A Proof that Square Root of 2 Plus Square Root of 3 is Irrational


In the previous post, we have learned that \sqrt{6} is irrational.  In this post, we will use that theorem to show that \sqrt{2} + \sqrt{3} is also irrational. This proof was explained by Ivan Niven in his book Numbers: Rational and Irrational.

A common argument that might arise is that since \sqrt{2} is irrational and \sqrt{3} is irrational, then their sum is irrational. Like multiplication, the set of irrational numbers is not closed under addition. The sum of two irrational numbers is not necessarily irrational. For instance, \sqrt{2} is irrational, -\sqrt{2} is irrational, but \sqrt{2} + (-\sqrt{2}) = 0 is rational.


\sqrt{2} + \sqrt{3} is irrational. 


Just like the previous post, we use proof by contradiction to prove the theorem above.

Suppose \sqrt{2} + \sqrt{3} is rational, say r so that

\sqrt{2} + \sqrt{3} = r.

Squaring both sides, we have 2 + 2 \sqrt{6} + 3 = r^2 which means that \sqrt{6} = r^2 - 5.

Since the set of rational numbers is closed under multiplication and addition, r^2 - 5 is therefore rational. However, as we have proved in the previous post, \sqrt{6} is irrational. A contradiction!

Therefore, \sqrt{2} + \sqrt{3} is irrational.

2 thoughts on “A Proof that Square Root of 2 Plus Square Root of 3 is Irrational

  1. It should be 2sqrt(6) = r^2 – 5. I think you forgot the 2 would need to be divided out so sqrt(6) = (r^2 – 5) / 2.

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