Find a simple proof that π is irrational

Question

I know there are many questions on the site about finding a proof that π is irrational, but I'm posting the question separately to discuss a particular proof further

We know that the Wallis Product is :

$$\frac{π}{2}=(\frac{2}{1}\cdot\frac{2}{3})(\frac{4}{3}\cdot\frac{4}{5})(\frac{6}{5}\cdot\frac{6}{7})(\frac{8}{7}\cdot\frac{8}{9})\cdots$$

This means that if $\pi$ is a rational number, its numerator will be an even number and its denominator will be an odd number

After that, all we have to do is find a formula for the number $\pi$ that gives a "reversed" fraction whose numerator is an odd number and whose denominator is an even number. Thus, we obtain a proof similar to the classical proof that $\sqrt{2}$ is irrational. Indeed, after some research, I found formula of this model that are attributed to Leonard Euler:

Assuming that $p_n$ is a notation that refers to the prime number $n$, the formula we want is :

$$\frac{π}{4}=\prod_{n=1}^∞ (\frac{p_n}{p_n+(-1)^{\frac{p_n+1}{2}}})=\frac{3}{3+1}\cdot\frac{5}{5-1}\cdot\frac{7}{7+1}\cdot\frac{11}{11+1}\cdot\frac{13}{13-1}\cdots$$

It represents an odd number divided by an even number as required. Thus, we obtain a contradiction showing that $\pi$ is irrational.

My question is : Is my proof valid or is there an error in this proof?
I don't know if I can deduce this from an infinite ratio or if it is invalid. If this is not true please give an example of a rational number that has two representations as an infinite product of the two opposite forms.

Answer 1

Here is an explicit example to show the proof doesn't work. Consider the infinite product $$\frac12\prod_{n=0}^\infty\frac{2^{2^n}+1}{2^{2^n}}=\frac12\cdot\frac32\cdot\frac54\cdot\frac{17}{16}\cdots$$ where all numerators are odd and all denominators even. We can check by induction that the partial products are $$\frac12\prod_{n=0}^N\frac{2^{2^n}+1}{2^{2^n}}=\frac{2^{2^{N+1}}-1}{2^{2^{N+1}}},$$ so the infinite product evaluates to $1$.

Then taking the reciprocals of all terms gives another representation of $1$ with opposite parities.

Answer 2

The proof is not valid. The key flaw is that properties that hold for all members of a sequence do not necessarily hold in the limit. For a simple example, consider that all members of the sequence $\dfrac11,\dfrac12,\dfrac13,\ldots,\dfrac1n,\ldots$ are greater than zero, a property that clearly fails for their limit, which is zero. As another example, relevant to the topic, we have found many sequences of rationals that converge to $\pi$, such as the truncations of its decimal expansion $3,3.1,3.14,\ldots$ Again, the property fails in the limit, as $\pi$ is irrational. For your specific example, how can we be sure that the numerator and denominator of the limit of the Wallis product keep their parity (oddness/evenness)? It could feasibly behave like the sequence $\frac{2}{2\times2+1},\frac{4}{2\times4+1},\frac{6}{2\times6+1},\ldots\to\frac12$, where, in the limit, the parities of the numerator and denominator are swapped!

This means, if we are to take a property into the limit in a valid proof, we would need to actually prove that the property transfers to the limit (which is effectively what we're seeking to prove here in the first place!). The idea of your proof is appealing and I think it's a great tactic to use a proof by contradiction, but it has such an intrinsically fatal flaw that it's unlikely to ever succeed. And, from a historical perspective, the first proof of the irrationality of $\pi$ by Lambert in 1761 came over 100 years after the 1656 publication of the Wallis product. It's relatively unlikely that research would have spent 100 years overlooking a simple implication of a significant property. So we can expect to need another idea or to break up the full proof into smaller chunks, as Lambert did by proving properties of the continued fraction expansion of $\tan x$.