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Two problems from chapter 3 of How to Prove It I just cannot get my head around

  1. Aug 3, 2011 #1
    Two problems from chapter 3 of "How to Prove It" I just cannot get my head around

    The thing I hate about proving things is that I can always think it through in my head, but don't know how to express it mathematically for a sound proof.

    1. The problem statement, all variables and given/known data

    7) Suppose a, b, c, and d are real numbers, 0 < a < b, and d > 0. Prove that if ac >= bd then c > d.

    8) Suppose x and y are real numbers, and 3x + 2y >= 5. Prove that if x > 1 then y < 1.


    2. Relevant equations

    none.


    3. The attempt at a solution

    7) We will prove the contrapositive. Suppose d > c. Thus, multiplying the inequality by any constant a will result in ad > ac. This is where I get stuck - something so simple, but I can't think of a sound way to prove it.

    8) Suppose x > 1. Thus, 3x > 3. Then, by the transitive property (?) 3x > 5 - 2y. Therefore, 5 - 2y < 3. Subtracting 5 from both sides yields -2y < -2. This is equivalent to y < 1. Therefore, assuming 3x >= 5 - 2y, if x > 1 then y < 1. But this is not sound, because there is no transitive property?

    Thanks

    PS. This actually isn't a homework question.. but it's in the form of one so I figured put it here.
     
    Last edited: Aug 3, 2011
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  3. Aug 3, 2011 #2

    PeterO

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    Homework Helper

    Re: Two problems from chapter 3 of "How to Prove It" I just cannot get my head around

    Taking the first example, we know a,b and d are positive - we were told so.

    if ac >= bd we could divide both sides by a to get c >=bd/a = d. b/a

    [Note if b was negative, we would have to have reversed the inequality]

    Second,

    Tranpose to make y the subject and see what happens.
     
  4. Aug 4, 2011 #3
    Re: Two problems from chapter 3 of "How to Prove It" I just cannot get my head around

    I assume you mean:

    if ac >= bd we can divide both sides by a to get c >= bd/a which equals dc >= b / a? but then what?

    So, proving the contrapositive with y instead of x? Okay...:

    We will prove the contrapositive. Suppose 3x + 2y <= 5, and y > 1. Thus, 2y > 2. Returning to the original inequality, this means that 3x < 3, thus x < 1. Therefore, if 3x + 2y >= 5 and x > 1, then y < 1.

    I hope I understood you correctly... thanks for all your help :)
     
  5. Aug 7, 2011 #4

    NascentOxygen

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    Re: Two problems from chapter 3 of "How to Prove It" I just cannot get my head around


    I don't like your working. I think the answer in the book is wrong. :cry:

    Given 3x + 2y >= 5
    Rearrange as: 3x - 5 >= -2y

    Given x > 1, then LHS is always greater than -2

    so the inequality becomes: -2 > -2y

    Now, divide both sides by -2 and reverse the inequality,
    (OR, instead, if you prefer, add 2 + 2y to both sides)

    and we end up with a solution different from what you were aiming for.

    That probably explains why you were having trouble getting the right answer. :smile:
     
  6. Aug 8, 2011 #5

    PeterO

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    Re: Two problems from chapter 3 of "How to Prove It" I just cannot get my head around

    I am not sure why you did what you did??????

    Have you forgotten you were trying to prove that c > d ??

    b/a >1, since 0 < a < b

    So what do you think the line c>= d (b/a) tells you??
     
  7. Aug 8, 2011 #6

    PeterO

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    Re: Two problems from chapter 3 of "How to Prove It" I just cannot get my head around

    WHy all this contra positive stuff?????

    Transpose to make y the subject.

    The subject is on the left hand side

    y is to be the subject. Not 2y, and certainly not -2y

    Once you have transposed it will either read y <= ....... or y >= ......

    by the way, are x and y restricted to integers/whole numbers? or might x > 1 mean x = 1.15 is a possible value?
     
  8. Aug 9, 2011 #7
    Re: Two problems from chapter 3 of "How to Prove It" I just cannot get my head around

    I'm going to differ from the other responders and say that I think your approach to 7) is fine. Proving the contrapositive seems like the natural approach to me as well; it is certainly simpler than a direct proof.

    This is almost sound; there is only one mistake so far. What is the contrapositive of "ac >= bd implies c > d?" (Hint: Be very careful of > versus >= signs. These will matter in the proof.)

    For the next step, consider the relationship between bd and ad. How does that fit into the chain of inequalities?

    It should always be sound to add or subtract something from both sides of an inequality. Try doing this to isolate the "2y" term, and then build a chain of inequalities.
     
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