prove: If $b\geq 0$ then $|b|>a$ iff $b>a$ or $b<-a$.

I want explanation on how to prove this theorem: If b is greater than or equal to zero then absolute value of b is greater than a iff b is greater than a or b is less than negative a.

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if $b\ge 0$ then $|b|=b.$ –  Xabier Domínguez Apr 17 '12 at 14:01
It is not a theorem. It is just the definition. $|b| \geq a$ is equivalent to the $b \geq a$ or $b \leq -a$, because of the definition of $|b|$ which is $b$ for $b \geq 0$ and $-b$ for $b \leq 0$. –  Beni Bogosel Apr 17 '12 at 14:06

3 Answers

It is (almost) trivially true. Note first that by definition $$\lvert b\lvert = \begin{cases} b & \text{if } b \geq 0 \\ -b &\text{if } b < 0. \end{cases}$$ This is true for any real number $b$. So when you assume that $b\geq 0$, then $\lvert b \lvert = b$. So you are really asking for a proof of :

If $b\geq 0$ then $b > a$ if and only if $b > a$ or $b < -a$.

$(\Rightarrow)$: This direction is trivial. Indeed if $b > a$ then $b> a$ (or $b< -a$)

$(\Leftarrow)$: So now assume that $b > a$ or $b < -a$. If $b >a$ then you are done, so lets say that $0 \leq b < -a$. But from this we see that $a$ is negative (because $-a$ is positive), so clearly it follows that $a< b$.

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Your last two lines each prove (in two different ways) that $a<b$... –  M Turgeon Apr 17 '12 at 15:31
@MTurgeon: I was trying to be careful with the logic. With $(\Leftarrow)$ we are assuming that $b> a$ or $b < -a$. So we have to treat each of those to cases differently (just being careful). –  Thomas Apr 17 '12 at 15:33
Fair enough; always good to be careful :) –  M Turgeon Apr 17 '12 at 15:37

I recommend you, look this sentences in real line. So you have $|b|\geq 0$ and $|b|>a$, these inequalitys, means, thes absolute value of b, could be: $-b$ if $b<0$ or $b$ if $b>0$. With these two cases, you have $-b>a$ or $b>a$. The first inequality $-b<a$ multiply for $-1$, you get $b>-a$, because $-1<0$ then change the changes the direction of inequality. And, the second inequality: $b>a$. So you have your proof.

Regards,

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The way you have stated it is trivial and redundant. You probably meant:

If a is greater than or equal to zero then absolute value of b is greater than a iff b is greater than a or b is less than negative a.

Am I right?

In this case, consider two cases: if b $\geq 0$ then $|b| = b$ and we want $b > a$. If $b\leq 0$ then $|b| = -b$ and we want $-b > a$, i.e. $-a > b$ (by adding $b-a$ to both sides). This can also be written as $b < -a$.

This proves the 'only if' condition.

For the 'if' condition, note that if $b > a$, it implies that $b > 0$ and thus $|b| = b > a$. On the other hand, if $b < -a$, it implies that $b < 0$, so $|b| = -b < a$ (by adding $a-b$ to both sides of $-a > b$).

So this proves the 'if' condition too.

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"By adding $a-b/b-a$ to both sides". I mean, you're (also) just multiplying by -1. –  M Turgeon Apr 17 '12 at 15:35
True, but it is a bit harder to keep track of inequality signs when we see it as multiplying by negative numbers. –  Wonder Apr 17 '12 at 15:45