now.

Just kind of threw me for a second that the series 1/2n + 1/2n-1 would be equal to 1/2n + 1/2n.

>>6513805
actually you don't need absolute convergence to do that. Every terms is positive, it's sufficient to rearrange the terms of a serie. So the assumption is not really used here.

>>6513807
I thought you weren't allowed to rearrange terms in a series?

>>6513763
I love this proof. It's absolutely beautiful. Your natural reaction is to go "But we don't know ((2^(1/2))^(2^(1/2)) Is irrational!" but then you see that if that is true then it's already proved!

>>6513808
like i said, if the terms of the series are all positive then you can rearrange the terms and you'll always get the same value, even if your starting series was divergent. It's the same for permutation of sries and integrals or permutations of two integrals, if everything is positive then you can always do it, even if it's infinite.

This is quite possibly the best thread on /sci/, ever.

As for my contribution, I know it has already been said, but Euclid's infinite primes proof is a favourite of mine.

>>6513822
It is a beautiful proof, but the idea itself is actually fairly common and shows up in all sorts of places. We don't know whether something is true, so we split it up into 2 (or more) cases and prove each case separately.

For a simple example, the following lemma in linear algebra forms part of the proof that the product of determinants = determinant of products:

"If det(A) = 0 then det(AB) = 0."

Suppose for a contradiction that det(AB) <span class="math">\neq[/spoiler] 0 so AB is invertible.

If det(B) = 0 then we can find a nonzero <span class="math">x\in V[/spoiler] such that Bx = 0. Hence ABx = 0 <span class="math">\Rightarrow x = (AB)^{-1}0 = 0[/spoiler], contradiction.
If <span class="math">\det(B) \neq 0[/spoiler] then B is invertible, so let <span class="math">y = B^{-1}x[/spoiler] where x is any nonzero vector in the kernel of A. Then <span class="math">ABy=ABB^{-1}x=Ax=0[/spoiler], so <span class="math">y=(AB)^{-1}0 = 0[/spoiler]. Finally, <span class="math">x = By = B0 = 0[/spoiler] and we have a contradiction.

I like the Γ(1/2)=√π proof with the switch to spherical coordinates. How come noone stated any of the proofs of the above? I once collected 11 or 12 of them, don't know if there are more.

>>6513392
>>6513395
>prove order density of irrationals and rationals.
Do you actually need that? I seem to be able to get that with just the Archimedean property, unless that's what you're referring to.

>Let a, b be any two real numbers such that a < b. Then the interval (a,b) contains infinitely many rationals and infinitely many irrationals.
Proof: Let k = 1. By the Archimedean property <span class="math">\exists Q\in\mathbb{N}[/spoiler] such that <span class="math">Q > k/(b-a)[/spoiler], i.e. Qb/k - Qa/k > 1.
Let P = <span class="math">\lfloor Qa/k + 1\rfloor[/spoiler] where the 'floor' of any real number is <span class="math">\lfloor x\rfloor = \max \{m\in \mathbb{Z} \mid m\leq x\}[/spoiler] so <span class="math">0\leq x-\lfloor x\rfloor < 1[/spoiler].

Then Qa/k < <span class="math">\lfloor Qa/k\rfloor + 1 = \lfloor Qa/k + 1 \leq [/spoiler] Qa/k + 1 < Qb/k
i.e. Qa/k < P < Qb/k so Pk/Q is our required rational (Q, k are both strictly positive). We get infinitely many rationals by setting k = 2, 3, ... etc., and infinitely many irrationals from k = <span class="math">\pi, 2\pi, 3\pi[/spoiler] ...
(Unless P = 0, i.e. aQ/k <span class="math">\in[/spoiler] [-1, 0) implying Q <span class="math">\leq k/|a|[/spoiler]. But we can simply use the Archimedean property to choose a bigger Q to get around this.)

>>6513416
NOICE ONE!
I like it.

>>6513884
I think they are equivalent, but I could totally be wrong

>>6513586
Awesome one.

>>6513586
I wish I could understand this.

>>6513992
It's not particularly hard, it's just shifting a few neutral elements around to show that the definition here can be relatively weak and still work.

>>6513992
you are def capable of this. (1) is associativity, (2) states left action by identity is trivial, (3) says a formal inverse left acting on a element cancels it to the identity.

so, we want to know the other direction: left action of an element on its inverse. we can act on the left with the identity, by (1). We can "expand" the identity with the "reverse" of (3) into an element which is an inverse, and the inverse of that inverse (we cannot assume this is an involution a priori). The "middle" terms cancel since we have a left action of an inverse on the element it is an inverse of. The identity then acts trivially on x^-1. We then have an inverse acting on the left on the element it is an inverse of, and we have reduced it such that we now now an inverse may multiply on either side, whether or not multiplication is commutative in general in the group, to reduce to the identity.

similarly, we want to know if the identity can act on the right the same as the left. we "expand" the identity into an element an it's inverse (which we can now do in either direction). The "middle term" step is now the first two terms, and we reduce, so we are now back at axiom (1), where we have left action of the identity. So left and right action by the identity on any element is trivial, even if multiplication is not in general commutative.

>>6513416
Wait a second, isn't that totally trivial just by the axioms? I mean you need the empty set for everything in set theory (at least the one I learned).

>>6514013
Thanks for the post. I'm a bigger math pleb than you think though. I never finished HS and I'm just now getting into some basic maths that I should know.

I kind of get what's going on, but I don't know what a left identity and left inverse are. I get associativity and stuff, but I don't understand why x*x^-1 is denoted as e and not just 1...sorry, I'm a huge pleb.

>>6514014
It's generally taken to be a definition; I was just showing that it could be proved.
It isn't an axiom in ZFC or NGB

>>6514022
I meant NBG not NGB goddammit!

>>6514022
>>6514024
Yeah after I had sent the post, I saw that it said subset of, not element of.

In Mechanics 2 last year there was a really neat little solution to the famous Brachistochrone problem using the Euler-Langrange equation.

>>6514044
with Snell's law?

http://www.proofwiki.org/wiki/Brachistochrone

>>6513554
Seeing how math is treated in certain disciplines is actually quite interesting.

Piketty's new economics treatise has caused quite a favourable stir amongst the leftwing media, partly because it (apparently) forgoes on the mathematics.

Elsewhere, John Maynard Smith has been criticised (or complimented) for bringing in so much math to Evolutionary Biology.

>>6514047
Indeed.

>>6513763
i have a simpler proof:
proposition: there exists p and q that are irrational such that p^q is rational
proof: take p=q=1 then p^q=1 QED

>>6514019
math pleb here too, though I think I can clarify on your last question. The proof is dealing with groups besides just the integers, and things like vectors or matricies have their own equivalent of 1 for multiplication. Using e is just covering all bases without explicity mentioing the number 1.

>>6514064
> 1 is irrational
How's high school going?

For some reason, I'm rather fond of this one:

>Suppose <span class="math">a^2 + b^2 = c^2[/spoiler] where a, b, c are natural numbers with no common factor. Then c must be odd.

Proof: Suppose c is even so <span class="math">c^2[/spoiler] is also even.
If a and b are both even then (a,b,c) have a common factor of 2, a contradiction.
So at least one of them is odd. If a is odd and b is even, then <span class="math">a^2[/spoiler] is odd and <span class="math">b^2[/spoiler] is even so <span class="math">a^2 + b^2[/spoiler] is odd, but <span class="math">c^2[/spoiler] is even, a contradiction. The same logic applies if a is even and b is odd.
So both a and b are odd.

Now let a = 2p + 1 and b = 2q + 1 for natural numbers p, q. Then <span class="math">a^2 = (2p+1)^2 = 4p^2 + 4p + 1[/spoiler], which is 1 greater than a multiple of 4. Likewise <span class="math">b^2 = 4q^2 + 4q + 1[/spoiler] is also 1 greater than a multiple of 4.
Finally, c is even so c = 2r for some natural number r, and <span class="math">c^2 = (2r)^2 = 4r^2[/spoiler] is divisible by 4.
But <span class="math">c^2 = a^2 + b^2[/spoiler], and <span class="math">a^2 + b^2[/spoiler] is 2 greater than a multiple of 4 (and is not divisible by 4). We have a contradiction.

So c cannot be even; it must be odd.

>>6514133
This is cool except it is so limited. When could it ever be useful?

>>6514064
>1 is irrational
What?

>>6513235
dayummmm

>>6513605
10/10

J.H. McKay has a brilliantly written proof for Cauchy's theorem of groups with order divisible by a prime.

I like the proof in Fraleigh that a finite integral domain is a field. It is similar to proof 2 here:
http://www.proofwiki.org/wiki/Finite_Integral_Domain_is_Field

>>6513235
lel

>>6513849
Hey, guy who wrote the first post of the chain here.

For some reason, your "\neq" symbols show as "=" for me. It's still easy to read the proof because I know where you're going but fuck, jsMath bugs are annoying.

Here's my favorite example of all of these types of proofs. It's the proof that if X is a set and P(X) is the power set of X, then there is no bijection between X and P(X). I think it's commonly viewed as a very beautiful proof.

Anyway, we assume that there is such a bijection, we call it f, and we show that it leads to a contradiction.

Define U the subset of X so that an element x of X is in U if and only if x is not in f(x). (Remember that f(x) is an element of P(X), so it's a subset of X, and since x is an element of X, it can belong to f(x) or not: if it doesn't, then x is in U, otherwise x is not in U).

Now, since f is a bijection and U is a subset of X (meaning U is an element of P(X)), there is a reverse image of U by f. Let's call u the element of X such that f(u)=U.

And here's the question we have no answer: Is u in U?

1) If u is in U=f(u), then by definition of U, u is not in f(u). This is a contradiction.

2) If u is not in U=f(u), then by definition of U, u is in f(u). This is a contradiction too.

In any case, there's a contradiction, so we're done.

What's even more amazing about this proof and yours compared to the first I posted is that when we split our argument into two choices and prove each of them wrong, we're already in a proof by contradiction.

So in the end, in the first proof, you can say "I don't know which of these two things is true, but one of them is, and the logical OR of these things is a sufficient condition for my result, and someday maybe I'll know whether it was the first or the second condition that was true". However in our case, both conditions are derived from a false statement. Is u in f(u)? This question doesn't even "exist" mathematically, because f doesn't exist.

>>6513235
Jokes aside, the proof with infinite descent is also cool.
http://en.wikipedia.org/wiki/Square_root_of_2#Proofs_of_irrationality

Simple, and a clever argument.

>>6514229
Where did my pic go?

>>6513416

Theorem: There is one and only one empty set.

By definition, the empty set is a subset of every set.

A = osub1 and B = osub2

A is a subset of B
B is a subset of A

Therefore A = B

There is one and only one empty set

>>6514236

I'm on a phone that o is an empty set.

>>6514228
Those prove different things. Infinite descent proves sqrt2 is irrational, while that proof shows that ANY integer nth root of 2 is irrational for n>2.

>>6513233
I don't get this
Does this means that there can be only one additive identity?

>>6514251
Exactly. It shows only one additive identity exists, and because we can show that 0 is an additive identity we know it can be the only one.

>>6513338
Is 0 divisible by 5?

>>6514251
>>6514258

Yes, but these two are stronger because they don't assume commutativity.
>>6513597
>>6513586

>>6513897
I think it's a good hint that two subsets of a set are dense if you can prove that in between any two elements of any of the two sets, you find elements of both sets, but I don't think it's enough.

Let me try and build a counter-example.

Let X be the set of negative rational numbers, and Y the set of negative irrational numbers.

In between any two distinct elements of X, there is an element of X and an element of Y (the proof from above still works). Similarly in between any two distinct elements of Y, there is an element of X and an element of Y.

Done. Right?

Unless I'm thinking poorly, when you take a set S that is a union of open intervals of R, and you define X as the rationals in R and Y as the rationals in Y, you have the 4 "in between any 2, there's 1 of each" properties, but you don't have density in R anymore (only density in S, which honestly no one cares about).

Obviously though, the equivalence you thought about is true in the other direction: density of both X and Y is sufficient for X and Y to meet all 4 "in between any 2, there's 1 of each" properties.

>>6514276
Why can't we assume commutativity? I mean it is easy enough to prove

>>6514292
The whole point is to show that the property still works even in non-commutative groups.

>>6514268
It is

>>6514331
Alright, I wasn't sure. Thanks

>>6514292
Because not all groups are abelian...
Are you seriously asking that question?

>>6514169
This is one of the coolest proofs I know. It leads to very natural proofs of Sylow theorems. Recommended (very short) reading for anyone unfamiliar with this top aesthetic proof:

http://www.cs.toronto.edu/~yuvalf/McKay%20Another%20Proof%20of%20Cauchy's%20Group%20Theorem.pdf
(tinyurl seen as spam, lel)

>>6513784
>what do you guys think of godels ontological proof?

"If there are supreme gods then there can only be one true god." It's pretty obvious that multiple supreme gods is illogical.

>has it been debunked?

Why would it need to be debunked?

You probably need to have seen some of the more hideously long proofs that π is irrational to truly appreciate this one page one.

>>6514135
That proof technique of showing inequality by checking moduli comes up a lot in number theory and crypto.

>>6514409
Very nice. I would need to really understand "where it came from" to appreciate it though. As it is, it really comes out of nowhere.

>>6514289
>two subsets of a set are dense if you can prove that in between any two elements of any of the two sets, you find elements of both sets, but I don't think it's enough.
For one set at least, the Cantor set on [0,1] seems like a counterexample, since between any two elements you can find another element, yet it's not dense in [0,1]. (The elements are precisely the reals whose base-3 decimal expansion contains no '1'. Strictly speaking, you'd have to remove the endpoints as well, to exclude things like (1/3, 2/3) = <span class="math">\emptyset[/spoiler].)
Not sure if this can be reworked to produce two sets, each of which contains elements of the other. Maybe changing the base, or working in a condition like "only finitely many digits can be 2" could produce something, but I'm not familiar enough with this branch of analysis to attempt a proper proof.

>>6514538
Just consider the set of rational number whose base 3 expansion is finite and contains no 1, and split it into those whose last digit is a 0 and those whose last digit is a 2.

>>6514409
Very nice one.

>>6514135
>This is cool except it is so limited.
Seems to sum up a lot of number theory proofs.

my favorite proof is the graph structure theorem:
https://en.wikipedia.org/wiki/Graph_structure_theorem

it is not "elegant," but that's why i like it
>things are actually complicated, who knew

>>6514819
the GST has far-reaching implications for defining the "dimension" of graph-structured data (which is pretty much all data) but is *not* beautiful or easy to understand precisely because it ties together a bunch of stupid branches of theory that have no business being together

I'll leave this one as an exercise, because it feels so good to find the proof.

Prove that if you have 5 points on a sphere, there lie at least four of them on a closed hemisphere.

>>6514923
That sounds like fun. I'll try and do it. My geometry is kind of trash, but I think it'll be interesting.

>>6514923
Seems like it would be true for any number of points on a sphere greater than 3. Unless my proof is fucked up.

>>6514933
The proof is really simple (once you see it): you need only trivial geometry and you don't have to do any algebraic stuff. It was a Putnam question (I think B2 but I don't know from what year).

>>6514940
There is indeed a generalization that can be proved in an analogous way. But I'm not sure what you mean by 'this' (of course there are 4 points on a closed semisphere if there are more than 5 points on the sphere, for this is weaker than the original statement).

>>6514923
Draw a circumference of the sphere through two points.
The circumference divides the sphere into two hemispheres that contain three other points in total.

One of them must contain at least two points, which combined with the two points on the circumference makes four.

Does that work?

>>6513242

Is Fermat the greatest troll in maths or what?

>>6513256

Was also going to post that.

>>6514958
That's what I did.
Seems really trivial...

>>6514958
This occurred to me but I'm bothered with the notion that points on the edge of hemisphere are contained within it. I mean, suppose you have a sphere with a hemisphere line drawn on it, now choose five points along the hemisphere line. Which hemisphere are those points in?

>>6513233
>ITT: favourite/most elegant or interesting math proofs. They don't necessarily have to be all that exciting, just some simple tricks that give you that "aha" moment when you realize how the proof works.

Cantor's diagonalization argument. It's a beautifully simple proof that could be understood by a third-grader.

>>6513235
> If nth root of 2 is rational then
> 2^(1/n)=a/b for some real a,b
> "for some real a,b"
I think you meant "for some positive integers a,b"

>>6514992
That's why the problem said 'closed hemisphere'. It means the border is included.

>>6514958
Yes, that works.

I am about to finish my first linear algebra class, so far this one is my favorite.

Prove that a square skew symmetric matrix (i.e. A^transpose= -A) of size n is singular if n is an odd number
A^t =-A
det(A^t) = det(-A)
det(A) = det(-A)
det(A)=(-1)^n*det(A)
if n is an odd number than det(A) = 0 ==> A is singular.

>>6514998
He has a couple diagonal arguments. I could totally see a 3rd grader understanding the 1-1 correspondence between the naturals and the rationals, but I doubt a 3rd grader could understand the uncountability one.

>>6515018
TeX isn't that hard people.

>>6515025
>CS major detected

>>6515018
This reminds me of another proof of the same result:
>Let A be a real n*n skew-symmetric matrix. If n is odd, A is singular.
Proof: <span class="math">\langle Ax, x\rangle = \langle x, A^Tx\rangle = -\langle x, Ax\rangle = 0[/spoiler], so x and Ax are orthogonal for every vector x.
Suppose A is singular. Then <span class="math">Ax \neq 0[/spoiler] whenever <span class="math">x \neq 0[/spoiler].
So we have a nonvanishing tangent vector field on the unit n-ball (or (n-1)-sphere).
This contradicts the hairy ball theorem, so A must be singular.

>>6515018
Another proof of this result:

>Let A be a <span class="math">n\times n[/spoiler] skew-symmetric matrix. If n is odd, then A is singular.
Proof: Writing <span class="math">\langle \cdot, \cdot\rangle[/spoiler] for scalar product: <span class="math">\langle Ax, x\rangle = \langle x, A^Tx\rangle = -\langle x, Ax\rangle = 0[/spoiler], so x and Ax are orthogonal for any vector x.
Suppose A is nonsingular. Then x != 0 <span class="math">\Rightarrow[/spoiler] Ax != 0.
So we have a nonvanishing tangent vector field on the unit n-ball (i.e. the unit [n-1]-sphere).
But this contradicts the hairy ball theorem, so A must be singular.

>>6514992
>Which hemisphere are those points in?
Both. The alternative case, where all hemispheres are open can easily be defeated:

Take an arbitrary great circle. Place all 5 points on the great circle, at the vertices of a regular pentagon. At most 3 of these 5 points lie in any open hemisphere. Proof:

Let C be the great circle defining your chosen hemisphere and D be the great circle containing all 5 points. If D=C, neither hemisphere contains any points. Otherwise, C intersects D at two points p and q, where pq is a diameter of D. but any diameter of D has at most 3 of the points on either side of it; thus, at most 3 points are in any given hemisphere.

>>6513425
Not the other anon, but if you want topological proofs, I like this one.

Property: The set of invertible real n-by-n matrices is not connected.

Proof: The image of a connected set by a continuous mapping is a connected set. Consider the image of the set of invertible real n-by-n matrices by the (continuous) determinant function: that image is the real line minus {0} (matrices are invertible iff their determinant is non-zero). Now that set is not connected. Therefore by contraposition, neither is the set of invertible real n-by-n matrices. QED.

The nice thing with this proof is that it's very visual: the reason the set of invertible real matrices isn't connected is because if you take a path among invertible matrices, your determinant is going to change continuously along that path, and either it's going to cross 0 (meaning your path went through a singular matrix), or it's not going to change sign, meaning you won't be able to reach matrices with determinant of opposing sign.

It also gives a pretty good intuition of why the set of invertible complex n-by-n matrices is connected: this time if you need to go from determinant -1 to determinant 1, you can just loop around 0 because you are now in 2D. The proof that it's actually connected requires a bit more than just this intuition though (but is still relatively simple, definitely undergrad level, at least with a few hints).

>>6513235
This had got me thinking, does Fermat only hold for two numbers to the nth power?

i.e, does
<span class="math">a^{n} + b{n} + c^{n} = {d^n}[/spoiler]
Have any solution for n>2 (...or I suppose even n=2)?

What got me thinking about this is that the proof I'm replying to essentially ended with.
<span class="math">2b^n=a^n[/spoiler]
Which (to me) isn't quite Fermat's Last Theorem because, as far as I'm aware, in Fermat, a, b and c must be different numbers (that's not to say that its false when a, b and c are the same, just that I don't believe Wiles proved it).

>>6515246
...obviously, on the third line, I intended to write.
<span class="math">a^{n} + b^{n} + c^{n} = d^n [/spoiler]

>>6515247
3^3 + 4^3 + 5^3 = 6^3

>>6515250
...beat me to it, I just began searching for a counter example and found exactly that.

>>6515252
So, this begs the question, does the last line of the proof hold?
i.e, is <span class="math">2b^n \neq a^n[/spoiler]
I feel like its obvious, but was it proved by Wiles?

>>6515256
It is obvious, in the prime factorisations 2 appears 1+kn times in the LHS and mn times in the RHS for some non-negative integers m,k which is only possible if n=1.

>>6515246
>>6515247
>>6515250
>>6515252
Tried running some simulations for n=4, assuming that you need n>=4 for the property to hold. No tuple for a,b,c<=1000 seems to work. Same for n=5.

Is the property true for n>=4? If so can we prove it (is it easier than Fermat?) and if not can we find a counter-example (or a proof)?

>>6515303
(Am now running in the backgrounds the tests for n=6 to n=20, but for n>=7 it'll overflow and I don't want to use big integers otherwise it'll be too slow, so I might get some false-positives if a^n+b^n+c^n=d^n mod 2^64 or something)

Cayley-Hamilton for modules is pretty cool. I like the determinant trick.

http://math.berkeley.edu/~mbaker/Cayley-Hamilton.pdf

>>6515303
Apparently not, see
http://en.wikipedia.org/wiki/Euler's_sum_of_powers_conjecture

>>6515246
>>6515247
That is a common question, but yes, FLT still holds when a=b. This special case was proven relatively early on, if I recall correctly.

>>6515469
I'd hope Fermat could have at least proved that since it takes one line of text to demonstrate. see >>6515267

>>6515479
Oh haha I should have noticed that.

>>6513452
It was used. It's circular.

>>6515494
Where was it used?

>>6515496
Indeed, we can assume <span class="math">n[\math] is prime (just like for FLT) and then the proof of FLT first passes from a hypothetical nontrivial solution to <span class="math">a^n+b^n=c^n[\math] for prime n>2[\math] to a suitable "Frey curve" y2=x(x−a^n)(x+b^n)[\math] where one has to rig certain congruential and gcd conditions on (a,b,c), including that a, b, and c are pairwise coprime.[/spoiler][/spoiler]

>>6513233
pls prove that 0!=1

>>6515671
n!=n((n-1)!)
1!=1(0!)
0!=1
This is slightly circular, but we can separately prove that 1!=1 and that n!=n*(n-1)! so it's all good.

1 + 2 + 3 + 4 = -1/12

google that. numberphile did a video on it. Ramanujan was a badass.

>>6515699
You incorrectly defined the equation, and either way, the proof is shitty and untrue and based on a huge assumption that a nonconvergent set converges to the average of its possible outcomes, which I guess is useful for some applications, but doesn't speak to reality.

>>6515671
This was posted here some time ago.
It proves that 1 != 2, but it can easily modified to match your needs (in particular, it proves that 1 != 0 at proposition 4)
www.scribd.com/doc/208378319/1%E2%89%A02

>>6515671
>>6515716
I thought you meant that as 0 "not equal to" 1, but reading >>6515691 I see that maybe you meant that as 0 "factorial". In that case, it's usually part of the definition of factorial:
<span class="math">0! = 1[/spoiler]
<span class="math">n \neq 0 \Rightarrow n! = n\cdot n![/spoiler]

>>6514409
Very nice. Herstein's proof in his Abstract Algebra book is quite long but it's essentially the same deal. Regardless, I like this one better.

>>6515691
<span class="math">\Gamma (1)[/spoiler]

>>6515671
<div class="math">\int_0^\infty t^n e^{-t} \mathrm{d}t = n!</div>
for all non-negative integers n
by direct evaluation, this holds for n = 0
suppose it holds for n = k, then
<div class="math">\int_0^\infty t^{k+1} e^{-t} \mathrm{d}t = (0 - 0) - (k+1)\int_0^\infty t^k e^{-t} \mathrm{d}t / (-1)</div>
<span class="math"> = (k+1)k! = (k+1)![/spoiler]
there's a lot more theory behind this integral, but most of that shit is too complex for me.

>>6515711

I realized I fucked up right after I hit 'post'.

B-b-but muh Riemann Zeta function

>>6515671
Look at the <span class="math">\Gamma[/spoiler] function. I don't know if it's a proof, but <span class="math">\Gamma (n) = (n - 1)! for integers n. Just evaluate \Gamma (1) = 1 = 0![/spoiler]

>>6515381
Cool, thanks.

>>6514958
This is what I got. The only part I was bothered by was the
>Draw a circumference of the sphere through two points.
part (apparently it is called a great circle). I wasn't sure that this could always be done, but I was able to convince myself.

>>6515241
Nice

>>6514923
>>6514958

Such a cute problem and proof!

>>6515319

We had a proof interpreting nxn matrices as n2 polynomials–that one was great as well

>>6515750
The only other thing you really need to mention is that it is the <span class="math">\bf{ ONLY }[/spoiler] function that is analytic (~basically infinitely differentiable or smooth) that is a continuation of the factorial on the positive integers so defining 0!=1 is the "nicest" choice you can pick.

>>6515722
I think you mean n!=n*(n-1)!

>>6515699
get out

I think all that mamikon shit is pretty cool, it's just a shame the websites are all such an incomprehensible mess.

http://www.its.caltech.edu/~mamikon/calculus.html
http://www.mamikon.com/

Seriously, half the shit on there looks like elementary school bad.
http://www.its.caltech.edu/~mamikon/BookCover.html

There's an article by Tom Apostol which sounds great until you realize all the math equations have been horribly butchered.
http://www.its.caltech.edu/~mamikon/VisualCalc.html

Apparently they wrote a book together on the topic but it's fairly new and I haven't looked at it yet.

>>6517025
Nice, thanks for sharing this.