Tuesday, April 12, 2016

Prove that if n is not the square of a natural number, then sqrtn is irrational.











I have this homework problem that I can't seem to be able to figure out:

Prove: If nN is not the square of some other mN, then n must be irrational.

I know that a number being irrational means that it cannot be written in the form ab:a,bN b0 (in this case, ordinarily it'd be aZ, bZ{0}) but how would I go about proving this? Would a proof by contradiction work here?



Thanks!!


Answer




Let n be a positive integer such that there is no m such that n=m2. Suppose n is rational. Then there exists p and q with no common factor (beside 1) such that



n=pq



Then



n=p2q2.



However, n is an positive integer and p and q have no common factors beside 1. So q=1. This gives that




n=p2



Contradiction since it was assumed that nm2 for any m.


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