Thread: 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

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.

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...

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 we say that is positive. But if is positive then is positive (by the properties of order on the rationals). This means that by definition.

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:

and

and

(using Lemma 1, replacing by and by )





Grandad