Is the set of irrational numbersclosedunder the operation of exponentiation? That is, for all irrational numbersaandb, isa^birrational? Why?

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- Sep 25th 2011, 02:16 PMthamathkid1729Irrational Numbers Question
Is the set of irrational numbers

**closed**under the operation of exponentiation? That is, for all irrational numbers*a*and*b*, is*a*^*b*irrational? Why? - Sep 25th 2011, 05:26 PMDevenoRe: Irrational Numbers Question
i do not know if $\displaystyle \sqrt{2}^{\sqrt{2}}$ is rational, but if it is not, then certainly $\displaystyle (\sqrt{2}^{\sqrt{2}})^{\sqrt{2}} = (\sqrt{2})^2 = 2$ is.

- Sep 26th 2011, 12:36 AMmr fantasticRe: Irrational Numbers Question
- Sep 26th 2011, 12:20 PMebainesRe: Irrational Numbers Question
Consider $\displaystyle e^{\ ln A} = A$. if $\displaystyle A$ is a rational number than$\displaystyle \ln A$ is irrational. So here you have an example of an irrational number raised to an irrational number yielding a rational number result.