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?

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