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Number Theory Level 4

The number $\sqrt{2}^\sqrt{2}$ , is it rational, an algebraic irrational or is it transcendental?

Recall that a rational number can be expressed as $p/q$ for integers $p, q$ with $q \neq 0$ . An algebraic irrational number $\alpha$ is a solution to the equation $P(x) = 0$ , where $P(x)$ is a polynomial with rational coefficients. A transcendental number is a real number that is irrational but not an algebraic irrational.

Rational Impossible to tell Algebraic Irrational Transcendental

1 solution

Krishnan Shankar
Mar 11, 2016

The solution appeals to a deep result in number theory called the Gelfond-Schneider theorem: if $a, b$ are both algebraic numbers (i.e., solutions of polynomial equations in $\mathbf{Q}$ ) and $b$ is irrational, then $a^b$ is transcendental. Now observe that

$(\sqrt{2}^\sqrt{2})^\sqrt{2} = 2$

so if $\sqrt{2}^\sqrt{2}$ were algebraic, then the Gelfond-Schenider theorem would imply that the number $2$ is transcendental, a contradiction.

You don't need the "Now observe ..." part; the first two lines make the point.

Otto Bretscher - 5 years, 3 months ago

Get real!

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