How can you define $x \in [-3, 5]$ using absolute values? I really want to say that $x > -3$ and $x < 5$ and then somehow get that into the form $x + a > -b, x + a <b$, where $a,b \in \mathbb N$, but I have no idea how to do that. Any help is appreciated, thanks.
 A: The interval $[-3, 5]$ can be written as $[1 - 4, 1 + 4]$. This means that it is the interval of radius 4 centered at the point 1, so this may be written as $|x - 1| \leq 4$.
A: $x>-3 \quad \iff \quad x-1<-4$
and
$x<5 \quad \iff \quad x-1<4$
A: Note that $\vert b-a\vert$ is the distance between $a$ and $b$. So $\vert x-1\vert \leq 4$ are all points that lie no more than a distance of $4$ from $1$. This gives you $[-3,5]$.
I works the other way around as well: $[a,b]$ has length $\vert b-a\vert$. Its midpoint is $\frac {b+a}2$. So $[a,b]$ can also be seen as the set of all point that lie no more than $\frac{\vert b-a\vert}{2}$ from the point $\frac {b+a}2$. Hence we can denote it by $\left\vert x-\frac{b+a}2\right \vert\leq\frac{\vert b-a\vert}2$. 
A: $a=-\dfrac{b-a}{2}+\dfrac{b+a}{2}$ and $b=\dfrac{b-a}{2}+\dfrac{b+a}{2}$ therefore
$$a\le x\le b\iff-\dfrac{b-a}{2}+\dfrac{b+a}{2}\le x\le\dfrac{b-a}{2}+\dfrac{b+a}{2}\iff\left|x-\dfrac{b+a}{2}\right|\le\dfrac{b-a}{2}$$
For the particular case $(a,b)=(-3,5)$ we have $|x-1|\le 4$
A: An interval $[a,b]$ can be characterised as the set of points $x$ such that the distance of $x$ to the mid-point $\dfrac{a+b}2$ of the interval is at most the radius ( = half length) of the interval. In formula:
$$\biggl\lvert\mkern1mu x-\frac{a+b}2\mkern1mu\biggr\rvert \le \frac{b-a}2.\quad\text{Here: }\enspace\lvert x-1\rvert\le 4.$$
