Greatest lower bound/least upper bound in Q

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Main Question or Discussion Point

I have the following question:

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

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

Answers and Replies

  • #2
Bacle2
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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.
 

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