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In this question it is stated that Somos' quadratic recurrence constant $$\alpha=\sqrt{2\sqrt{3\sqrt{4\sqrt{\cdots}}}}$$ is an irrational number. [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 $\sqrt{2}$, $\sqrt{2\sqrt{3}}$, $\sqrt{2\sqrt{3\sqrt{4}}}$, $\dots$ do not converge quickly enough to $\alpha$, so one cannot reuse the proof of Liouville's theorem in this case.

Approximation arguments do not seem a good way, since $$ \sqrt{2\sqrt{2\sqrt{2\sqrt{\cdots}}}}=2$$ is rational instead!

What am I missing?

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    $\begingroup$ Another way to think about your number is:$$\alpha=\prod_{k=1}^{\infty}(k+1)^{\frac{1}{2^k}} $$ $\endgroup$ – Elaqqad May 1 '15 at 12:46
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    $\begingroup$ It would not shock me if the irrationality of this number is only mathematical folklore - someone said it once, and everyone else thought, "That's not too surprising" and repeated it - without anyone ever actually proving it. $\endgroup$ – Milo Brandt Jun 9 '15 at 0:13
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    $\begingroup$ It's definitely irrational, but is there any proof of this? $\endgroup$ – GEdgar Jun 11 '15 at 14:48
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    $\begingroup$ Sources, sources, we want reliable sources... $\endgroup$ – Did Jun 20 '15 at 17:55
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    $\begingroup$ Is $\sqrt {2 \sqrt {3 \sqrt {4 \ldots}}}$ algebraic or transcendental? $\endgroup$ – user217174 Jun 21 '15 at 20:31
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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 $\alpha$ in the following way: $$\alpha=\prod_{n=1}^\infty\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}.$$

We will also define the partial product

$$\alpha_N=\prod_{n=1}^N\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}.$$

Then, for some integers $p$, $q$:

$$\left|\alpha-\frac{p}{q}\right|\leq\left|\alpha-\alpha_N\right|+\left|\alpha_N-\frac{p}{q}\right|$$.

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

\begin{eqnarray}\alpha-\alpha_N&=&\alpha_N\left(\prod_{n=N+1}^\infty\left(1+\frac{1}{n}\right)^{2^{-(n-1)}}-1\right)\\ &\leq& \alpha_N\left(\prod_{n=N+1}^\infty\left(2\right)^{2^{-(n-1)}}-1\right)\\ &\leq& \alpha_N\left(2^{2^{-(N-1)}}-1\right) \\ &\leq& 4\alpha_N\cdot 2^{-N} \\ &\leq& 4\alpha\cdot 2^{-N} \end{eqnarray}

However, I can't see how to bound the other expression, $\left|\alpha_N-\frac{p}{q}\right|$. Because the $\alpha_N$ 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.

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  • $\begingroup$ Yes, that's exactly what I tried before asking the question. Maybe some expert in diophantine approximation can solve this problem using similar techniques.. $\endgroup$ – Mizar Nov 16 '15 at 13:38
  • $\begingroup$ Okay; the reorganisation of the terms here make it converge faster than what you said you had tried in the OP. I think that Liouville's theorem shows we can't approximate the final term better than $\frac{A}{q^{2^n}}$, but we only need an approximation better than $\frac{A}{q^n}$ which could exist. I tried looking for rational approximations of $n$th roots but found nothing helpful. $\endgroup$ – A Simmons Nov 16 '15 at 13:41
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I may be completely out of line here and my thinking doesn't seem to be at the level of most of the other proofs presented, but I wonder if the following could lead to a proof. Please do not consider this a rigorous proof by any stretch.

First, use the distributive property of roots to separate $\sqrt{2}$ from $\sqrt{\sqrt{3\sqrt{4\sqrt{...}}}}$ which gives us $\sqrt{2}*\sqrt{\sqrt{3\sqrt{4\sqrt{...}}}}$ Now, most have seen the proof from contradiction that $\sqrt{2}$ is irrational. Would it not then just be an issue of showing that no matter the remaining sequence, when multiplied by $\sqrt{2}$, it can never produce a rational number? Essentially, an irrational number cannot be multiplied by any such number as to produce a rational number $\frac{p}{q}$. Am I completely off base here?

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  • $\begingroup$ Well, $\sqrt{2}^2 = 2$ so you'd have to have a good reason for the product of 2 irrational numbers to necessarily not be rational. $\endgroup$ – Chantry Cargill Dec 12 '15 at 1:55
  • $\begingroup$ I realized that shortly after I posted what I laid out. I wonder if such is an exception or a rule. Such that if one can show that the only number sufficient to make $\sqrt{2}$ rational is itself (no other number can do so) then one only needs to show that the remainder of the sequence does not equal $\sqrt{2}$. Does that hold? $\endgroup$ – Chris Thomas Dec 12 '15 at 2:04
  • $\begingroup$ Take $(\sqrt{2} + 1)(\sqrt{2} - 1) = 1$. For root 2, I can only think of multiples of it off the top of my, but you certainly would need a proof of that. Then you would need to show that the rest isn't a multiple of root 2. $\endgroup$ – Chantry Cargill Dec 12 '15 at 4:08
  • $\begingroup$ Isn't the issue there a problem of identity? Such that $\frac{a}{b}=c\sqrt{2}$ which means $\frac{a}{bc}=\sqrt{2}$. Since root 2 is irrational, it cannot be expressed as a quotient of two integers. Meaning that c can only be a multiple of $\sqrt{2}$. Stating that because root 2 multiplied by itself is rational is harkening to the definition of a root, and thus a possible exception to the case that an irrational number multiplied by any other number produces a rational number. Maybe I am not stating it clearly. $\endgroup$ – Chris Thomas Dec 12 '15 at 23:28

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