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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?

John Douma
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wantToLearn
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  • Looks fine to me but learn to format on the website please. – K. Y May 26 '19 at 19:16
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    I tried to fix the format. One thing to note is that an expression like $x^2+y^2$ can be rendered by wrapping the entire expression in dollar signs. You don't have to place dollar signs around each symbol of the expression. And yes, your proof looks like a correct application of "proof by contradiction". – John Douma May 26 '19 at 19:21

2 Answers2

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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.

Charly
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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 $$x+y>h$$

AsdrubalBeltran
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