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
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...