Using contrapositive how to prove $x,y \in \mathbb R  \wedge x \lt 0 \implies \nexists y$ such that $x=y^2$? Consider the following implication,  $x,y \in \mathbb R  \wedge x \lt 0 \implies \nexists y$ such that $x=y^2$.
Question asks to use contrapositive, so here is my proof:
Let $x=y^2$ (since it's negation of conclusion). I want to show that $x \ge 0$.
So from new hypothesis we know that x is positive real number greater than or equal to 0, since $x=y^2$ (x is equal to y square) this means the new conclusion is correct. 
So contrapositive is true meaning implication is true. 
Is my proof correct, or am I missing something?
Thanks!
Edits:
Could someone please give me correct proof for this problem? Thanks.
 A: :-) look what youve done: you just put "new hypothesis" and said "if the new hypothesis is true, then the new hypothesis is true". And from this you concluded that the implication is true :-)
The correct proof would depend on the axiom set. Suppose we know that if $a \ge b$ and $c > 0$, then $ac \ge ab$, that $a \cdot 1 = a$, $a + 0 = a$, $a + (-a) = 0$, $a + (-b) = a - b$, and $a \cdot (b + c) = ab + ac$.
You want to prove that for $y \in \mathbb{R}$, $y^2 \ge 0$
1) if $y \ge 0$, then $y \cdot y \ge 0 \cdot y$, thus $y^2 \ge 0$, QED
2) if $y < 0$, then it is more complicated, first prove that $-y = (-1) y$:
$0 = (-1 + 1) y$, thus $0 = (-1) y + y$, thus $-y = (-1) y$
Then prove that $(-1)(-1) = 1$: we know that $1 + (-1) = 0$, thus $-1 \cdot 1 + (-1)(-1) = 0$, and thus $(-1)(-1) = 1$
Finally, from 1) we know that $(-y)^2 \ge 0$, so:
$0 \le (-y)^2 = ((-1) y)^2 = (-1)(-1) y^2 = 1 \cdot y^2 = y^2$ 
QED
A: forall x, x^2 > 0
therefore (by taking the contrapositive)
there does not exist x, x^2 < 0
now you can introduce 'y' if you want but it's redundant.
A: Proof by contradiction. 
Suppose not. Therefore, $x = y^2$ where $x,y \in \mathbb{R}$ and $x < 0$. Since $y \in \mathbb{R}$, $y^2 \ge 0$ and thus $ x \ge 0$ (by assumption). But this contradicts the fact that $x<0$ , which implies that $x \neq y^2$.
Hence, since $x \neq y^2$ for $x,y \in \mathbb{R}$ and $x < 0$, $\nexists y$ such that $x = y^2$.
A: I don't know what everyone's going on about, as your proof is completely correct.  You could have been a little clearer in your wording, though.  
You want to prove that $x,y \in \mathbb R  \wedge x \lt 0 \implies \nexists y$ such that $x=y^2$, using the contrapositive.  
Quite rightly, you start by assuming that $\exists y$ such that $x=y^2$; i.e., the negation of the conclusion you wnat to prove.  You then set out to prove the negation of the premise, namely that $x \ge 0$.  This is exactly how the contrapositive works.  
You prove this by noting that $x$ is a square and must therefore be positive.  So you win!  
Don't listen to these people telling you that your proof is wrong.  I'm not quite sure why so many people think that it's wrong, but they're the ones who are wrong, not you.  
