hw4-solutions

# But this takes up the entire repeating part thus the

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times must appear. But this takes up the entire repeating part; thus the repeating part must consist entirely of 2s, which contradicts the way that the number is constructed. 4. Given two real numbers, either there’s an integer between them, or there isn’t. If there is an integer between them, that’s the rational number we’re looking for. If there is not an integer between them, then we can write them as n.c 1 c 2 c 3 . . . , n.d 1 d 2 d 3 . . . where n is some integer, and the c s and d s are decimal digits; assume n.c 1 c 2 c 3 . . . < n.d 1 d 2 d 3 . . . . Then there’s some k such that c 1 = d 1 , c 2 = d 2 , . . . , c k - 1 = d k - 1 , and c k < d k ; that is, the first k - 1 digits after the decimal point are the same in both real numbers, and the k th digit is smaller in the first one. Then take n.d 1 d 2 . . . d k ; that is, cut off the second number just after the first digit that disagrees with the first number. This is the number we want. 1

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For example, say we have the two real numbers 3 . 141592 . . . and 3 . 142857 . . . ; these differ for the first time at the third digit, so take 3 . 142 to be the rational number between them. (Strictly speaking, this doesn’t work if n.d 1 d 2 d 3 . . . ends in infinitely many zeroes, because then truncating it doesn’t change it. So we should adopt the rule that we use the represen-
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• Summer '09
• Lugo
• Math, Real Numbers, Decimal, Natural number, Rational number, Countable set

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