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Math Help - proof of existence of z in x < z < y

  1. #1
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    proof of existence of z in x < z < y

    If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying x < z < y.

    At my disposal, the least-upper-bound axiom - every nonempty set S of reals bounded above has a supremum. And a handful of theorems including x > sup S - h where h > 0 and set S has a supremum.

    Is my proof OK?

    Let the set S be bounded above by y, so y = sup S. Let x \in S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

    Thanks,
    Scott
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  2. #2
    Senior Member Sampras's Avatar
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    Why does the fact that  S is bounded above by  y imply that  y = \sup S ?

    The least upper bound property says the following: Suppose  E \subset S where  S is ordered,  E \neq \emptyset , and  E is bounded above. Then  \sup E exists in  S .
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  3. #3
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    Quote Originally Posted by Sampras View Post
    Why does the fact that  S is bounded above by  y imply that  y = \sup S ?
    A little loose with my wording... you're right, it doesn't.

    How's this, instead?

    Let y be the least upper bound of set S, so y = sup S. Let x \in S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

    Scott
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  4. #4
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    Quote Originally Posted by ScottO View Post
    If x and y are arbitrary real numbers with x < y, prove that there is at least one real z satisfying x < z < y.

    At my disposal, the least-upper-bound axiom - every nonempty set S of reals bounded above has a supremum. And a handful of theorems including x > sup S - h where h > 0 and set S has a supremum.

    Is my proof OK?

    Let the set S be bounded above by y, so y = sup S. Let x \in S and x + h = y = sup S. Then by the above theorem, there exists a z > sup S - h, therefore sup S = y > z > sup S - h = x. Q.E.D.

    Thanks,
    Scott
    you want to show that real Nos are dense. This you can show in the following way:

    Let x<y, then by adding x to both sides you get :

    2x<y+x............................................ .......................................1

    by adding y to both sides you get:

    x+y<2y............................................ ........................................2

    from (1) AND (2) we have :


    x<(x+y)/2 <y ,so there exists z,such that x<z<y and z=(x+y)/2
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