# Thread: [SOLVED] Stumped - Help get me started

1. ## [SOLVED] Stumped - Help get me started

Ok, so I have the following to prove:

If a and b are real numbers s.t. 0<a<b, then there is a natural number n s.t. a<b-1/n<b.

I am confused as to what I should do to prove this. Is this a two part proof? If so, what parts do I prove? Thanks.

2. Originally Posted by spearfish
Ok, so I have the following to prove:

If a and b are real numbers s.t. 0<a<b, then there is a natural number n s.t. a<b-1/n<b.

I am confused as to what I should do to prove this. Is this a two part proof? If so, what parts do I prove? Thanks.
Consider the set of natural numbers

$\left\{n \in \mathbb{N}: n > \frac{1}{b-a} \right\}$ This is a non-empty set of natural numbers that is bounded from below.

So by the well ordering of the postive integers this set has a least element $n_0$

Obviously $b > b-\frac{1}{n_0}$

Now we just need to show that $b-\frac{1}{n_0}> a$

we know that

$n_0> \frac{1}{b-a} \iff b-a > \frac{1}{n_0} \iff b-\frac{1}{n_0}> a$

Putting the above two inequalities together we get

$a < b-\frac{1}{n_0}< b$

3. This makes sense. Thanks a bunch!