Why pure mathematics is beyond me

[cragwolf]

I've always been of the opinion that mathematicians, especially pure mathematicians, are the most intelligent segment of our society, and also that pure mathematicians are born rather than made. Anyone can improve their mathematical skills and knowledge, but blood, sweat and tears are not enough to become a professional mathematician: you need the gift. Here's a little anecdote that might serve as an example of what I mean.

Yesterday I went to a local bookshop and bought Shilov's "Elementary Real and Complex Analysis". When I got home, I started flicking through it, and noticed a problem in the first chapter: prove that [tex]\sqrt{3}[/tex] is irrational. So I decided to get a pen and paper and prove it myself. Having seen before the proof that [tex]\sqrt{2}[/tex] is irrational, I decided immediately that "proof by contradiction" was probably the best way to go. So I wrote down [tex]\sqrt{3}[/tex] as a/b, where a and b are whole numbers which do not share any common factors. This produced the following equation:

[tex]3b^2=a^2[/tex]

We'll call this equation 1.

So how did I proceed from this point? I will only provide a sketch of my proof, rather than bore you with all the steps. I decided to look at oddness and evenness. I showed that if a was even, then b was even, but we can't have that because then they would share a common factor, namely, 2. Then I showed that if a was odd, then b was odd. This meant that I could write a (or b) in the form 2m + 1, where m is a whole number; but substituting this form for a (and b) into equation 1 leads to a contradiction. Hence, [tex]\sqrt{3}[/tex] is irrational, QED.

So how would Shilov solve this? In the back of the book, he provides a hint. His solution goes like this (again in sketch form): If [tex]a^2[/tex] is divisible by 3, then a is divisible by 3. But by equation 1, this would mean that [tex]b^2[/tex] is divisible by 3, and hence b is divisible by 3. But a and b can't share common factors (in this case, 3). Contradiction! So, [tex]\sqrt{3}[/tex] is irrational, QED.

Now clearly, Shilov's proof is more elegant than mine. I think it illustrates the difference between a mediocre mind (mine) and a gifted mind (his). The elegant solutions, the ones that are short and sweet, are usually beyond the capacity of the average mind to independently discover, except, perhaps, rarely and in a roundabout way. This is why I usually find myself at a loss when trying to solve the harder problems, which can't be cracked by the ugly, brute-force methods I use to solve the simpler problems.

It's not very nice to have the passion for mathematics but lack the aptitude for it. On the other hand, there's still a lot of mathematics that's accessible to mediocre minds like mine, and for that I suppose I should be grateful. But it is frustrating to know that over the horizon there's a sh1tload of beautiful mathematics permanently beyond my reach.

 

[Hurkyl]

On the other hand, the more elegant proofs you see, the easier it is to tweak them to make new elegant proofs.

For instance, I bet that now you could prove √7 is irrational without breaking a sweat!

 

[Bob3141592]

cragwolf quoting Shilov
If [itex]a^2[/itex] is divisible by 3, then a is divisible by 3.

Forgive me for being dense, but I don't see this. Doesn't that assertion implicitely assume that [itex]\sqrt{3}[/itex] is irrational, only in different terms? The statement isn't convincing, certainly not like stating "if n is even then [itex]n^2[/itex] is even".

 

[matt grime]

If a^2 is divisible by three then a is divisible by 3 is a simple deduction from the properties of primes in Z (and not in other rings which is what makes Z so useful), so that implies 9 divides a^2, this implies 9 divides 3b^2, and that 3 must divide b^2 and hence b, so either by reductio ad absurdum or by contradiction on some unstated hypothesis, root three is irrational

 

[HallsofIvy]

Originally posted by Bob3141592
Forgive me for being dense, but I don't see this. Doesn't that assertion implicitely assume that [itex]\sqrt{3}[/itex] is irrational, only in different terms? The statement isn't convincing, certainly not like stating "if n is even then [itex]n^2[/itex] is even".

You're right. saying "if a² is divisible by 3 then a is divisible by 3" is not nearly as convincing as "if a is divisible by 3 then a² is divisible by 3" (which is more to the point than "even").

However, one can prove it: If a is not divisible by 3 then it is either of the form 3n+1 or of the form 3n+2.

(3n+1)²= 9n²+ 6n+ 1= 3(3n²+ 2n)+ 1 and so the square is not divisible by 3.

(3n+2)²= 9n²+ 12n+ 4= 9n²+ 12n+ 3+ 1= 3(3n²+ 4n+ 1)+ 1 which not divisible by 3.

That is, n² is divisible by 3 only if n is divisible by 3.-

 

[Hurkyl]

Also, recall that because 3 is prime, if 3 divides m*n, then 3 divides m or n (or both).

If 3 divides k², then 3 divides k*k, then 3 divides k or k, and thus 3 divides k.

 

[Jin314159]

Sure, innate ability is important. But I think you're vastly under-estimating the power of passion.

 

[Janitor]

Mathematicians

In the few weeks I have been here, I have been really impressed with the insight of some the members. I won't name names out of fear of leaving some worthy members off my list.

 

[cookiemonster]

I'm afraid of math. I can barely manage plain arithmetic and algebra, so I always kind of thought the more complicated stuff would kill me. That being said, I still jump into the higher math pools before having any idea how to swim.

cookiemonster

 

[cragwolf]

Look at HallsofIvy's proof that [itex]n^2[/itex] is divisible by 3 only if n is divisible by 3, and look at Hurkyl's proof of the same thing. Hurkyl's proof is more elegant than HallsofIvy's proof (no offence, HallsofIvy). Well, guess what? It was HallsofIvy's method that I used when I had a go at proving this. Once again, at least two ways to prove it, and I chose the least elegant and most tedious way. This is yet another demonstration of the thread title.

I may be underestimating passion or hard work. Certainly no one did any great or difficult mathematics without it. But the same can be said of natural talent or aptitude. Alright, I should stop whinging, and get back to the mathematics.

 

[NateTG]

Originally posted by cragwolf
Look at HallsofIvy's proof that [itex]n^2[/itex] is divisible by 3 only if n is divisible by 3, and look at Hurkyl's proof of the same thing. Hurkyl's proof is more elegant than HallsofIvy's proof (no offence, HallsofIvy). Well, guess what? It was HallsofIvy's method that I used when I had a go at proving this. Once again, at least two ways to prove it, and I chose the least elegant and most tedious way. This is yet another demonstration of the thread title.

Now I'm stuck wondering what methods you're referring to, and what you mean by elegance, and where you see such a big difference between the Newman twins. (I usually have some trouble telling them apart.)

Someone more eloquent than I described mathematical elegance as when there's nothing left to add and nothing left to remove.

Realistically, most ot the time the goal is to provide insight, and proofs are often only a byproduct or a waypoint of developing understanding. Thus when proofs are presented here, I always wonder whether it's good to go for the slick refined proofs - most of the time, when I prove something, the first, second, heck even the tenth version is usually ugly as all getout. Getting to the top of the mountain isn't just about the view after all.

[Edit: Fixed some punctuation]

 

[KingNothing]

Originally posted by NateTG
(ommitted for shortness)

I have to agree, I think Hurkyl's response is more elegant. I don't feel the need to define elegant as I feel the normal meaning of the word works too.

 

[matt grime]

We all aim for elegance and brevity, but it often takes many attempts by many people to get it in this 'perfect' form.
I think we can all agree Grothendieck is a little special when it comes to mathematics. One of his results concerns Grothendieck Duality. His (long) proof requires you to glue together local information (it's a result on manifolds) to get a global result with painstaking care taken to show that nothing goes wrong. About 15 years later Neeman uses another unrelated technique to prove the result in a paragraph. If Grothendieck can produce something so inelegant I think we can too.
This is how it goes. I'm sure when we all prove root 3 is irrational we use the less elegant proof first. Then with hindsight, or perhaps someone else tells us of the method, we realize there is a better way (better in the sense that it generalizes to show that the only natural numbers with rational square roots are perfect squares).
You will (if you stick with it) come across algebraic numbers where thinking about things differently means you can easily and elegantly prove this fact.

 

[Hurkyl]

Hurkyl's proof is more elegant than HallsofIvy's proof (no offence, HallsofIvy).

Presentation is an art form too. My first draft wasn't nearly as pretty; it looked like:

Write n as 3x + y.
Therefore, n² = 9x² + 6xy + y².
So if 3 does not divide y, then 3 does not divide n²

And that was when I realized that what I really wanted to do was say that if 3 doesn't divide y then 3 doesn't divide y * y, and that reminded me of the above fact about primes.

(And to be honest, at the very beginning I had to spend effort resisting falling back on the fact that Z mod p is a field when p is prime, a proof which is only elegant if you like fields a lot. )

 

[verty]

Here's my attempt to make an 'elegant' proof, as it were. I don't have any expertise in this. For what it's worth, when it comes to mathematics I have aptitude and some passion, but little knowledge.


if (x^(1/n), n>1) is rational, it can be written as a fraction a/b, with a and b relatively prime to each other.

x = a^n/b^n

For a fraction to be integral, the denominator must divide the numerator. Since gcd(a, b)=1, and therefore gcd(a^n, b^n)=1, for a^n/b^n to be integral, b^n = 1. Since x is integral, the equation above then simplifies to:

x = a^n

This tells us that the only rational n'th roots (where n>1) are the integral ones, which come from integers of the form a^n.