This proof will probably seem like something very basic to most of you, but I am just starting so please bare with me.
Let a and b be real numbers. Prove that the absolute value of a is less than or equal to b iff -b is less than or equal to a is less than or equal to b.
Every other absolute value problem, we used the method of exhaustion. I'm not sure if it applies here and if so, how to work it to the problem.
Thank you in advanced for any help.
first: i have to assume here that $\displaystyle b>0$Originally Posted by zhupolongjoe
Suppose $\displaystyle |a|\leq b$.
1) if $\displaystyle a>0$, then $\displaystyle a = |a| \leq b$ and $\displaystyle -a \leq a = |a| \leq b \Rightarrow -a \leq b \Rightarrow -b \leq a$. combining the two facts, we have $\displaystyle -b \leq a \leq b$ for $\displaystyle a>0$.
2) if $\displaystyle a<0$, then $\displaystyle -a = |a| \leq b \Rightarrow -b \leq a$ and $\displaystyle a \leq |a| \leq b $. combining the two facts again, we have $\displaystyle -b \leq a \leq b$ for a<0.
3) if a=0, then it is trivial.
Thus, for any $\displaystyle a\in \mathbb{R}$, $\displaystyle -b \leq a \leq b$
next:
Suppose $\displaystyle -b \leq a \leq b$
1) if $\displaystyle a>0$, $\displaystyle |a|=a\leq b$
2) if $\displaystyle a <0$, use the fact that $\displaystyle -b\leq a \Rightarrow -a \leq b$. thus, $\displaystyle |a| = -a \leq b$
3) if $\displaystyle a=0$, trivial.
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