In this question it is stated that Somos’ quadratic recurrence constant

α=√2√3√4√⋯

is anirrationalnumber.[update: the author of that question is no longer claiming to have a proof of this]This fact seems by no means trivial to me. The algebraic numbers √2, √2√3, √2√3√4, … do not converge quickly enough to α, so one cannot reuse the proof of Liouville’s theorem in this case.

Approximation arguments do not seem a good way, since

√2√2√2√⋯=2

is rational instead!What am I missing?

**Answer**

Your original post implies you have thought of, or know of, a good way of using Liouville’s theorem to solve the problem. Here’s the start of some ideas on the issue that perhaps you will be able to finish off.

We can write α in the following way:

α=∞∏n=1(1+1n)2−(n−1).

We will also define the partial product

αN=N∏n=1(1+1n)2−(n−1).

Then, for some integers p, q:

|α−pq|≤|α−αN|+|αN−pq|.

Now, the first of these values can be quickly bounded:

α−αN=αN(∞∏n=N+1(1+1n)2−(n−1)−1)≤αN(∞∏n=N+1(2)2−(n−1)−1)≤αN(22−(N−1)−1)≤4αN⋅2−N≤4α⋅2−N

However, I can’t see how to bound the other expression, |αN−pq|. Because the αN are algebraic, we can’t instantly point to the existence of some rationals that approximate them “well enough” for our purposes here, as far as I can see.

**Attribution***Source : Link , Question Author : Mizar , Answer Author : Mizar*