Question 447435
<pre>
Given hypothesis:  {{{sqrt(a)*sqrt(b)}}} is irrational.

To prove {{{sqrt(a)+sqrt(b)}}} is irrational

For the sake of contradiction, let's assume

{{{sqrt(a)+sqrt(b)}}} is rational and equal to {{{p/q}}} where

p and q are relatively prime positive integers.

{{{sqrt(a)+sqrt(b)=p/q}}}

Square both sides:

{{{(sqrt(a)+sqrt(b))^2=(p/q)^2}}}

{{{a + 2sqrt(a)sqrt(b) + b=p^2/q^2}}}

Solve for {{{sqrt(a)sqrt(b)}}}

{{{2sqrt(a)sqrt(b)=p^2/q^2-a-b}}}

{{{sqrt(a)sqrt(b)=(p^2/q^2-a-b)/2}}}

The expression on the right is rational
by the closure properties of addition and 
multiplication of rational numbers.  There is
no need to simplify it unless you just want to.

So we have reached a contradiction to the
given hypothesis.

Therefore the assumption that the sum of the
square roots of a and b is rational is false.

Therefore the sum of the two square roots is irrational.

Edwin</pre>