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I want to prove directly that the following statement is correct:

$$\forall x \in \mathbb{R^+} \exists y \in \mathbb{R^-}: y^2 = x$$

By having a plain look I can say that it is true. However, what is hard for me is the formal prove of this statement.

I tried:

Let $x \in \mathbb{R^+}$ and $y \in \mathbb{R^-}$

Then $-y ^2$ is in $\mathbb{R^+}$. Therefore, the statement is true!

Is this a correct direct proof. I appreciate your answer!!!

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    $\begingroup$ How are you defining multiplication in $\Bbb R$? Or at least squaring? $\endgroup$ – Git Gud Sep 8 '13 at 14:51
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nono at first $-y^2$ is still in $\mathbb{R}^-$.

The second point is you don't have to show that the square of a negative number is positive, but you have to show that every positive number is the square of a negative one.

Furthermore given the existence of a root function which maps every $x$ to it's square root you have:

$$y^2=x\iff y^2-x=0$$

and via the third binomial theorem you have

$$y^2-x=(y-\sqrt{x})\cdot (y+\sqrt{x})$$

So you just have to check that there is a $y\in \mathbb{R}^-$ such that $$y+ \sqrt{x}=0$$

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    $\begingroup$ If $\sqrt x$ is defined, by it's very definition $y=-\sqrt x$ will work immediately. $\endgroup$ – Git Gud Sep 8 '13 at 14:58
  • $\begingroup$ Third binomial theorem? $\endgroup$ – Alraxite Sep 8 '13 at 14:59
  • $\begingroup$ @Alraxite $(a+b)\cdot (a-b)=a^2-b^2$. $\endgroup$ – Dominic Michaelis Sep 8 '13 at 15:00
  • $\begingroup$ @DominicMichaelis I mean, I've never heard of it. Is it really called the third binomial theorem? $\endgroup$ – Alraxite Sep 8 '13 at 15:02
  • $\begingroup$ @GitGud for sure, but do you think they should show that $\sqrt{x}$ is well defined? We did use Banachs Fix point theorem for that if I remember correct $\endgroup$ – Dominic Michaelis Sep 8 '13 at 15:03

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