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Trying to prove inequality

  1. Nov 27, 2016 #1
    1. The problem statement, all variables and given/known data
    Prove the following facts about inequalities. [In each problem you will have to consider several
    cases separately, e.g. ##a > 0## and ##a = 0##.]
    (a) If ##a \leq b##, then ##a + c \leq b + c##.
    (b) If ##a \geq b##, then ##a + c \geq b + c##.
    (c) If ##a \leq b## and ##c \geq 0##, then ##ac \leq bc##.
    (d) If ##a \leq b## and ##c \leq 0##, then ##ac \geq bc##.

    2. Relevant equations


    3. The attempt at a solution
    So, I've tried to prove the first one (the second is basically the first one but with inequalities inverted) the following way:
    If ##a > 0, b > 0## or ##a < 0, b < 0## (do I really have to say that ##b> 0## too?)
    ##a \leq b \implies a + c \leq b + c##
    ##a + (c + (-c)) \leq b + (c + (-c))##
    ##(a + c) + (-c) \leq (b + c) + (-c)##
    ##a + c \leq b + c##
    If ##a = 0, b > 0##
    ##0 \leq b \implies c \leq b + c##
    ##(c + (-c)) \leq b + (c + (-c))##
    ##c + (-c) \leq (b + c) + (-c)##
    ##c \leq b + c##
    So it's proved, right?
    Now, going to (c), we have this ##c \geq 0## that stops me from using ##c^{-1}## to prove them because it's not ##c > 0##.
    Which other way can I prove the last two?

    EDIT: Sorry, I didn't notice the inequalities were wrong.
     
    Last edited: Nov 27, 2016
  2. jcsd
  3. Nov 27, 2016 #2

    Ray Vickson

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    Looking at several cases for each part is a waste of time; in every part there is no need at all for "cases". Just use an appropriate definition of "≤" or "≥".
     
  4. Nov 27, 2016 #3
    An appropriate definition of ##\leq## and ##\geq##? Wouldn't that be:
    ##a \leq b## means ##a < b## or ## a = b##
    ##a \geq b## means ##a > b## or ## a = b##
     
  5. Nov 27, 2016 #4

    Ray Vickson

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    What does ##a < b## mean? What is a very simple way to test if ##a < b##?

    Anyway, I did not claim that what you are doing is wrong; I am just saying it is not the fastest way.

    To deal with your original question: in your argument you started with ##a \leq b##, then got ##(a+c) + (-c) \leq (b+c) + (-c)##, then cancelled the "##(-c)##' on both sides to end up with ##a+c \leq b+c##. However, that last step is invalid: it basically assumes what you are trying to prove, because if you set ##A = a+c## and ##B = b+c## you are going from ##A -c \leq B-c ## to ##A \leq B##---but you have not proved that yet.
     
    Last edited: Nov 27, 2016
  6. Nov 27, 2016 #5
    ##a < b## means that ##a## is a number smaller than ##b##.
    To see if ##b - a## is still a positive number? If it's negative it means that actually ##a > b##.
     
  7. Nov 27, 2016 #6

    Ray Vickson

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    Now you are getting it (in your second sentence above).
     
  8. Nov 27, 2016 #7
    Oh right. I can't think of any way to proceed from that part right now.

    So I should simply see if ##(b + c) - (a + c) > 0##?
     
  9. Nov 27, 2016 #8

    Ray Vickson

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    That is the way I would do it (although I would use "##\geq##" instead of "##>##").
     
  10. Nov 27, 2016 #9
    Oh yes.
    Doing like that I would end up with ##b + c - a - c \geq 0## and so ##b - a \geq 0##. But does this really prove what I was trying to prove? Isn't this the opposite way since I started directly with ##a + c \leq b + c##, which was what I had to end up while proving?
     
  11. Nov 27, 2016 #10

    Ray Vickson

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    No: you assume ##a \leq b##, which means that ##b-a \geq 0##. However, ##(b+c) - (a+c) = b-a,## so we have ##(b+c) -(a+c) \geq 0##, which means (by definition) that ##b+c \geq a+c##. No circular arguments are involved; just elementary algebra.
     
  12. Nov 27, 2016 #11
    Oh, now I get it.
    Thank you very much.
     
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