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Proof of identity $\sqrt {xy} = \sqrt x \sqrt y$ for $x,y \in \mathbb R^+$

I've been looking at the stated identity, which makes sense in $\mathbb R^+$ but fails in $\mathbb R$, since $\sqrt {-1 \cdot -1} \neq \sqrt {-1} \sqrt {-1}$.

How does one prove this identity ?

Suppose we have $x,y \in \mathbb R^+$ then $\sqrt{xy}^2 = xy = (\sqrt x \sqrt y)^2$.

Idea 1: The square-root function is bijective (monotonic increasing) and has inverse $X^2$. This in turn implies $\sqrt {xy} = \sqrt x \sqrt y$

Idea 2: The equation $X^2 = xy$ has at most $2$ solutions in $\mathbb C$, which implies $\sqrt x \sqrt y$ must be equal to $\pm \sqrt {xy}$.

Are these ideas rigorous enough ? Is there some simpler way of proving this ?

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  • $\begingroup$ If $x = 0$ or $y = 0$, then it is trivially true. Else, $(\sqrt{xy}-\sqrt{x}\sqrt{y})(\sqrt{xy}+\sqrt{x}\sqrt{y})=0$ so $\sqrt{xy}=\sqrt{x}\sqrt{y}$ because the other factor is $> 0$. (I admit this is somewhat similar to your Idea 2) $\endgroup$
    – Siméon
    Jan 5 '14 at 13:32
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    $\begingroup$ Your first equation in combination with either of the two "ideas" will be enough. I would have phrased it this way: by definition, $\sqrt{xy}$ is the unique nonnegative number whose square is $xy$. Let $c = \sqrt{x}\sqrt{y}$, and show that it satisfies the definition of $\sqrt{xy}$. $\endgroup$
    – user119191
    Jan 5 '14 at 13:34
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Hint: If you have proved that $\sqrt{x}\sqrt{y}$ is a non-negative number such that $(\sqrt{x}\sqrt{y})^2 = xy$, then this means by definition that $\sqrt{x}\sqrt{y} = \sqrt{xy}$.

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  • $\begingroup$ It only means this by definition, because we know the square root function is bijective ? $\endgroup$
    – Shuzheng
    Jan 5 '14 at 16:08
  • $\begingroup$ Yes, but first and foremost that it is a well-defined function $[0,\infty) \to [0,\infty)$ at all, i.e. that for every $a \geq 0$ there exists an $x \geq 0$ such that $x^2 = a$. This can be done by using the intermediate value theorem on the function $g(x) = x^2 - a$. Then one has to show that it is bijective. $\endgroup$
    – Ulrik
    Jan 5 '14 at 22:11
  • $\begingroup$ Showing it is injective is it enough to say: suppose $x^2 = a$ then for every $z < x < y$ we have $z^2 < x^2 < y^2$ hence $x$ is the only value for which $x^2 = a$ ? $\endgroup$
    – Shuzheng
    Jan 6 '14 at 16:28
  • $\begingroup$ If you assume that $z \geq 0$ then yes. You can also do the following: Let $x^2 = y^2 = a$. Then $0 = x^2 - y^2 = (x+y)(x-y)$ which implies that $x=y$ or $x=-y$. If both $x$ and $y$ were positive to begin with, then the only possibility is that $x=y$. $\endgroup$
    – Ulrik
    Jan 6 '14 at 17:57
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This is based on user119191's comment:

From definition, let $a,b\in \mathbb R^+ :a=\sqrt {xy}, b=\sqrt x \sqrt y$

$a^2=xy,b^2=xy\rightarrow a^2=b^2\rightarrow a=b \to\sqrt {xy}=\sqrt x \sqrt y$

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