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

Question

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...

Answer 1

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

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 $$x^2=y^2\iff x^2-y^2=0\iff (x-y)(x+y)=0$$

Answer 2

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.

Answer 3

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

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

Answer 4

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