# Contradiction: for any $(x,y)\in\mathbb{R}^2$, $x=y \Leftrightarrow-x=y \vee x=-y$

I'm pretty sure my error is quite simple but I can't find it. For any $$(x,y)$$ of $$\mathbf R^2,$$

$$x=y \iff x^2=y^2 \iff \sqrt{x^2}=\sqrt{y^2} \iff -x=-y \text{ or} -x=y \text{ or } x=-y \text{ or } x=y.$$

Finally I find $$x=y \iff -x=y \text{ or } x=-y$$ between the two other good solutions.

I think it has something to do with the fact that I square $$x$$ and $$y$$ before square-rooting them (if you do $$\sqrt{x}^2=\sqrt{y}^2$$ all is ok), but idk why it doesn't work...

$$x=y\iff -x=y\lor x=-y\tag{1}$$

Doesn't makes sense since $$-x=y\iff x=-y$$, that is, they're equivalent. This reduces $$(1)$$ to $$x=y\iff -x=y\tag{2}$$ Which is only true when $$x=y=0$$ (since it implies that $$-x=x$$). In your approach, the first step:

$$x=y\iff \sqrt{x^2}=\sqrt{y^2}\tag{3}$$

is false. This is because $$\sqrt{x^2}=|x|$$, so $$(3)$$ becomes $$x=y\iff |x|=|y|$$

This is only true when both $$x,y>0$$ in which case $$|x|=x$$. For instance $$|2|=|-2|$$, but $$2\neq -2$$.

Do you mean

$$x^2=y^2\Rightarrow x=y\lor x=-y\tag{4}$$

?

In which case, notice that you can write $$x^2=y^2\iff x^2-y^2=0\iff (x-y)(x+y)=0$$

• Yes I know (1) isn't true (or only for 0) and that's why I was disappointed to find those 2 solutions among the 4 solutions I found. Did you read my question entirely? Sep 8, 2019 at 11:13
• @sebastienfinor I have to admit I don't understand your question, given that you know $(1)$ is false. Sep 8, 2019 at 12:25
• It's just that I tried some manipulations to x=y and I found it implies x=-y or -x=y, which isn't possible. So my question is why did I find those wrong answers, given the developpement that seemed right to me. Sep 8, 2019 at 12:34
• @sebastienfinor I see, I've updated my answer. Sep 8, 2019 at 12:40
• Thank you. I see that I confused implication and equivalence. Because $$x=y \iff x^2=y^2.$$ is false, instead it's $$x=y \implies x^2=y^2.$$ Sep 8, 2019 at 13:01

You made your very first mistake in the first line. The statements $$x=y$$ and $$\sqrt{x^2}=\sqrt{y^2}$$ are not equivalent as claimed, since the latter means $$|x|=|y|.$$ Thus, if you take $$x$$ to be negative, the equivalence is seen to be false. In particular, although $$x=y\implies |x|=|y|,$$ is true, it's not in general true that $$|x|=|y|\implies x=y$$ is also true.

• Oh okay! When i try to do x=y ⟺ x²=y² ⟺ √(x²)=√(y²), x=y ⟺ x²=y² is false. Instead, I have to write x=y ⟹ x²=y². Thank you Sep 8, 2019 at 12:53
• @sebastienfinor You're welcome! Sep 8, 2019 at 12:57

$$\sqrt{x^2}= \sqrt{y^2}$$ doesn’t imply $$x=y$$!!!

Take $$x=2$$ and $$y=-2$$.

You are wrong at the first step: $$x=y \nLeftrightarrow \sqrt{x^2} =\sqrt{y^2}$$