Meaning and example(s) of Qiaochu's quote.

I happen to come across this page http://math.uchicago.edu/~chonoles/quotations.html which contains some beautiful quotes by various mathematicians and I came across Qiaochu's quote as claimed by the site which seemed intriguing.

"I believe that in mathematics nothing is a trick if seen from a sufficiently high level." - Qiaochu Yuan

Now I was wondering if anyone could interpret (maybe even Qiaochu himself) and give examples in mathematics that would convey the meaning of this quote.

NB: Hopefully this question isn't too off topic? Can a moderator turn this into a wiki if deemed appropriate? Also, any appropriate tags?

Edit: Since my question wasn't clear as I would have liked, I'd prefer this question to be example based. So I'd like as much examples from different areas of matheamtics as possible. Since a lot of users on this site are at different levels in terms of the amount of mathematics one has learned, maybe anyone can contribute by giving examples say at a high school level, undergraduate level, graduate level, or research level etc.

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As great as the quote is, it seems that there is at least one trick -the Poisson integral trick for evaluating $\int_0^{\infty}e^{-x^2}dx$: unf.edu/~dbell/Poisson.pdf But then again, I wonder if we can see it from a higher level so that it stops being just a trick? –  Aleks Vlasev Oct 17 '12 at 6:21
Interestingly, I believe it is the other way around. Everything is a trick and then the trick is then used to create an abstraction, which in turn gives a better insight into the trick. But the trick is the one which kick starts the process. –  user17762 Dec 8 '13 at 17:51

"An idea which can be used only once is a trick. If one can use it more than once it becomes a method." - George Pólya, Gábor Szegö, "Problems and Theorems in Analysis I" http://books.google.ca/books?id=b9l2NqGEFzgC&pg=PR8&lpg=PR8

From a higher level, you may be able to see the "trick" as an instance of a larger idea that has wider applicability. In the case of $\int_0^\infty e^{-x^2}\ dx$, while this specific trick does not work for other integrals, the larger ideas are (1) the fact that $e^{-x^2} e^{-y^2}$ is rotationally invariant (which corresponds to the fact that independent normal random variables have a joint distribution that is multivariate normal) and (2) change of variables in multiple integrals. Both of those are very widely applicable.

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Although very close to $\int_0^\infty e^{-x^2}\,\mathrm{d}x$, the integrals $\int_0^\infty\sin(x^2)\,\mathrm{d}x$ and $\int_0^\infty\cos(x^2)\,\mathrm{d}x$, as derived here, uses the same "trick". –  robjohn Oct 17 '12 at 23:40

One example of such a "trick" is rationalizing the denominator. You are taught a pattern to follow, long before you understand the general theory. For example, $$\frac{1+\sqrt{3}}{5-2\sqrt{3}} = \frac{1+\sqrt{3}}{5-2\sqrt{3}} \left( \frac{5+2\sqrt{3}}{5+2\sqrt{3}} \right) = \frac{(1+\sqrt{3})(5+2\sqrt{3})}{25-12} = \frac{11+7\sqrt{3}}{13}.$$

If you have a denominator of $a+b\sqrt{c}$, you can multiply the top and bottom by $a-b\sqrt{c}$ to get rid of the square root in the denominator.

If you learn Galois theory, you can explain exactly why this works, and you know how to get rid of algebraic numbers in the denominator for any example, like $$\frac{1}{1+\sqrt{2}+\sqrt{3}}, \qquad \frac{1}{3+\sqrt[3]{5}}, \quad \text{or} \quad \frac{1}{1+\alpha}$$ where $\alpha$ is the real root of $x^5+x+1=0$. We need to multiply the denominator by all of its Galois conjugates. That means that for the first example, we multiply the top and bottom by $$(1+\sqrt{2}-\sqrt{3})(1-\sqrt{2}+\sqrt{3})(1-\sqrt{2}-\sqrt{3}).$$ In the second example, we multiply the top and bottom by $$\left(3+\sqrt[3]{5}\left(\frac{-1+i\sqrt{3}}{2}\right)\right) \left(3+\sqrt[3]{5}\left(\frac{-1-i\sqrt{3}}{2}\right)\right).$$ And in the final example, we multiply the top and bottom by $$(1+\alpha_2)(1+\alpha_3)(1+\alpha_4)(1+\alpha_5),$$ where the $\alpha_i$ are the other 4 roots of $x^5+x+1=0$.

Further, we can do this without introducing any new algebraic numbers. For example, $(1+\alpha_2)(1+\alpha_3)(1+\alpha_4)(1+\alpha_5)=2-\alpha+\alpha^2-\alpha^3+\alpha^4.$ Pretty cool!

Once you understand the symmetries of algebraic numbers, the high school algebra "trick" of rationalizing the denominator isn't much of a trick. It's just a routine use of the basic ideas of Galois theory!

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Specifically, if $P(x)$ is a polynomial over $\mathbb{Q}$, Galois theory gives a nice explanation and intuition for why $P(\alpha)P(\alpha_2)P(\alpha_3)P(\alpha_4)P(\alpha_5)$ is in $\mathbb{Q}$, and why $P(\alpha_2)P(\alpha_3)P(\alpha_4)P(\alpha_5)$ is in $\mathbb{Q}[\alpha]$. –  Jonas Kibelbek Oct 18 '12 at 1:04
Thanks you for your answer! Exactly the types of examples I'm looking for. I haven't studied Galois theory yet and have only done basic group theory and a bit on rings and ideals. So hopefully I can fully appreciate this after I have done some Galois theory. –  tcmtan Oct 18 '12 at 5:58

I think the intended meaning is that in reality there are no tricks. If you think something is a "trick", probably you don't understand well enough...

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