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Unformatted text preview: MATH 410: Homework 2 Solutions Section 1.2 1. For each of the following statements, determine whether it is true or false and informally justify your answer. (a) Z is dense in R . Solution: False. For example (0 , 1 , . 2) contains no integers. (b) R > is dense in R . Solution: False. For example ( 4 , 3) contains no elements of R > . (c) Q \ Z is dense in R . Solution: True. Informally for ( a, b ) we pick q Q ( a, b ) by density of Q . If we accidentally picked an integer we pick a slightly larger rational noninteger less than b . 2. Suppose that S Z is nonempty and bounded below. Show S has a minimum. Solution: According to the CA the set S has a greatest lower bound. Define a = inf S . Since a is the greatest lower bound for S we know a + 1 is not a lower bound for S and so m S such that m < a +1. Hence m 1 < a and since a is a lower bound for S we have the set inclusion S ( m 1 , ). Since S is a set of integers, m is an integer and moreover ( m 1 , m ) contains no elements of S . Therefore we have the improved set inclusion S [ m, ). Thus since m S we have m as a minimum for S . 3. Show that c R , ! x Z with x ( c, c + 1]. Solution: There are two ways to approach this. Method 1: Modify the proof of Theorem 1.8 in the book: Define S = { n  n Z , n > c } . First observe that n is nonempty since the AP asserts there is a natural number n greater than c and hence in S . Now, since c is a lower bound for S and S is a set of integers, Proposi tion 1.7, mutatis mutandi, asserts there is a smallest element intion 1....
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 Spring '08
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 Integers

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