Question 964707
Because,
{{{sqrt(x)*sqrt(y)=sqrt(xy)}}}
but
{{{sqrt(x)+sqrt(y)<>sqrt(x+y)}}}
.
.
.
Let's assume it did,
{{{sqrt(x)+sqrt(y)=sqrt(x+y)}}}
Square both sides,
{{{(sqrt(x)+sqrt(y))^2=x+y}}}
{{{sqrt(x)sqrt(x)+2sqrt(x)sqrt(y)+sqrt(y)sqrt(y)=x+y}}}
{{{x+2sqrt(xy)+y=x+y}}}
{{{2sqrt(xy)=0}}}
The left side is only equal to the right when {{{x=0}}} or {{{y=0}}} or {{{x=y=0}}}.
So it's not an identity, so,
{{{sqrt(x)+sqrt(y)<>sqrt(x+y)}}}