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Prove: There exists $a \in \mathbb{Q}$ and $b \in \mathbb{R}\smallsetminus \mathbb{Q}$ such that $a^b \in \mathbb{R} \smallsetminus \mathbb{Q}$.

I've tried using $\log_23$, $\sqrt 2$, and $\frac{1}{\sqrt 2}$ for the irrational number, but couldn't find a way to prove $a^b$ was irrational.

Is there a way to prove this without using Gelfond–Schneider theorem?

Ian
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    Hint: Just take $a=2$. Can we have $2^x$ rational for all $x \in \mathbb{R}$? For all irrational $x$? – Michael May 15 '18 at 15:35

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Well, either $2^{\sqrt{2}}$ is irrational and we are done, or $(2^\sqrt{2})^{\sqrt{2}/4}=\sqrt{2}$ is an irrational, which is a rational to an irrational power.

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fix $a=7.$ The set of $b$ is uncountable.

Will Jagy
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Try $a=2$ and $b=\log_2(\sqrt{3})$

QC_QAOA
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Let $p$ be a rational in the form $\frac{a}{b}$ and $q$ be an irrational number, now,
$$p^q=1+\frac{\left(q\ln p\right)}{1!}+\frac{\left(q\ln p\right)^2}{2!}...$$
For all $p \gt 1$, Each term in series is irrational thus $p^q$ must be irrational.

Abhishek Choudhary
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  • With you argument the terms of $\sum_{n=0}^{\infty} (-1)^n \dfrac{\pi^{2n+1}}{(2n+1)!}$ are all irrational, hence the sum is irrational? Oh, but wait, the sum is $\sin \pi=0$. It works with only two terms too: $\sqrt2$ and $1-\sqrt2$ are irrational, what can you guess about the sum? – Jean-Claude Arbaut May 17 '18 at 22:21
  • It's because they cancel each other and hence sum is zero while if they won't the sum is bound to be irrational and in your second example its just two numbers and my argumen't isn't that sum of two irrationals is irrational but that if irrational terms don't cancel out then sum of any number of irrationals is irrational. – Abhishek Choudhary May 18 '18 at 10:49
  • You don't seem to understand the argument. You don't know in advance what will cancel. How can you, just looking at the series, know that $\sin\pi/2$ is rational while $\sin\pi/3$ is not? How do you decide which of $\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)\pi}$ and $\sum_{n=1}^{\infty} \dfrac{(-1)^n}{(2n+1)\sqrt2}$ may be rational? – Jean-Claude Arbaut May 18 '18 at 11:06
  • The claim that "each term in the series is irrational thus the sum must be irrational" is blatantly wrong. Don't make a fool of yourself. If you claim that "it's sometimes wrong but we can explain", then it's just wrong, period. – Jean-Claude Arbaut May 18 '18 at 11:12
  • Maybe my explanation is wrong but a rational raised to irrational must be irrational – Abhishek Choudhary May 18 '18 at 11:21
  • You don't have proved that, and it's not even the purpose of the question. – Jean-Claude Arbaut May 18 '18 at 11:24
  • the question must be prove that a rational number raised to a surd is irrational which is true – Abhishek Choudhary May 18 '18 at 11:29
  • No, the question requires to prove that there exists at least one rational number $a$ and an irrational $b$ such that $a^b$ is irrational. You claim more, and you have proved nothing. Now, I am suspecting that you are wrong on purpose, so I'll leave this discussion as it's leading nowhere. – Jean-Claude Arbaut May 18 '18 at 11:43
  • Let us continue this discussion in chat. – Abhishek Choudhary May 18 '18 at 11:45