Show that there exist irrational numbers \(m,n\) such that \(m^n\) is rational.

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Show that there exist irrational numbers \(m,n\) such that \(m^n\) is rational.

Mathematics
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let p,q be primes then \(\Large n=\sqrt {q}, m=\sqrt {p}\) i'd like p=2 it would make it much easy. so let \( \Large u=\sqrt{q} ^{\sqrt{2}}\) either way if its rational then done if its not rational then \(\large u^\sqrt 2 =(\sqrt{q} ^{\sqrt{2}})^\sqrt{2}=\sqrt q^2=q\) which is rational , how ever its ok to have p not equal 2 we would do these steps p times in general but as long we are talking about existence then its fine.
Brilliant!
do u have some other idea ?

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i think we can generate this to \( \Large \text{if } n,m \\\Large \text{are irrational then there exist rational r s.t }\\ \Large r=\sqrt[x]{m}^{\sqrt[y]{n}} \)
for any integers x,y

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