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

Printable View

- December 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 - December 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.

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... - December 18th 2008, 08:55 AMThePerfectHacker
- December 20th 2008, 07:26 AMGrandadOrdering relations