To prove that a number is irrational, show that it is almost rational

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To prove that a number is irrational, show that it is almost rational

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[QUICK DESCRIPTION]
If $p/q$ and $r/s$ are two distinct rational numbers written in their lowest terms, then $ps\ne qr$, which implies that $|ps-qr|\geq 1$, which implies that $|p/q-r/s|=|(ps-qr)/qs|\geq 1/qs$. Therefore, if $\alpha$ is a real number and we can find a sequence of rational numbers $r_n/s_n$ (in their lowest terms, with denominators tending to infinity) such that $|\alpha-r_n/s_n|=o(1/s_n)$ (which is equivalent to saying that $s_n|\alpha-r_n/s_n|\rightarrow 0$), then $\alpha$ cannot be rational.<comment thread="438" /> Loosely speaking, if you can approximate $\alpha$ well by rationals, then $\alpha$ is irrational. This turns out to be a very useful starting point for proofs of irrationality.

[EXAMPLE]
Let us construct inductively a sequence of rationals that approximate $\sqrt{2}$. (This is not necessarily the best proof of the irrationality of $\sqrt{2}$ but it gives an easy illustration of the technique.) We begin with $r_0=s_0=1$, and observe that $2s_0^2=r_0^2+1$. If $r_1/s_1$ were $\sqrt{2}$, then we would have $2s_1^2=r_1^2$, so the "$+1$" at the end of this is our error term in the first approximation. Now suppose we have defined $r_n$ and $s_n$ in such a way that $2s_n^2=r_n^2+(-1)^n$. Then set $r_{n+1}=r_n+2s_n$ and $s_{n+1}=r_n+s_n$. (The justification for this choice is that if $r_n/s_n$ were $\sqrt{2}$ then $r_{n+1}/s_{n+1}$ would be too, as can easily be checked.) Then
[maths] 2s_{n+1}^2-r_{n+1}^2=2(r_n+s_n)^2-(r_n+2s_n)^2=r_n^2-2s_n^2=(-1)^{n+1}.[/maths]
Thus, we have constructed a sequence of rationals $(r_n/s_n)_{n=1}^\infty$, with denominators tending to infinity, such that $2s_n^2-r_n^2=(-1)^n$ for every $n$. But from this we deduce that $|2-r_n^2/s_n^2|=1/s_n^2$, and therefore that $|\sqrt{2}-r_n/s_n||\sqrt{2}+r_n/s_n|=1/s_n^2$. Since $r_n/s_n\geq 1$ (as may easily be checked), this implies that $s_n|\sqrt{2}-r_n/s_n|$ tends to $0$ (at roughly the same rate as  $1/s_n$), and therefore that $\sqrt{2}$ is irrational.

[EXAMPLE]
To prove the irrationality of $e$, we start with the power-series expansion:
[maths]e=\frac 1{0!}+\frac 1{1!}+\frac 1{2!}+\frac 1{3!}+\dots [/maths]
We then set $r_n/s_n$ to be $\frac 1{0!}+\frac 1{1!}+\frac 1{2!}+\dots+\frac 1{n!}$. This is a fraction with denominator $s_n$ that divides $n!$. It differs from $e$ by 
[maths]\frac 1{(n+1)!}+\frac 1{(n+2)!}+\dots\leq \frac 1{(n+1)!}(1+(n+1)^{-1}+(n+1)^{-2}+\dots)\leq \frac 2{(n+1)!}.[/maths]
Also, this difference is strictly positive and not zero. Therefore, $s_n|e-r_n/s_n|\leq 2/(n+1)\rightarrow 0$, so $e$ is irrational.

[GENERAL DISCUSSION]
The two proofs given so far can easily be, and usually are, presented in other ways that do not mention the basic principle explained in the quick description. However, sometimes that basic principle plays a much more important organizational role: one is given a number $\alpha$ to prove irrational, and one attempts to do so by finding a sequence of good rational approximations to $\alpha$.

Another point is that if $\alpha$ is irrational then such a sequence always exists: one can take the convergents from the continued-fraction expansion of $\alpha$. But this observation is less helpful than it seems, since for many important irrational numbers (such as $\pi$) there does not seem to be a nice formula for the continued-fraction expansion. The point of the method explained here is that it is much more flexible: ''any'' sequence of good approximations will do (and sequences are indeed known that prove the irrationality of $\pi$).

[note attention] It would be very nice to have a simple example of a number that has an explicit sequence of good rational approximations but for which there is no known formula for the continued-fraction expansion. [/note]<comment thread="1213" />

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