How to prove that an irrational number to an irrational number could be irrational. - SOLVED

Question

There exists $a,b \in \mathbb{R}-\mathbb{Q}$ such that $a^b \in \mathbb{R}-\mathbb{Q}$.

I am lost as to where to begin. In my class, the only irrational numbers we've really covered is $\sqrt{2}$ and cubed root of two. Any help would be appreciated!

Edit : How my professor did it.

Proof: Consider $\sqrt{2}^{\sqrt{2}}\cdot\sqrt{2}^{1-\sqrt{2}} = \sqrt{2}$. Since a rational multiplied by an irrational is irrational, either $\sqrt{2}^{\sqrt{2}}$ is irrational or $\sqrt{2}^{1-\sqrt{2}}$ is irrational. We will proceed by cases.

Case 1: $\sqrt{2}^{\sqrt{2}}$ is irrational. Then $a=\sqrt{2},b=\sqrt2$.

Case 2: $\sqrt{2}^{1-\sqrt{2}}$ is irrational. Then $a=\sqrt2,b=1-\sqrt2$.

Thus there exists irrationals a,b where $a^b$ is irrational.

Answer 1

There are several ways to approach this problem at an elementary level. But it depends a lot on the context and the material you are expected to know. Here is a suggestion. Consider the function $f(x)=e^x$. I assume you know that $e$ is irrational and that $f$ is injective on the reals. If it were the case that for all irrational $x$ the value $f(x)$ were rational, then you would have an injection from the irrationals to the rationals, thus violating cardinality.

Answer 2

@IttayWeiss's answer is good, but we don't even need to know irrationals are uncountable. Since $x:=\sqrt{2}^{\sqrt{2}}>1$, even if $x$ is rational its prime factorization ensures all sufficiently large $n\in\Bbb N$ satisfy $x^{1/n}\notin\Bbb Q$, whence we can take $a=\sqrt{2},\,b=\sqrt{2}/n$. (The Gelfond-Schneider theorem shows $n=1$ works anyway.)

Answer 3

Consider $x=\sqrt{2}^\sqrt{2}$. It's either rational or irrational. If it's irrational, you have your answer. If it's rational $x=\frac{p}{q}$, then using the decompositions of $p$ and $q$ into prime factors it's easy to show that there exist $n\in\mathbb{N}$ such that $$ \sqrt[n]{x} = \sqrt{2}^{\sqrt{2}/n} $$ is irrational, and then your answer is the pair $(\sqrt{2},\sqrt{2}/n)$.

Answer 4

Here's the simplest way. $\sqrt2$ is irrational, as you know. $\log_23$ is irrational, since $\log_23=a/b$ is $2^a=3^b$ which contradicts the Unique Factorization Theorem. And $\sqrt2^{\log_23}=\sqrt3$, which can be shown to be irrational by the techniques that work for $\sqrt2$.