Question

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

Answer 1

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.

Answer 2

Brilliant!

Answer 3

do u have some other idea ?

Answer 4

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

Answer 5

for any integers x,y