My book presents the following: $$7 \le x \le 9 $$ so $$ -1 \le x - 8 \le 1 $$ and $$ |x-8| \le 1$$

I usually get confused with the way that taking the absolute value of an expression works. Could anybody explain why the inequalities above are equivalent? I understand how the first and the second one are equivalent but not how the third one is equivalent to the second one.

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    $\begingroup$ The absolute value of $x-a$ is the distance between $x$ and $a$. Think geometrically, with a "number line." I Don't like the first transition, from $7\le x\le 9$ to the next line. Oh, it is absolutely correct. But the real point is that $8$ is halfway between $7$ and $9$. So if you sit at $8$, everybody is at distance $\le 1$ from you. $\endgroup$ – André Nicolas Apr 23 '13 at 17:37

On a number line, $|x-a|$ represents the (non-negative) distance between $x$ and $a$. So $|x-8| \leq 1$ can be interpreted as: The distance between $x$ and $8$ is no more than $1$ unit. This immediately gives you $7 \leq x \leq 9$.

To analyze this the other direction, if $a \leq x \leq b$, then think of the midpoint: $\frac{a+b}{2}$, and half the distance from $a$ to $b$: $\frac{b-a}{2}$. You obtain:

$$ \left| x - \frac{a+b}{2} \right| \leq \frac{b-a}{2}.$$

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  • $\begingroup$ I think your approach here is very good: what's happenin' comes first, then, maybe, manipulation. $\endgroup$ – André Nicolas Apr 23 '13 at 17:41

Suppose you had $|x| < a$ (note: the idea below holds the same for $\leq$ as well). Let's break this into two cases: $x \ge 0$ and $x < 0$.

If $x \ge 0$, then $|x| = x$ since $x$ is already nonnegative so what the inequality $|x| < a$ says in that case is that $x < a$.

If $x < 0$, then $|x| = -x$ since $x$ is negative. What the inequality then says is that $-x < a$, or equivalently that $x > -a$.

Putting these two pieces together we get that $-a < x < a$. In your case we have that $-1 \leq x - 8 \leq 1$. Well using the above expression, we have that this is equivalent to $|x-8| \leq 1$.

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First of all, inequality relation is unaffected by

$(i)$ the addition (hence, subtraction) of finite real numbers

$(ii)$ the multiplication (hence, division) by finite positive real numbers

Again if $y$ is real and $|y|\le b$ Clearly, $b\ge0$

$\implies y^2\le b^2\implies (y-b)\{(y-(-b)\}\le0$

$\implies$ either $y\ge b$ and $y\le -b$ which is impossible as $b\ge0$

or $y\le b$ and $y\ge -b\implies -b\le y\le b$

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