1. ## infimums and supremums

Find and Prove the inf and sup:

{x is a rational number: x^2 < 2 }

2. If you were taking x in the real numbers, then your supremum would be $\sqrt 2$ and your infimum would be $- \sqrt 2$.

However, these are both irrational numbers. There does not exist a supremum (least upper bound) in the rational numbers. Say it did exist, and call it a. Then ${a^2} > 2$. Define $b = \frac{{2a + 2}}{{a + 2}}$. Then b is also an upper bound of your set, and $b < a$, which contradicts our assumption that a was our LEAST upper bound.

A similar argument works for the infimum.