# An Irrational Sum

I spent two straight days last week marking midsemester exams with my fellow tutors and the two lecturers for MATH1115. The exam contained a couple of questions on performing operations with irrational numbers, such as “is the sum of two irrational numbers an irrational number?” The answer here is a resounding ‘not always’, the common counterexample being

$\sqrt{2} + (-\sqrt{2}) = 0.$

Of course, sometimes we do have that the sum of two irrational numbers is irrational. One of the exercises in the first assignment had the students prove that $\sqrt{2}+\sqrt{5}$ is irrational, under the assumption that $\sqrt{10}$ is. The solution is as follows: Assume to the contrary, that $\sqrt{2}+\sqrt{5}$ is rational and so can be written as $a/b$ for two non-zero integers $a$ and $b$. We then have, upon squaring both sides and manipulating, that

$\sqrt{10} = \frac{a^2-7b^2}{2b^2}.$

That is, we have expressed $\sqrt{10}$ as the quotient of two integers and so $\sqrt{10}$ must be rational. This is a contradiction, and so we can conclude that our assumption was incorrect and thus $\sqrt{2}+\sqrt{5}$ is irrational.

Halfway through the second day, whilst marking the irrational numbers section of the exam no doubt, one of the tutors proposed a question to the group: Is $\sqrt{2} +\sqrt{3}+\sqrt{5}$ irrational?

The answer given by the group was ‘most probably’. This was followed by a brainstorming session on how to prove such a result using only elementary methods. After twenty minutes of ‘fiddling around’ on the board with no success, we all went back to marking.

I don’t doubt that the other people in the room did something much more interesting with their Sunday night, but I managed to knock out a proof that $\sqrt{2} +\sqrt{3}+\sqrt{5}$ is indeed irrational.

As usual, assume to the contrary, that $\sqrt{2} +\sqrt{3}+\sqrt{5}$ is rational. Squaring and rearranging gives that $\sqrt{6} +\sqrt{10}+\sqrt{15}$ is also rational. We save this for now.

Under our assumption, $1/(\sqrt{2} +\sqrt{3}+\sqrt{5})$ must also be rational. We multiply the top and bottom of this fraction by  $\sqrt{2}+\sqrt{3}-\sqrt{5}$ to get that

$(\sqrt{2}+\sqrt{3}-\sqrt{5})/(2 \sqrt{6})$

is also rational. Squaring this we conclude that $\sqrt{6}-\sqrt{10}-\sqrt{15}$ is also rational. Thus adding this to $\sqrt{6} +\sqrt{10}+\sqrt{15}$ should yield a rational number. Instead we get $2 \sqrt{6}$, and this is our contradiction.

Anybody know another proof?