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

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- Dec 3rd 2008, 02:52 PMpeguindollarrational 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 - Dec 17th 2008, 10:49 PMlarley
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... - Dec 18th 2008, 08:55 AMThePerfectHacker
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.

- Dec 20th 2008, 07:26 AMGrandadOrdering relations
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