Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Appendices - Section A.2 - Mathematical Induction - Exercises A.2 - Page AP-9: 1


See proof below.

Work Step by Step

1. (base) Check that the formula holds for n = 1. LHS$=|x_{1}|,\qquad $RHS$=|x_{1}|$ $|x_{1}|\leq|x_{1}|\qquad $... yes, it is true when $n=1.$ 2. (step) Prove that IF the formula holds for any positive integer $n=k$, THEN it also holds for the next integer, $n=k+1$. Let us assume that for $n=k$, $|x_{1}+x_{2}+...+x_{k}|\leq|x_{1}|+|x_{2}|+...+|x_{k}|$ (we call this the step hypothesis) For $n=k+1$, the LHS equals $|x_{1}+x_{2}+...+x_{k}+x_{k+1}|$ If we set $a=x_{1}+x_{2}+...+x_{k}$ and $b=x_{k+1}$, the LHS can be rewritten as $|a+b|$. We now apply the triangle inequality $|a+b|\leq|a|+|b|$ $|x_{1}+x_{2}+...+x_{k}+x_{k+1}|\leq|x_{1}+x_{2}+...+x_{k}|+|x_{k+1}|$ By the step hypothesis, $|x_{1}+x_{2}+...+x_{k}|\leq|x_{1}|+|x_{2}|+...+|x_{k}|$ so we can say that $|x_{1}+x_{2}+...+x_{k}+x_{k+1}|\leq|x_{1}|+|x_{2}|+...+|x_{k}|+|x_{k+1}|,$ which makes the formula true for $n=k+1.$ By mathematical induction, the formula is true for any integer $n.$
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