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Math Help - rational numbers and order relations

  1. #1
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    Lightbulb 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
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  2. #2
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    Quote Originally Posted by peguindollar View Post
    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.
    \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...
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  3. #3
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    Quote Originally Posted by peguindollar View Post
    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 b>a we say that b-a is positive. But if c is positive then c(b-a) = bc - ac is positive (by the properties of order on the rationals). This means that bc > ac by definition.
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  4. #4
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    Ordering relations

    Hi -

    Quote Originally Posted by peguindollar View Post
    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:

    a<0 and b<0

    \implies a<0 and -b>0

    \implies a\times (-b) < 0 \times (-b) (using Lemma 1, replacing b by 0 and c by -b)

    \implies -ab <0

    \implies ab >0

    Grandad
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