# Thread: rational numbers and order relations

1. ## rational numbers and order relations

i'm need to prove the following:

for all a,b,c in Q if a<b and c>0 the ac<bc

for all a,b in Q if a<0 and b<0 then ab<0

any help would be much appreciated

2. Originally Posted by peguindollar
i'm need to prove the following:

for all a,b,c in Q if a<b and c>0 the ac<bc

for all a,b in Q if a<0 and b<0 then ab<0

any help would be much appreciated
Well, for the first one, we will assume ac > bc (the opposite). Since c>0, we can divide both sides by c and the greater-than does not become a less than.
$\frac{ac}{c} > \frac{bc}{c}$
which gives us a > b . However, this goes against the previous statement that a < b, which is a contradiction. Therefore, ac < bc.

For the second one it would have to be ab > 0, because two negatives multiplied always make a positive. Not entirely sure how to prove that, though. Seems so elementary...

Hope this helped a bit...

3. Originally Posted by peguindollar
i'm need to prove the following:

for all a,b,c in Q if a<b and c>0 the ac<bc
I do the first one. When we say $b>a$ we say that $b-a$ is positive. But if $c$ is positive then $c(b-a) = bc - ac$ is positive (by the properties of order on the rationals). This means that $bc > ac$ by definition.

4. ## Ordering relations

Hi -

Originally Posted by peguindollar
i'm need to prove the following:

for all a,b,c in Q if a<b and c>0 the ac<bc

for all a,b in Q if a<0 and b<0 then ab<0

any help would be much appreciated

For the second one (which should be, of course, ab > 0), use the result of the first one, which I've called Lemma 1 as follows:

$a<0$ and $b<0$

$\implies a<0$ and $-b>0$

$\implies a\times (-b) < 0 \times (-b)$ (using Lemma 1, replacing $b$ by $0$ and $c$ by $-b$)

$\implies -ab <0$

$\implies ab >0$