Prove that there is an irrational number x such that x^(sqrt2) is rational?

Answer 1

This proof can proceed by finding one.
Let y=(sqrt2)^(sqrt2).
Then y^(sqrt2)
=( (sqrt2)^(sqrt2) )^(sqrt2)
=(sqrt2)^( (sqrt2) * (sqrt2) )
=(sqrt2)^2
=2.
Now y must be irrational, for suppose y is rational,
y^2 must also be rational,
y^2
=[ (sqrt2)^(sqrt2) ]^2
= [(sqrt2)^2 ]^ (sqrt2)
= 2^(sqrt2) is irrational which contradicts the supposition that y is rational.
Conclude that y is irrational.

Answer 2

First, there is no such number if x is algebraic.
Gel'fond's theorem states that x^√2 is transcendental
in that case.
So let's look at (√3)^√2.
This number is transcendental by Gel'fond's theorem.
Therefore it is irrational.
Now raise it to the √2 power and we get
√3^2 = 3, which is rational.
Here is the statement of Gel'fond's theorem:
(Source: Wikipedia)
Gelfond's theorem, also called the Gelfond-Schneider theorem, states that a^b is transcendental if

1. a is algebraic a != 0,1 and

2. b is algebraic and irrational.

Answer 3

Let x = (5)^(1/sqrt(2))

x will be irrational, but x^(sqrt 2) = 5

Answer 4

ok, um what does that mean