- #1

karseme

- 15

- 0

I know how to prove that, for example, \(\displaystyle \sqrt{2} \) is irrational, so I tried to do something similar with this expression.

So, let's assume otherwise, that \(\displaystyle \sqrt{n+\sqrt{n}} \) is not irrational, then it must be rational ie. \(\displaystyle \sqrt{n+\sqrt{n}}\in\mathbb{Q} \). Then we can write \(\displaystyle \sqrt{n+\sqrt{n}}=\dfrac{a}{b}, \, a\in\mathbb{Z}, b\in\mathbb{N} \). After squaring the expression, we have: \(\displaystyle \sqrt{n}=\dfrac{a^2}{b^2}-n \). Since \(\displaystyle \mathbb{N}\subset\mathbb{Q} \), then also \(\displaystyle n\in\mathbb{Q} \). And since there is closure of subtraction in \(\displaystyle \mathbb{Q} \), then \(\displaystyle \dfrac{a^2}{b^2}-n\in\mathbb{Q} \). So it is obvious that it must be \(\displaystyle \sqrt{n}\in\mathbb{Q} \). If n is not a perfect square then it is obvious that

\(\displaystyle \sqrt{n}\notin\mathbb{Q} \), so we have a contradiction. Then, let's assume that n is a perfect square then \(\displaystyle \sqrt{n}\in\mathbb{Q} \), and we can write \(\displaystyle n=k^2, \, k\in\mathbb{Q} \). Let's also assume that k is not a perfect square, then we have: \(\displaystyle \dfrac{a}{b}=\pm\sqrt{k(k+1)} \). But we assumed that k is not a perfect square so \(\displaystyle \sqrt{k} \) is irrational and therefore \(\displaystyle \dfrac{a}{b} \) is irrational which is in contradiction with assumption that \(\displaystyle \dfrac{a}{b} \) is rational. But also, if k is a perfect square, then it is obvious that \(\displaystyle k+1 \) is not a perfect square, so \(\displaystyle \sqrt{k+1} \) is irrational. Then again, \(\displaystyle \dfrac{a}{b} \) is irrational which is in contradiction with assumption that \(\displaystyle \dfrac{a}{b} \) is rational. So, \(\displaystyle \sqrt{n+\sqrt{n}} \) can't be rational, it must be irrational.