For the second step we consider x 0 Choose an integer k such that k x Then we

For the second step we consider x 0 choose an integer

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For the second step, we consider x 0. Choose an integer k such that k > | x | . Then we apply the same argument above to the positive number x + k and y + k to conclude that x + k < q < y + k for some rational number q . Then r := q - k satisfies x < r < y . Corollary 0.4 Let x, y R with x < y . There exists an irrational number w such that x < w < y. Proof: By Theorem 0.3, there exists a rational number r such that x 2 < r < y 2 . It follows that x < r 2 < y , and w := r 2 is irrational. Now we can show Theorem 0.5 Let p be a prime number. Then p R , i.e., there is a positive x R such that x 2 = p . 27
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Proof: Step 1. Define x Let S := { r R | r > 0 , r 2 < p } . The set S is non-empty. Also, since r 2 < p < p 2 , it implies that r < p , i.e., S is bounded above. By the completeness axiom, x := sup ( S ) exists in R . It remains to show that x 2 = p . It suffices to show that neither x 2 < p nor x 2 > p can hold. Step 2. Show x 2 < p cannot hold We want to find a large integer n such that 3 x 2 < x + 1 n 2 < p, i.e., to find an integer n such that x 2 + 2 x n + 1 n 2 < p, i.e., x 2 + 1 n 2 x + 1 n < p, i.e., to find an integer n such that x 2 + 1 n 2 x + 1 < p, i.e., 1 n < p - x 2 2 x + 1 . Here we use p > x 2 to imply p - x 2 2 x +1 > 0. In other words, n > 2 x + 1 p - x 2 . We have found the integer n . Step 3. Show x 2 > p cannot hold We want to find a large integer m such that p < x - 1 m 2 < x 2 , i.e., to find an integer m such that p < x 2 - 2 x m + 1 m 2 < p, 3 If this is true, then sup ( S ) x + 1 n , which is a contradiction with the fact = sup ( S ). 28
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i.e., p < x 2 - 1 m 2 x - 1 m < p, i.e., to find an integer m such that p < x 2 - 1 m 2 x - 1 < p, i.e., 1 m < x 2 - p 2 x - 1 , i.e., m > 2 x - 1 x 2 - p . We have found the integer m . Appendix. Problems did in the class 3.3.1. True or false: (a) If a non-empty subset of R has an upper bound, then it has a least upper bound. (b) If a non-empty subset of R has an infimum, then it is bounded. (c) Every non-empty bounded subset of R has a maximum and a minimum. (d) If m is an upper bound of S and m 0 < m , then m 0 is not an upper bound of S . (e) If m = inf ( S ) and m 0 < m , then m 0 is a lower bound of S . (f) For each real number x and each > 0 , there exists n N such that n > x . Solution: (a) Yes. By the definition of least upper bound. (b) False. For example, S = N . (c) False. For example, S = (0 , 1). (d) False. For example, S = [0 , 2], m = 4 and m 0 = 3. Here m 0 < m but m 0 still is an upper bound. (e) True. By the definition of inf ( S ). (f) True. If x 0, we are done. If x > 0, by Theorem 3.3.10(a), there is n N such that n > x , i.e., n > x . 3.3.6b. Suppose that m and n are both maxima of a set S . Prove m = n . Proof: We have m n 29
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because n = max ( S ) and m S , and n m because m = max ( S ) and n S . Then m n and n m ⇐⇒ m = n.
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