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.
$\displaystyle \frac{ac}{c} > \frac{bc}{c}$
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...
I do the first one. When we say $\displaystyle b>a$ we say that $\displaystyle b-a$ is positive. But if $\displaystyle c$ is positive then $\displaystyle c(b-a) = bc - ac$ is positive (by the properties of order on the rationals). This means that $\displaystyle bc > ac$ by definition.
Hi -
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:
$\displaystyle a<0$ and $\displaystyle b<0$
$\displaystyle \implies a<0$ and $\displaystyle -b>0$
$\displaystyle \implies a\times (-b) < 0 \times (-b)$ (using Lemma 1, replacing $\displaystyle b$ by $\displaystyle 0$ and $\displaystyle c$ by $\displaystyle -b$)
$\displaystyle \implies -ab <0$
$\displaystyle \implies ab >0$
Grandad