Proof by contradiction to show two equations are not equal

Question

I apologize for the terrible formatting. I'm reading the guide and trying stuff out but its not working. Extremely sorry, but please bear with me.

Prove the following: If $x$ and $y$ are positive real numbers then $(x + y)^2 \neq x^2 + y^2$

I attempted to prove this statement by contradiction.

Assume $x$ and $y$ to be positive real numbers and $(x + y)^2 = x^2 + y^2$

Then $x^2 + 2xy + y^2 = x^2 + y^2$

$2xy = 0$ and $xy = 0$ so either $x = 0$ or $y = 0$ or both. In either case this contradicts the fact that $x$ and $y$ are positive real numbers.

Am I using proof by contradiction correctly?

Answer 1

Your proof by contradiction is correct. If you want a direct proof here is one.

Let $x$ and $y$ be positive real numbers, then

$\left(x+y\right)^2=x^2+y^2+2xy>x^2+y^2$, so $\left(x+y\right)^2>x^2+y^2$ which implies they can't be equal.

Answer 2

Yes is correct your proof. Other way is show that $x$, $y$ are the lengs of a rigth triangle, then by Pitagora's Theorem $x^2+y^2=h^2$ but $h^2=(x+y)^2\implies h=x+y$ but this contradicts the triangle inequality $$x+y>h$$