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

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 is irrational, under the assumption that is. The solution is as follows: Assume to the contrary, that is rational and so can be written as for two non-zero integers and . We then have, upon squaring both sides and manipulating, that

That is, we have expressed as the quotient of two integers and so must be rational. This is a contradiction, and so we can conclude that our assumption was incorrect and thus 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 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 is indeed irrational.

As usual, assume to the contrary, that is rational. Squaring and rearranging gives that is also rational. We save this for now.

Under our assumption, must also be rational. We multiply the top and bottom of this fraction by to get that

is also rational. Squaring this we conclude that is also rational. Thus adding this to should yield a rational number. Instead we get , and this is our contradiction.

Anybody know another proof?

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