- #1
cragwolf
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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.
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.