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...


4 Answers 4


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

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

  • $\begingroup$ 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? $\endgroup$ Sep 8, 2019 at 11:13
  • $\begingroup$ @sebastienfinor I have to admit I don't understand your question, given that you know $(1)$ is false. $\endgroup$ Sep 8, 2019 at 12:25
  • $\begingroup$ 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. $\endgroup$ Sep 8, 2019 at 12:34
  • $\begingroup$ @sebastienfinor I see, I've updated my answer. $\endgroup$ Sep 8, 2019 at 12:40
  • $\begingroup$ Thank you. I see that I confused implication and equivalence. Because $$x=y \iff x^2=y^2.$$ is false, instead it's $$x=y \implies x^2=y^2.$$ $\endgroup$ 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.

  • 2
    $\begingroup$ 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 $\endgroup$ Sep 8, 2019 at 12:53
  • $\begingroup$ @sebastienfinor You're welcome! $\endgroup$
    – Allawonder
    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: $$ x=y \nLeftrightarrow \sqrt{x^2} =\sqrt{y^2} $$


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