Exercise 1045Logic, sets and proofs › Proof techniquesL1Added 3 Nov 2026

Existence proofs

#existence#constructive#non-constructive

Statement

Give a constructive proof that there exist irrational $x,y$ with $x^y$ rational.

Hint

Consider $x=\sqrt2$ , $y=\sqrt2$ ; if $x^y$ is rational, done. Otherwise try $x=(\sqrt2)^{\sqrt2}$ .

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Solution

Attempt 1. $(\sqrt2)^{\sqrt2}$ : if this is rational, done with $x=\sqrt2$ , $y=\sqrt2$ .

Attempt 2. If not, set $x=(\sqrt2)^{\sqrt2}$ , $y=\sqrt2$ . Then $x^y=((\sqrt2)^{\sqrt2})^{\sqrt2}=(\sqrt2)^2=2$ , rational. And $x$ is irrational (by assumption).

Conclusion. In all cases, there exist irrational $x,y$ with $x^y$ rational. This proof is non-constructive in determining which case holds.

Existence proofs

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