An old (math) chestnut

Theorem: \sqrt[n]{2} is irrational for any integer n \geq 3.

Proof: Suppose the number is rational, and let \sqrt[n]{2} = a/b where a and b are integers. Then 2 = (a/b)^n, which can be written as

b^n + b^n = a^n.

This contradicts the Fermat’s Last Theorem. QED

________________________________

ADDENDUM: This is a perfectly correct proof, although it is akin to killing a mosquito using a bazooka.

Which reminds me of a story that happened 12 years ago. I was in Form 5 and had been called to attend the IMO selection camp at UPM. At the time, I knew much less mathematics compared to the other students, so I wasn’t so sure whether I could make the team.

During an introductory lecture, Prof. Abu Osman gave us this problem:

Given positive integers a, b, c, n \geq 2 such that a^n + b^n = c^n. Prove that a > n.

Being a wiseass, I immediately shouted out, “Solved it! Using the FLT we must have n=2 and we know that the smallest Pythagorean triple is (3,4,5) so we’re done.”

One of the students in the camp quickly shot me down, and explained that we were not supposed to use any advanced results since we were doing high school contest problems. I protested, “but the FLT is true, isn’t it? Tanya prof kalau tak percaya.” He countered, “that is probably not the solution the author had in mind.” I then turned away and muttered in disgust, “who the hell cares what the author had in mind…” .

The intended solution goes like this: We can assume that a \leq b. From the equation, we have

a^n = c^n - b^n = (c-b)(c^{n-1} + c^{n-2}b + \dotsb + b^{n-1}).

Since c>b\geq a, we have c-b \geq 1, and the second bracket is > na^{n-1}. Therefore, a^n > na^{n-1}, and so a>n.

4 Responses to An old (math) chestnut

  1. Adeeb says:

    nice! =)

  2. Adeeb says:

    Sir, can you explain more on the last paragraph of the proof. Because I cant understand why the second bracket is >na^(n-1)?

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