Thinking Mathematically (6th Edition)

Published by Pearson
ISBN 10: 0321867327
ISBN 13: 978-0-32186-732-2

Chapter 2 - Set Theory - 2.2 Subsets - Exercise Set 2.2 - Page 68: 11


{$\frac{4}{7}$,$\frac{9}{13}$} ⊄ {$\frac{7}{4}$,$\frac{13}{9}$}

Work Step by Step

As no element of first set is present in the second set, Therefore {$\frac{4}{7}$,$\frac{9}{13}$} ⊄ {$\frac{7}{4}$,$\frac{13}{9}$}
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