{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

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

This preview shows pages 1–2. Sign up to view the full content.

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 . Thus, x < r 0 < r 1 < y . By part (c), there exists t 6∈ Q such that r 0 < t < r 1 . Therefore x < r 0 < t < r 1 < y , or x < t < y , so there is an irrational number

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}