Domains and setbuilder notation

Question

We're learning about domains and setbuilder notation in school at the moment, and I want to make sure what I did was right.

My thought process: \begin{align*} -\frac12|4x - 8| - 1 &< -1 \\ -\frac12|4x - 8| &< 0 \\ |4x - 8| &> 0 \end{align*} $x =$ all real numbers.

{real numbers} :

<||||||||||[0]|||||||||>

{x| x is any real number}

{whole numbers}

... <----[-2]---[-1]---[0]---[1]---[2]---> ...

{x|...-2,-1,0,1,2...}

Answer 1

First let's consider how absolute-value is defined:

$$ |a| = \begin{cases} a, & \text{if } a \geq 0, \\ -a, &\text{if } a \lt 0. \end{cases} $$

Therefore,

$$4x-8 > +0\phantom{.}$$

$$4x-8 < -0.$$

Now, solve for $x$ to get the answer:

$$x > 2\phantom{.}$$

or $$x < 2.$$

Note: This is same as $$ x \neq 2.$$

Answer 2

You get $|4x - 8| > 0$, which I agree with; now you want to find all $x$ satisfying this inequality. It's true that for any number $y$ we have $|y| \geq 0$, but equality can hold: $|y| = 0$ if and only if $y = 0$. Use this fact to find the single value $a$ of $x$ for which $|4x - 8| = 0$. In set-builder notation, I would write this as \[ \{x \mid x \neq a\} \qquad \text{or, more carefully,} \qquad \{x \in \mathbb R \mid x \neq a\}, \] replacing $a$ by the number you find.

Your representations of the real numbers look fine to me. There is always controversy over what "whole numbers" should mean, and I would call \[ \{x \mid x = \ldots, -2, -1, 0, 1, 2, \ldots\} \] the set of integers. Note the slight difference between your expression and mine.