More rational/irrational junk

• Sep 25th 2008, 12:41 PM
Sinuyen
More rational/irrational junk
Here we go again:

If a,b are real numbers with a < b prove:

1) There is a rational r within (a,b).
I am given the hint to start by selecting an interger q > 0 such that
1/q < b-a. I am not sure how it applies.
I am trying to go the route a < q < b and somehow show that q is a rational.

2) There is a irrational c within (a,b).
Says to look at part 1)

3) (a,b) contains infinitely many rationals and irrationals.

Seems like once it is shown to me how to do it, I can grasp the concept, but it seems kind of difficult to assign homework that is not lectured on how to do. Oh well, thanks in advance MHF people. Off to class.
• Sep 26th 2008, 04:04 PM
kalagota
Quote:

Originally Posted by Sinuyen
Here we go again:

If a,b are real numbers with a < b prove:

1) There is a rational r within (a,b).
I am given the hint to start by selecting an interger q > 0 such that
1/q < b-a. I am not sure how it applies.
I am trying to go the route a < q < b and somehow show that q is a rational.

2) There is a irrational c within (a,b).
Says to look at part 1)

3) (a,b) contains infinitely many rationals and irrationals.

Seems like once it is shown to me how to do it, I can grasp the concept, but it seems kind of difficult to assign homework that is not lectured on how to do. Oh well, thanks in advance MHF people. Off to class.

1) WLOG, let $\displaystyle 0<a<b$. From the hint, it follows that $\displaystyle 1 < qb-qa$. then $\displaystyle qa+1 < qb$

$\displaystyle qa \in \mathbb{R}^+$. thus, there is an integer $\displaystyle m$ such that $\displaystyle m-1 \leq qa < m$

in particular, $\displaystyle m\leq qa+1 < qb$. from here, $\displaystyle qa < m < qb$ implying that $\displaystyle a < \frac{m}{q} < b$

for other items, use density property..