# Simple proof that $\pi$ is irrational - using prime factors of denominator

Simple proof that $\pi$ is irrational

Consider the Gregory - Leibniz series for $\pi/4$: $$\frac \pi 4 = 1 - \frac 1 3 + \frac 1 5 + \cdots$$

Let $A_n/B_n$ be the irreducible fraction given by partial sum $S_n$ up to the $n$th term $\pm 1/(2n-1)$.

It can be shown that largest prime number $p_\max$ in the individual term denominators of $S_n$ satisfies $n < p_\max \leq 2n-1$. (Bertrand's postulate).

It can be shown that $p_\max$ must be a prime factor of $B_n$, and therefore $p_\max$ is a lower bound on $B_n$.

It follows that $n$ is a lower bound on $B_n$.

Suppose $A/B$ is the irreducible fraction $\pi/4$.

Assumption: $B \geq \liminf_{n \to \infty} (B_n)$

Given this assumption, $B$ cannot be finite and so $\pi/4$ is irrational.

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I think to be a real proof, the assumption needs to be proven. Would that be difficult?

• In general, such the assumptions do not hold. For instance, for the series $\sum_{i=1}^\infty \frac 1{2^n}$ which has a rational sum $1$ – Alex Ravsky Dec 21 '14 at 20:32
• @CraigHicks After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. – AlexR Dec 21 '14 at 20:51
• Thank you for the edit Micheal Hardy – Craig Hicks Dec 24 '14 at 2:54

You proof idea breaks because it could prove that $0\notin\mathbb Q$ by considering the sequence $$a_n = 2^{-n}$$ We have $$\lim_{n\to\infty} a_n = 0$$ but the denominator is exactly $2^n$, wich diverges.
Generally, you assume $$\lim_{n\to\infty} \frac{a_n}{b_n} = \frac ab \Rightarrow \lim_{n\to\infty} a_n = a \wedge \lim_{n\to\infty} b_n = b$$ wich is false. Only the converse holds ($\lim a_n = a \in \mathbb R \ni \lim b_n = b \ne 0 \Rightarrow \lim \frac{a_n}{b_n} = \frac ab$)
Consider the primes, written in order: $p_1, p_2, p_3,\ldots$
That is, $p_1=2, p_2=3, p_3=5, p_4=7, p_5=11, \ldots$. Now consider the sequence $$\frac{p_1-1}{p_1}, \frac{p_2-1}{p_2}, \frac{p_3-1}{p_3},\ldots$$
Much like in the OP, this is a sequence of fractions, all irreducible, with an increasingly larger prime dividing the denominator. Yet the limit of this sequence is $1$, which is rational.