A soft question concerning the number $p_{\infty}\#$

Question

Here is an apparent contradiction involving an infinitely large integer, and my (soft) question is: Can anyone point out how or where my thinking has gone off the tracks?

Let $p_n$ denote the $n$th prime number, $p_n\#$ be the primorial, or product of the first $n$ prime numbers, and $\phi(x)$ be the totient of $x$.

Consider the number denoted by $p_{\infty}\#$. It seems to me that this number is a member of $\mathbb N$ (what else?). It further seems to me that no $n>1 \in \mathbb N$ can be relatively prime to $p_{\infty}\#$ because every such $n$ will have at least one prime factor, and that prime factor will also be a factor of $p_{\infty}\#$.

Accordingly, only the number $1$ is coprime to $p_{\infty}\#$, so $\phi(p_{\infty}\#)=1$. On the other hand, formulaically $\phi(p_{\infty}\#)=\prod_{i=1}^{\infty}(p_i-1)$ which is very much larger than $1$.

Either I am not thinking correctly about what is meant by $p_{\infty}\#$, or perhaps that object is not a proper member of $\mathbb N$. Or there is another explanation that will resolve the apparent contradiction. I would be most interested to hear the opinions of others.

Answer 1

"Infinitely large integers" do not exist. Every integer (say, every member of $\mathbb N$) is a finite number, even though there are infinitely many of them. So, $p_\infty\#$ is not a concept that makes sense inside $\mathbb N$, and so your deductions don't really hold water.