Greatest lower bound/least upper bound in Q

[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...

 

[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.