Let $n\ge3$. Prove that $\sqrt[n]2\notin\Bbb Q$.
Let us suppose that $\sqrt[n]2=p/q$, that is $2q^n=p^n$, so $q^n+q^n=p^n$, against FLT.
Do you know similar examples, in which simple problems are solved using huge weapons (maybe in a elegant way)?
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this communityLet $n\ge3$. Prove that $\sqrt[n]2\notin\Bbb Q$.
Let us suppose that $\sqrt[n]2=p/q$, that is $2q^n=p^n$, so $q^n+q^n=p^n$, against FLT.
Do you know similar examples, in which simple problems are solved using huge weapons (maybe in a elegant way)?