Irrational power

Number Theory Level 2

Let $i_1$ and $i_2$ be irrational numbers, can $i_1^{i_2}$ be rational?

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3 solutions

A Former Brilliant Member
Jan 8, 2018

Let $i_1$ and $i_2$ be irrational numbers. Then we want to know if it is possible that $i_1^{i_2}$ is rational. Let's try $i_1 = i_2 = \sqrt{2}$ : is $\sqrt{2}^{\sqrt{2}}$ rational? We don't know, but if it is, the problem is solved. If it is not, then take $i_1 = \sqrt{2}^{\sqrt{2}}$ and $i_2 = \sqrt{2}$ . Then

$i_1^{i_2} = \left( \sqrt{2}^{\sqrt{2}} \right)^{\sqrt{2}} = \sqrt{2}^{\sqrt{2} \cdot \sqrt{2}} = \sqrt{2}^{2} = 2,$

is certainly rational, which makes possible to take an irrational number to the power of an irrational number and have as result a rational number.

In fact $\sqrt{2}^\sqrt{2}$ is irrational. This is a consequence of the Gelfond-Schneider Theorem, which says that if $a,b$ are algebraic numbers with $a^2 \neq a$ and if $b$ irrational, then $a^b$ is transcendental.

Mark Hennings - 3 years, 4 months ago

This is simply beautiful

Stephen Mellor - 3 years, 4 months ago

Will van Noordt
Jan 11, 2018

Let $i_1 = e$ and $i_2 = \ln(q)$ where $q \in \mathbb{Q}$ . Then, $i_1^{i_2} = e^{\ln(q)} = q \in \mathbb{Q}$ .

Nicely done! We know that if $q \in \mathbb{Q}$ then $\ln(q)$ is irrational. But can you prove it?

A Former Brilliant Member - 3 years, 4 months ago

Suppose $\ln(q) \in \mathbb{Q}$ , then for coprime nonzero integers $m$ and $n$ , then $\ln(q) = \frac{m}{n}$ . This implies that $e^{\frac{m}{n}} = q$ , or, raising both sides to the $n^{th}$ power, $e^m = q^n \in \mathbb{Q}$ . We know that $e$ is transcendental, so therefore, for nonzero integers $m$ , $e^{m} \in \mathbb{I}$ , contradicting $e^{m} \in \mathbb{Q}$ . Therefore, $\ln(q) \in \mathbb{I}$ .

Will van Noordt - 3 years, 4 months ago

I should clarify: if $e^m \in \mathbb{Q}$ , then for nonzero coprime integers $p$ and $q$ , $e^m = \frac{p}{q}$ , or equivalently, $qe^m-p = 0$ , which would imply that $e$ is a root of the integer-coefficient polynomial $qx^m-p = 0$ , contradicting the fact that $e$ is transcendental, so $e^m \in \mathbb{I}$ .

Will van Noordt - 3 years, 4 months ago

Michael Boyd
Dec 1, 2018

By definition, irrational numbers are members of the set complementary to rational numbers, thus e, i=√(-1), and π are all irrational. π*i is also irrational, however, e^(πi)=-1, which is rational.

Irrational power

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