SOLUTION: Assuming the truth of the theorem that states that √n is irrational whenever n is a positive integer that is not a perfect square, prove that √2 + √3 is irrationa

For contradiction, assume √2 + √3 is rational

Then there exist two positive integers, p and q such that

            √2 + √3 = p/q 

Square both sides:

                  (√2 + √3)² = p²/q²

Both sides are rational since the square of a rational number is
rational.

          (√2 + √3)(√2 + √3) = p²/q²

Use FOIL

   √2√2 + √2√3 + √3√2 + √3√3 = p²/q²        

Using properties of radicals:
            
             2 + √6 + √6 + 3 = p²/q²

Combining like terms:

                     5 + 2√6 = p²/q²

Subtracting 5 from both sides

                         2√6 = p²/q² - 5

Both sides are rational because the difference of two rational
numbers is rational
                         
Multiply both sides by rational number 1/2

                          √6 = (p²/q² - 5)/2

Both sides are rational since the product of two rational
numbers is rational.

                           √6 is rational

We assume √n is irrational whenever n is a positive integer 
that is not a perfect square.

6 is not a perfect square, so √6 is irrational.

So we have reached a contradiction.  Therefore the assumption

                      √2 + √3 is rational

is false.  Therefore √2 + √3 is irrational.

Edwin