Question 123519
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 irrational.
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For contradiction, assume &#8730;2 + &#8730;3 is rational

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

            &#8730;2 + &#8730;3 = p/q 

Square both sides:

                  (&#8730;2 + &#8730;3)² = p²/q²

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

          (&#8730;2 + &#8730;3)(&#8730;2 + &#8730;3) = p²/q²

Use FOIL

   &#8730;2&#8730;2 + &#8730;2&#8730;3 + &#8730;3&#8730;2 + &#8730;3&#8730;3 = p²/q²        

Using properties of radicals:
            
             2 + &#8730;6 + &#8730;6 + 3 = p²/q²

Combining like terms:

                     5 + 2&#8730;6 = p²/q²

Subtracting 5 from both sides

                         2&#8730;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

                          &#8730;6 = (p²/q² - 5)/2

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

                           &#8730;6 is rational

We assume &#8730;n is irrational whenever n is a positive integer 
that is not a perfect square.

6 is not a perfect square, so &#8730;6 is irrational.

So we have reached a contradiction.  Therefore the assumption

                      &#8730;2 + &#8730;3 is rational

is false.  Therefore &#8730;2 + &#8730;3 is irrational.

Edwin</pre>