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Thread: 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 $\displaystyle a<b \implies a-\sqrt{2} < b-\sqrt{2}$ so there exists rational $\displaystyle x$ so that $\displaystyle a-\sqrt{2} < x < b-\sqrt{2}$ thus $\displaystyle a < x +\sqrt{2} < b$ that means $\displaystyle x+\sqrt{2}$ is irrational. Q.E.D.

    2. Prove that if a > 0, there is an integer n such that 1/n < a < n.
    Given $\displaystyle a>0,1>0$ by Archimedean ordering there is nartual $\displaystyle n_1$ so that $\displaystyle a< n_1\cdot 1 = n_1$. Since $\displaystyle 1/a>0,1>0$ by Archimedean ordering there is natural $\displaystyle n_2$ so that $\displaystyle 1/a < n_2\cdot 1 = n_2$ thus $\displaystyle 1/n_2 < a$. Let $\displaystyle n=\max\{n_1,n_2\}$ then $\displaystyle 1/n \leq 1/n_2 < a < n_1 \leq n$. 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 $\displaystyle a > b$ that means $\displaystyle a-b>0$ (by definition). Thus by Archimedian ordering with $\displaystyle 1/(a-b) > 0,1>0$ there is natural $\displaystyle n$ so that $\displaystyle 1/(a-b) < n$ thus $\displaystyle a-b > 1/n \implies a > b+1/n$ 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) $\displaystyle S_a$ has an upper bound, i.e. $\displaystyle a$. Argue that this is the least upper bound.
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