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.
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
$\displaystyle \left\{n \in \mathbb{N}: n > \frac{1}{b-a} \right\}$ 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 $\displaystyle n_0 $
Obviously $\displaystyle b > b-\frac{1}{n_0}$
Now we just need to show that $\displaystyle b-\frac{1}{n_0}> a$
we know that
$\displaystyle n_0> \frac{1}{b-a} \iff b-a > \frac{1}{n_0} \iff b-\frac{1}{n_0}> a$
Putting the above two inequalities together we get
$\displaystyle a < b-\frac{1}{n_0}< b$