For which natural numbers $n$ is $\sqrt n$ irrational? How would you prove your answer? [duplicate]

Solution 1:

In fact, every integer which is not a square has an irrational square root. Can you see how to generalize the result for $\sqrt{2}$ to prove this?

Solution 2:

Once again, here is a proof that if $n$ is a positive integer that is not a square of an integer, then $\sqrt{n}$ is irrational.

Let $k$ be such that $k^2 < n < (k+1)^2$. Suppose $\sqrt{n}$ is rational. Then there is a smallest positive integer $q$ such that $\sqrt{n} = p/q$.

Then $\sqrt{n} = \sqrt{n}\frac{\sqrt{n}-k}{\sqrt{n}-k} = \frac{n-k\sqrt{n}}{\sqrt{n}-k} = \frac{n-kp/q}{p/q-k} = \frac{nq-kp}{p-kq} $.

Since $k < \sqrt{n} < k+1$, $k < p/q < k+1$, or $kq < p < (k+1)q$, so $0 < p-kq < q$. We have thus found a representation of $\sqrt{n}$ with a smaller denominator, which contradicts the specification of $q$.

Note: This is certainly not original - but I had fun working it out based on the proof I know that $\sqrt{2}$ is irrational.

Note 2: It is interesting that this does not use any divisibility properties.