# $\pi$ is irrational.
It is a known fact. But how can one prove it?
Below is known as Mary Cartwright's proof, a simplification of Hermite's proof. The general outline is as follows.
1. Consider the integral for each nonnegative integer $n$, $I_{n}(t) = \int_{-1}^{1}(1-x^{2})^{n}\cos(tx)dx$.
2. Use integration by parts to derive a recurrence between $I_{n}, I_{n-1},$ and $I_{n-2}$.
3. Show by induction that $t^{2k+1}I_{k}(t)=k! P_{k}(t) \sin(t) + k! Q_{k}(t) \cos(t)$ for some integer polynomials $P_{k},Q_{k}$ whose degree are less than $2k+1$.
4. Let $t = \frac{\pi}{2}$ and suppose to the contrary that $t$ is a rational number. Derive a contradiction.
Consider the integral $$
I_{n}(t) = \int_{-1}^{1}(1-x^{2})^{n}\cos(tx) dx.
$$
For $n \ge 2$, by integration by parts we get recurrence $$
\begin{align*}
I_{n}(t) &= \frac{1}{t}\int_{-1}^{1}(1-x^{2})^{n} d(\sin(tx)) \\
\implies t I_{n}(t) &= [(1-x^{2})^{n}\sin(tx)]_{x=-1}^{1} + \int_{-1}^{1}n(1-x^{2})^{n-1}(2x) \sin(tx)dx \\
\implies t^{2} I_{n}(t) & = -2n \int_{-1}^{1}x(1-x^{2})^{n-1} d(\cos(tx)) \\
\implies t^{2}I_{n}(t) &= 2n \int_{-1}^{1} ( (1-x^{2})^{n-1} -2x^{2}(n-1) (1-x^{2})^{n-2} ) \cos(tx)dx \\
\implies t^{2} I_{n}(t) & =2nI_{n-1}(t) + 4n(n-1)\int_{-1}^{1}(-1+1-x^{2})(1-x^{2})^{n-2} \cos(tx)dx \\
\implies t^{2}I_{n}(t) & = 2n I_{n-1}(t) +4n(n-1)I_{n-1}(t)
- 4n(n-1)I_{n-2}(t) \\
\implies t^{2}I_{n}(t) &= 2n(2n-1)I_{n-1}(t) - 4n(n-1)I_{n-2}(t)
\end{align*}
$$
Note $I_{0}(t) = \int_{-1}^{1} \cos(tx)dx= \frac{1}{t}\sin(tx)]_{x=-1}^{1}=\frac{2}{t}\sin(t)$ and $$
I_{1}(t) = \int_{-1}^{1}(1-x^{2})\cos(tx) dx = \frac{4 \sin(t) - 4 t \cos(t)}{t^{3}}.
$$
So we have base cases $$
\begin{align*}
tI_{0}(t) &=\sin(t)\\
t^{3}I_{1}(t) &=4\sin(t)-4t \cos(t)
\end{align*}$$
Let us see what $I_{2}$ satisfies:$$
\begin{align*}
t^{2}I_{2}(t) = 12I_{1}(t)-8I_{0}(t) \\
\implies t^{2} I_{2}(t) = 12 \frac{4 \sin(t)-4t \cos(t)}{t^{3}} -8 \frac{\sin(t)}{t} \\
\implies t^{5} I_{2}(t) = (48- 8 t^{2}) \sin(t) - 48t \cos(t)
\end{align*}$$
So we see that $t^{2k+1} I_{k}(t)$ is some linear combination of $\sin(t)$ and $\cos(t)$ with polynomial coefficients. In particular $$
t^{2k+1} I_{k}(t) = k! (P_{k}(t) \sin(t)+ Q_{k}(t) \cos(t))
$$where $P_{k},Q_{k}$ are integer polynomials whose degree are strictly less than $2k+1$. Indeed, by induction on $k$, we see that $$
\begin{align*}
t^{2}I_{k+1}(t) &= 2(k+1)(2k+1)I_{k}(t) - 4(k+1)k I_{k-1}(t) \\
\implies t^{2}I_{k+1}(t) &= 2(k+1)(2k+1)\left[ \frac{k!(P_{k}(t)\sin(t) + Q_{k}(t) \cos(t))}{t^{2k+1}} \right] \\&- 4(k+1)k \left[ \frac{(k-1)!(P_{k-1}(t)\sin(t) + Q_{k-1}(t)\cos(t))}{t^{2k-1}} \right] \\
\implies t^{2k+3} I_{k+1}(t) &= 2(2k+1)(k+1)! [P_{k}(t) \sin(t) + Q_{k}(t) \cos(t)] \\
& - 4(k+1)! [ t^{2}P_{k-1}(t) \sin(t) + t^{2} Q_{k-1}(t) \cos(t)] \\
\implies t^{2k+3} I_{k+1}(t) & = (k+1)![2(2k+1)P_{k}(t)-4t^{2}P_{k-1}(t)] \sin(t) \\ &+ (k+1)![2(2k+1)Q_{k}(t)-4t^{2}Q_{k-1}(t)] \cos(t)
\end{align*}$$and we see that $P_{k+1}(t):=2(2k+1)P_{k}(t)-4t^{2}P_{k-1}(t)$ and $Q_{k+1}(t):=2(2k+1)Q_{k}(t)-4t^{2}Q_{k-1}(t)$ are both integer polynomials with degrees less than $2k+3$ by induction.
Now suppose $\pi$ is rational, then so is $\frac{\pi}{2}$. Say $\frac{\pi}{2} = \frac{a}{b}$ for some integers $a,b$. Then substitute $t = \frac{\pi}{2}$ we get for all $k$ $$
\frac{a^{2k+1}}{b^{2k+1}} I_{k}(\pi / 2 ) = k!P_{k}(a / b)
$$
Since $P_{k}$ is an integer polynomial of degree less than $2k+1$, $$
b^{2k+1} P_{k}(a / b)
$$is an integer. So $$
\frac{a^{2k+1}I_{k}(\pi /2)}{k!} = b^{2k+1} P(a /b)
$$is an integer, for all positive integer $k$.
But note $0\le \cos( \frac{\pi}{2} x) \le1$ for all $x\in [-1,1]$, so $$
0 < I_{k}\left( \frac{\pi}{2} \right) = \int_{-1}^{1}(1-x^{2})^{k}\cos\left( \frac{\pi}{2} x \right) dx \le \int_{-1}^{1} (1-x^{2})^{k} dx \le 2
$$for all positive integer $k$.
Hence $$
0 < \frac{a^{2k+1} I_{k}(\pi / 2)}{k!} \le 2 \frac{a^{2k+1}}{k!}
$$for all positive integer $k$.
But $\frac{2a^{2k+1}}{k!} = 2a \frac{(a^{2})^{k}}{k!}\to 0$ as $k \to \infty$. So for some $k$, we have $$
0 < \frac{a^{2k+1}I_{k}( \pi/ 2)}{k!} < \frac{1}{2},
$$implying that an integer is strictly between $0$ and $\frac{1}{2}$, a contradiction.
Hence, $\pi$ is not rational!