Number System

Number Theory Level 3

True or False?

$\quad$ If $a$ and $b$ are irrational numbers , then $a^b$ is always an irrational number.

False True

3 solutions

Naren Bhandari
Jan 30, 2017

The answer is $False$ .

We have $b = \sqrt{2}$ as an irrational number, and $a = \large \sqrt{2}^{\sqrt{2}}$ is also irrational.

But $\large a^{b} = (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \times \sqrt{2}} = \sqrt{2}^{2} = 2$ ,

which is rational.

Brother , use Latex.

Md Zuhair - 4 years, 4 months ago

I am typing you the code, Edit it like this Just erase all the text and just copy paste .

Is that true that Irrational raised to irrational is irrational, Which is $(Irrational)^{Irrational}$ $=$ Irrational

Type in 1 for True and 2 for False

Md Zuhair - 4 years, 4 months ago

I cleaned up the Latex in Naren's solution. Hope it looks o.k. to you now.

Brian Charlesworth - 4 years, 4 months ago

Another counterexample is $\large e^{\ln(2)} = 2$ .

Note that $\large \sqrt{2}^{\sqrt{2}}$ is known as the Gelfond-Schneider constant , which is not only irrational but transcendental as well by Gelfond's Theorem .

Brian Charlesworth - 4 years, 4 months ago

Vijay Simha
Jan 16, 2019

Both log(4) and √10 are irrational. However

√10 ^log(4) = 10^log(2) = 2.

Hjalmar Orellana Soto
Jan 31, 2017

If you let $a=-\sqrt{2}$ and $b=\sqrt{2}$ . Using De Moivre's formula, $a^b$ is gonna be a complex number