Doubt with Absolute Value Inequality

Question

Problem:

Find all values of $x$ for which $\dfrac{|x-2|}{x-2}>0$

My incorrect attempt:

Using the definition the Modulus, $|x-2|=x-2$ for all $x\ge2$ and $|x-2|=-x+2$ for all $x\le2.$ Splitting into 2 cases:

$$\text{CASE } 1:x\in [2,\infty)\Rightarrow |x-2|=x-2$$

$\dfrac{x-2}{x-2}>0$ $$\Rrightarrow x\in [2,\infty)\cap \mathbb{R}-\text{{2}}$$ $$\Rightarrow x\in(2,\infty)$$ $$$$ $$\text{CASE } 2:x\in(-\infty,2)\Rightarrow |x-2|=2-x$$ $$\Rightarrow \dfrac{2-x}{x-2}>0$$$$$$ On drawing the 'Wavy Curve Method' (also known as the Method of Intervals) for this, I got $x=-2,2$ as the critical points where the function changes its sign. With this, I got that the function $\dfrac{2-x}{x-2}$ is greater than $0$ in the interval $(2,2)$ and is less than $0$ for $x\in (-\infty,-2)\cup(2,\infty).$ $$$$Also, if we multiply $\dfrac{2-x}{x-2}>0$ by $-1$, then we get $\dfrac{x-2}{x-2}<0$ which is an obvious contradiction with the first case.$$$$I would be truly grateful if somebody could please clear my doubts and show me my errors. Many thanks in advance!

Answer 1

You're way over thinking this. Since $|x-2|\geq 0$ for all $x$, we just need to find when $x-2>0$. Add $2$ to both sides to get the answer:

$$x\gt 2$$

Answer 2

Note: $+$ is positive, $-$ is negative, $0$ is zero. $$\begin{array}{c|c|c|} & \text{|x-2|} & \text{x-2} & \frac {|x-2|}{x-2}\\ \hline (-\infty,2) & + & - & (+)\div(-)=-\\ \hline x=2 & 0 & 0 & 0\div0=\text {undefined}\\ \hline (2,\infty) & + & + & (+)\div(+)=+ \end{array}$$

Thus, $\frac {|x-2|}{x-2}$ is positive in $(2, \infty)$. See $\text {sgn}(x-2)$.