Pi is rational or irrational

Theorem

Pi ($\pi$) is irrational.

Proof that e is irrational

Proof 1

Aiming for a contradiction, suppose $\pi$ is rational.

Then from Existence of Canonical Form of Rational Number:

$\exists a \in \Z, b \in \Z_{>0}: \pi = \dfrac a b$

Let $n \in \Z_{>0}$.

We define the polynomial function:

$\forall x \in \R: \map f x = \dfrac {x^n \paren {a - b x}^n} {n!}$

We differentiate this $2 n$ times, and then we build:

$\ds \map F x = \sum_{j \mathop = 0}^n \paren {-1}^j \map {f^{\paren {2 j} } } x = \map f x + \cdots + \paren {-1}^j \map {f^{\paren {2 j} } } x + \cdots + \paren {-1}^n \map {f^{\paren {2 n} } } x$

That is, $\map F x$ is the alternating sum of $f$ and its first $n$ evenderivatives.

First we show that:

$(1): \quad \map F 0 = \map F \pi$

From the definition of $\map f x$, and our supposition that $\pi = \dfrac a b$, we have that:

$\forall x \in \R: \map f x = b^n \dfrac {x^n \paren {\pi - x}^n} {n!} = \map f {\pi - x}$

Using the Chain Rule for Derivatives, we can apply the Principle of Mathematical Induction to show that, for all the a

simplest proof that pi is irrational

how pi is irrational