Answers by moto

For positive integers $p$ and $q$, with $\frac{p}{q} \neq 1,\left(\frac{p}{q}\right)^{\frac{p}{q}}=p^{\left(\frac{p}{q}-1\right)}$. Then,$q^{p}=p^{q}$$q^{p}=p^{2 q}$$\sqr...