sA - MAT 137Y 2008-2009 Winter Session, Solutions to...

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MAT 137Y 2008-2009 Winter Session, Solutions to Problem Set Supplement #1 2 (a) Suppose x and y are numbers such that y - x > 1. Consider b x c . Then b x c ≤ x < y < y + 1. So x < b x c + 1 < y , so k = b x c + 1 is an integer such that x < k < y . (b) Since x < y , consider the number 1 / ( y - x ) . Pick any positive integer n such that n > 1 / ( y - x ) . Then y - x > 1 n . (c) Since there exists n N such that y - x > 1 n , then ny - nx > 1. By part (a), there exists k Z such that nx < k < ny , or x < k n < y . But k and n are integers, and n 6 = 0, so k n is rational. So there exists a rational number r = k n such that x < r < y , thereby showing that the rationals are dense. (d) Now suppose we are given two rational numbers r and s such that r < s . Since 2 / 2 is irrational, then 0 < 2 2 < 1 = 0 < 2 2 ( s - r ) < ( s - r ) = r < r + 2 2 ( s - r ) < s . Then r + 2 2 ( s - r ) is irrational (otherwise 2 2 ( s - r ) would be rational, which would imply that 2 2 is irrational since s - r is rational, a contradiction). Therefore, between any two rational numbers there exists an irrational number. (e) If x < y , then by part (b) there exists r 0 Q such that x < r 0 < y . But r 0 R , so by part (b) there exists r 1 Q such that r 0 < r 1 < y
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This note was uploaded on 04/09/2010 for the course MAT 137 taught by Professor Uppal during the Spring '08 term at University of Toronto- Toronto.

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sA - MAT 137Y 2008-2009 Winter Session, Solutions to...

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