## ganeshie8 one year ago Show that there exist irrational numbers $$m,n$$ such that $$m^n$$ is rational.

1. ikram002p

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.

2. ganeshie8

Brilliant!

3. ikram002p

do u have some other idea ?

4. ikram002p

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

5. ikram002p

for any integers x,y