Thomas' Calculus 13th Edition

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

Appendices - Section A.1 - Real Numbers and the Real Line - Exercises A.1 - Page AP-6: 29


See proof below.

Work Step by Step

"If and only if" statetments are equivalence statements. Proving an equivalence $A\Leftrightarrow B$ is done by proving $A\Rightarrow B$ and $B\Rightarrow A$ $1.\quad A\Rightarrow B$ Let $\quad |x|\gt a.$ Then, a positive number $k$ exists such that $|x|=a+k$ By definition of absolute value, $|x|=\left\{\begin{array}{ll} x, & x\geq 0\\ -x, & x \lt 0 \end{array}\right.$ If x is nonnegative, then $x=a+k$, that is $x \gt a.$ If x is negative, then $x=-(a+k)=-a-k$, which is less than $-a$, because we are subtracting a positive number from $-a.$ The implication $ |x|\gt a\quad \Rightarrow \quad x\gt a \quad $or$\quad x \lt a.$ has been proved. $2.\quad B\Rightarrow A$ Now, let $x \gt a$. Since $a$ is positive, it follows that x is also positive. By definition, $|x|=x, $and since $x \gt a$, it follows that $|x| \gt a.$ Finally, let $x \lt -a$. x is negative, so by definition, $|x|=-x$, and when we multiply the last inequality with $(-1)$, the inequality direction changes: $-x\gt a$ $|x|\gt a$ This implication has also been proved. Thus, $A\Leftrightarrow B$, and the statement has been proved.
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