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Help with algebraic manipulation of an inequality

  1. Oct 3, 2008 #1
    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.
     
  2. jcsd
  3. Oct 3, 2008 #2
    ok , here it is what i think


    since y>x=> y-x>0, now since 0<t<1, we have

    0<t(y-x)<y-x add an x on both sides and we get

    x<t(y-x)+x<y-x+x

    x<ty-tx+x<y

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

    what we actually need to show
     
  4. Oct 3, 2008 #3

    Dick

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    If x<y then xt+x(1-t)<yt+x(1-t)<yt+y(1-t).
     
  5. Oct 3, 2008 #4
    thanks for all the replies. i will try all these and reply back when i've gotten more work done on my own.

    thank you very much!
     
  6. Oct 5, 2008 #5
    Awesome! Thank you very much.:smile: That was very clear. I totally understood it and get it now! :cool:
     
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