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Math Help - analysis midterm

  1. #1
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    analysis midterm

    Sorry for all the questions, but I have a midterm coming up and I'm having problems with these questions:

    1. Prove that if a < b, then there is an irrational number x such that
    a < x < b. Hint: first show that if q not equal to zero is rational, then q+ root(2) and q*root(2) are irrational.

    2. Prove that if a > 0, there is an integer n such that 1/n < a < n.

    3. For real numbers a and b, prove that if a <= b + 1/n for all positive integers n, then a <= b.

    4. For each real number a, let the set S_a be the set of all rational numbers less than a. Prove that for each real number a, sup S_a = a.

    Thanks.
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  2. #2
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    Quote Originally Posted by eigenvector11 View Post
    1. Prove that if a < b, then there is an irrational number x such that
    a < x < b. Hint: first show that if q not equal to zero is rational, then q+ root(2) and q*root(2) are irrational.
    If a<b \implies a-\sqrt{2} < b-\sqrt{2} so there exists rational x so that a-\sqrt{2} < x < b-\sqrt{2} thus a < x +\sqrt{2} < b that means x+\sqrt{2} is irrational. Q.E.D.

    2. Prove that if a > 0, there is an integer n such that 1/n < a < n.
    Given a>0,1>0 by Archimedean ordering there is nartual n_1 so that a< n_1\cdot 1 = n_1. Since 1/a>0,1>0 by Archimedean ordering there is natural n_2 so that 1/a < n_2\cdot 1 = n_2 thus 1/n_2 < a. Let n=\max\{n_1,n_2\} then 1/n \leq 1/n_2 < a < n_1 \leq n. Q.E.D.

    3. For real numbers a and b, prove that if a <= b + 1/n for all positive integers n, then a <= b.
    Assume not, then a > b that means a-b>0 (by definition). Thus by Archimedian ordering with 1/(a-b) > 0,1>0 there is natural n so that 1/(a-b) < n thus a-b > 1/n \implies a > b+1/n which is a contradiction. Q.E.D.

    4. For each real number a, let the set S_a be the set of all rational numbers less than a. Prove that for each real number a, sup S_a = a.
    The set (non-empty) S_a has an upper bound, i.e. a. Argue that this is the least upper bound.
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