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Math Help - [SOLVED] Help with algebraic manipulation of an inequality

  1. #1
    Member ilikedmath's Avatar
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    Question [SOLVED] Help with algebraic manipulation of an inequality

    Let R denote the set of all real numbers and Q the set of all rational numbers.


    Statement to prove:
    If x and y are in R with x < y, show that x < ty + (1-t)x < y for all t, 0 < t < 1.

    My work on the proof so far:

    Given x and y are real numbers with x < y. By theorem we know there exists an r in Q such that x < r < y. Take r = t. So x < t < y.


    That's all I have so far.


    My professor said this proof was more of algebraic manipulation. I am stuck as to how I can algebraically manipulate the inequality to get to x < ty + (1-t)x < y.
    Last edited by ilikedmath; October 3rd 2008 at 01:51 PM. Reason: Typo
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  2. #2
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    Given x < y:

    (1-t)x < (1-t)y

    tx < ty

    x = tx + (1-t)x < ty + (1-t)x

    y = ty + (1-t)y > ty + (1-t)x
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  3. #3
    Member ilikedmath's Avatar
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    Clarification

    Given x < y:

    (1-t)x < (1-t)y -- I see here you just multiplied both sides by (1 - t)

    tx < ty -- Multiplying through I get x - tx < y - ty. I don't 'see' how you got to this part tx < ty.
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  4. #4
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    Quote Originally Posted by ilikedmath View Post
    Given x < y:

    (1-t)x < (1-t)y -- I see here you just multiplied both sides by (1 - t)

    tx < ty -- Multiplying through I get x - tx < y - ty. I don't 'see' how you got to this part tx < ty.
    He's used the property that x<y and multiplied through by t.
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  5. #5
    Member ilikedmath's Avatar
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    I think I got it! :)

    Thanks to everyone's help, I think I got it:

    My final proof:
    Let x, y be in the reals with x , y; and t in the reals with 0 < t < 1.
    Since y > x, that implies y - x > 0. And since 0 < t < 1, we have
    0 < t(y-x) < y - x. Add an x to both sides to get
    x < t(y-x) + x < y - x + x
    x < ty - tx + x < y
    x < ty + (1 - t)x < y.
    QED.
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  6. #6
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    I like that proof. Very elegant, given the scope of the problem.
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