Is $2^e$ irrational or rational?

Question

Please help me with the following question.

Prove or disprove that $2^e$ is an irrational number, where $e$ is the Neipper number.

Thanks.

Answer 1

Read the comments to know more

just two lines: (its all i could think and write)

let $$2^e=p$$ taking log of base 2 both sides

$$\log_22^e=\log_2p\implies e=\log_2p$$ and lets say that e is irrational

now assuming that $\log_2p=\frac ab, b\in N-\{1\}$ where $a$ is an irrational number and $N$is the set of all natural numbers and $2^a\ne n^mb$ where $ m$ being non-zero positive integers

so $$\log_2p=\frac ab\implies\log_2p^b=a\implies p^b=2^a\implies p=(2^a)^{\frac 1b}\ldots (1)$$

so from (1) it is clear that $p$ is always an irrational and from the very first statement we can say that a number power to an irrational number is an irrational number

Answer 2

Determining whether $2^e$ is irrational or not is an open problem. And more generally, the task of determining whether $(\text{rational})^{\text{irrational}}$ is rational or not appears to be extremely difficult.