# Prove or disprove that $2^{\frac{\pi}{2}}$ is an irrational number.

### Problem

Prove or disprove that $$2^{\frac{\pi}{2}}$$ is an irrational number.

### My Try

According to our mathematical intuition, we may want to apply Gelfond–Schneider theorem, which states that

if $$\alpha,\beta$$ are two algebraic numbers, where $$\alpha$$ dose not equal $$0$$ or $$1$$ and $$\beta$$ is not a rational number, then $$\alpha^\beta$$ is a transcendental number.

But in fact, it can't work here, because $$\beta=\dfrac{\pi}{2}$$ here is not an algebraic number at all, which doesn't satisfy the premise of the theorem.

Besides, I have tested this on WolframAlpha. It outputs the consequence as follows

Now, how to go on?

• Oh, this problem doesn't need any particular source, Maybe, it naturally comes from our mathematical thinking? Commented May 25, 2018 at 1:20
• Commented May 25, 2018 at 1:23
• @mengdie1982 The way the question was phrased when I initially wrote my reply was 'prove that $2^{\pi/2}$ is irrational', suggesting that you were being asked to find a proof; that, and the seemingly-arbitrary choice of $\pi/2$ rather than $\pi$ in the exponent, are why I asked to the source, because AFAIK (and as Michael's link suggests) the problem is open. If you were inquiring about the status, then I would suggest a title along the lines of 'Is it known whether $2^{\pi/2}$ is rational or not?' Commented May 25, 2018 at 1:31
• @SK19 bless your heart brother, just wanted to make sure in case you missed it. Side note: I have spent a lot of time, in the past, trying to solve Goldbach's conjecture, so I can relate. Commented May 25, 2018 at 3:12
• @Jepsilon: Unfortunately that's the opposite knowledge of what would be useful. If $\pi \ln(2)$ happened to be rational (or even algebraic) then the Lindemann-Weierstrass theorem would tell us that $2^\pi$ is transcendental, but if $\pi\ln(2)$ is transcendental, we're no further. Do you know that $\pi\ln(2)$ is irrational, though? It's not obvious to me. (Multiplying the Leibniz series and the alternating harmonic series might give something that's nice enough to work with. But since they're only conditionally convergent, multiplying term by term is not automatically valid). Commented May 25, 2018 at 11:20

As noted in the comment section, the irrationality of $$2^\pi$$ is unknown, so is that of $$2^{\pi/2}$$.