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 :


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.

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    $\begingroup$ Both products you mention are infinite. In particular, this holds true for the Wallis product. If $\pi$ were rational, then it would have a representation as a (finite) fraction. You would not be able to compare the numerator/denominator to the Wallis product, which would only work if the latter terminated after a finite number of terms. $\endgroup$
    – DominikS
    Commented Dec 4, 2023 at 10:30
  • 4
    $\begingroup$ Interesting Method , +1 , which requires some thought to decide whether it is valid or not. At high-level , it looks right , though at the same time , I can see 3 or 4 unresolved issues : thinking ! $\endgroup$
    – Prem
    Commented Dec 4, 2023 at 10:54
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    $\begingroup$ Your proof is not valid. It's possible to find similar products that converge to $1$. So your argument would prove that $1$ is irrational. $\endgroup$
    – jjagmath
    Commented Dec 4, 2023 at 11:13
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    $\begingroup$ For this to work, you would basically need to know that $\pi$ is a $2$-adic rational and the product converges to it in the $2$-adic topology. But that is not true, since the product obviously converges to $0$ in the $2$-adic topology. $\endgroup$
    – tomasz
    Commented Dec 5, 2023 at 2:15
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    $\begingroup$ Consider the sequence: $$\frac25,\quad\frac49,\quad\frac{6}{13},\quad\frac{8}{17},\quad\frac{10}{21},\quad\ldots$$ whose $n$th entry is $\frac{2n}{4n+1}$. Each member in the sequence is rational (in lowest terms) with even numerator and odd denominator. Yet the sequence converges to a rational limit whose numerator is odd and denominator is even. So such behavior is not preserved "in the limit". So you cannot say in advance that the partial products from Wallis's product (all of whose members have even numerators and odd denominators) do not converge in a similar way. $\endgroup$ Commented Dec 6, 2023 at 10:19

2 Answers 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 $\tan x$.


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.

  • 1
    $\begingroup$ What might be considered a counterexample should be of the form: $\frac{a}{b}=\prod_{n=0}^∞ \frac{odd}{even}=\prod_{n=0}^∞ \frac{even}{odd}$ or $\frac{a}{b}=\prod_{n=0}^∞ \frac{odd}{even}=\prod_{n=0}^∞ \frac{odd}{odd}$. $\endgroup$ Commented Dec 4, 2023 at 21:53
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    $\begingroup$ @زكرياحسناوي this is the first case, with $a=b=1$. It's more convenient to write the first term of the product separately, but it fits the parity pattern. $\endgroup$ Commented Dec 4, 2023 at 22:03
  • $\begingroup$ Can you find a counterexample where the infinite product doesn't evaluate to 1, though? 1 is a special case here because it's its own reciprocal. $\endgroup$
    – zwol
    Commented Dec 7, 2023 at 16:28
  • $\begingroup$ @zwol - just drop the $\frac 12$ from the front. Then the product is $2$. Any other reduced fraction with numerator not divisible by $4$ can be obtained by a similar change, while preserving the odd over even nature of the sequence. $\endgroup$ Commented Dec 7, 2023 at 17:47
  • $\begingroup$ @zwol to get something else in both ways, note that you can tweak this sequence to get either $3$ or $1/3$ as a product of odd/even (just changing the numerator of one of the first two terms). Thus you can also get either $1/3$ or $3$ as product of even/odd by taking the reciprocal. $\endgroup$ Commented Dec 8, 2023 at 9:38

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