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Math Help - proving rational number inequalities

  1. #1
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    proving rational number inequalities

    Suppose a,b,x, and y are positive real numbers with a/b < x/y.
    Explain why a/b < (a+x)/(b+y) < x/y.
    I guess I know the fact is true however how can I prove this?
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  2. #2
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    Quote Originally Posted by mathsohard View Post
    Suppose a,b,x, and y are positive real numbers with a/b < x/y.
    Explain why a/b < (a+x)/(b+y) < x/y.
    I guess I know the fact is true however how can I prove this?
    From the given if follows at once that
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  3. #3
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by mathsohard View Post
    Suppose a,b,x, and y are positive real numbers with a/b < x/y.
    Explain why a/b < (a+x)/(b+y) < x/y.
    I guess I know the fact is true however how can I prove this?
    a/b < x/y
    so ay < bx (since y and b are positive)
    so ab+ay < ab + bx
    so a(b+y) < b(a+x)
    so a/b < (a+x)/(b+y)

    EDIT: sorry didn't see Plato's post.
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  4. #4
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    I would have been inclined to do an indirect proof:
    Suppose a/b < (a+x)/(b+y) < x/y is NOT true. Then either a/b>= (a+ x)/(b+y) or (a+ x)/(b+y)>= x/y.

    If a/b>= (a+ x)/(b+ y) then a(b+y)>= b(a+ x) which is the same as ab+ ay>= ab+ bx and so ay>= bx which (assuming b and y are non-zero which is implied by the hypothesis) a/b>= x/y, contradicting the hypothesis.

    Similarly, if (a+ x)/(b+ y)>= x/y, then y(a+ x)>= x(b+y) which is the same as ay+ xy>= bx+ xy and so ay>= bx again which gives the same contradiction.
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