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Math Help - Show there is at least one n in Z

  1. #1
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    Show there is at least one n in Z

    (a) If x-y>1, show that there is at least one n \inZ (integers) such that y<n<x.

    For this should I try showing that there are no n's in Z? Essentially a proof by contradiction.

    (b) if y<x, show there is a rational number z such that y<z<x.

    At first i was thinking the transitive property (i think thats the name of the one i''m thinking of) but I don't seem to confident in it and i'm only recalling about relations like (a,b) (c,d) (a,d) etc.
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  2. #2
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    Quote Originally Posted by tn11631 View Post
    (a) If x-y>1, show that there is at least one n \inZ (integers) such that y<n<x.
    (b) if y<x, show there is a rational number z such that y<z<x.
    For the first one, note that y+1<x.

    The floor function has the property that \left\lfloor y \right\rfloor  \leqslant y < \left\lfloor y \right\rfloor  + 1 \leqslant y + 1


    For the second one you know that \left( {\exists n\in \mathbb{Z}^+ \right)\left( {n > \frac{1}{{x - y}}} \right). WHY is that?

    Notice that means nx-ny>1. Apply the first part to that.
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  3. #3
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    Fundamentally, you are using the "Archimedean principle"- if z is any real number, there exist an integer, N, such that N> z- and the fact that the set of positive integers is "well ordered"- any non-empty set of positive integers contains a smallest integer.

    Certainly if x- y> 1, x- y is positive. Let z= x- y. Then there exist a positive integer N> z= x- y so the set of "all integers N, greater than x- y" is non-empty and so has a smallest integer, n. Since n is the smallest integer in the set, n- 1 is NOT in the set. Consider the two cases:
    1) n- 1= x- y

    2) n-1< x- y
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