# Greatest lower bound/least upper bound in Q

1. Nov 11, 2011

### autre

I have the following question:

Let $n\in\mathbb{Z}^{+}$ st. $n$ is not a perfect square. Let $A=\{x\in\mathbb{Q}|x^{2}<n\}$. Show that $A$ is bounded in $\mathbb{Q}$ but has neither a greatest lower bound or a least upper bound in $\mathbb{Q}$.

To show that $A$ is bounded in $\mathbb{Q}$ I have to show that it has a infimum and a supremum in $\mathbb{Q}$, right? Not sure where to start...

2. Nov 11, 2011

### Bacle2

No, only thing you need to show boundedness is that there are rationals q,q' , such that

q2<n and n<q'2. For the rest, I'm not sure what material you're familiar with. If you know the proof of the irrationality of √2 , you can show that this extends to the irrationality of √n , when n is not a perfect square.