Question 1208358
<pre>
To prove {{{sqrt(x*y)=sqrt(x)*sqrt(y)}}}

Let x and y be non-negative real numbers. 

lemma: square roots of non-negative real numbers are unique.

Proof:
For contradiction, suppose for positive real number " a ", p and q 
are two different non-negative square roots of " a ".  Then,

{{{p^2=q^2=a}}} => {{{p^2-q^2=0}}} => {{{(p-q)(p+q)=0}}} 
since {{{p+q>=0}}}, p-q=0, so p=q.  So we have reached a contradiction 
namely, that p and q are NOT different.

We show that the squares of {{{sqrt(x*y)}}} and {{{sqrt(x)*sqrt(y)}}} are equal.

{{{(sqrt(x*y))^2}}}{{{""=""}}}{{{x*y}}}

{{{(sqrt(x)*sqrt(y))^2}}}{{{""=""}}}{{{(sqrt(x)*sqrt(y))(sqrt(x)*sqrt(y))}}}{{{""=""}}}{{{sqrt(x)*sqrt(x)*sqrt(y)*sqrt(y)}}}{{{""=""}}}{{{(sqrt(x))^2(sqrt(y))^2}}}{{{""=""}}}{{{x*y}}}

Therefore {{{(sqrt(x*y))^2=(sqrt(x)*sqrt(y))^2}}}

and by the lemma, {{{sqrt(x*y)=sqrt(x)*sqrt(y)}}}

PROVED.

Edwin</pre>