Originally Posted by

**Matt Westwood** In this context I don't believe they do - but it has not been specified in what set they are allowed to be. In this context it ought to be $\displaystyle \mathbb Q$ (the rationals).

So I would say that for b the sup is 1 and inf is 0 as you suggest.

For the sup, set m=0 and let n be anything you like, you've got 1 and nothing you can do can make it bigger.

Second, set n = 0 and let m = anything you like, you've got 0.

Are you using the convention that $\displaystyle \mathbb N = \{1, 2, 3, ...\}$ or $\displaystyle \mathbb N = \{0, 1, 2, ...\}$? In the first case the answers will be the same but you won't have 0 and 1 actually in the sets described.