# Doubt with Absolute Value Inequality

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!

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:

$$x\gt 2$$

• @BetterWorld, along these lines, what are the solutions of $$\frac{|y|}{y} > 0$$ They are $y > 0$. Hence avid19's answer. – Simon S Jun 28 '15 at 19:07
• Thanks a lot Sir! I'm really sorry for my approach. But Sir, could you please tell me where I went wrong in my attempt? – Ishan Jun 28 '15 at 19:08
• @BetterWorld Please don't call me sir, and don't apologize for your attempt. Your mistake is $\frac{2-x}{x-2}>0$ which is never true. – user223391 Jun 28 '15 at 19:12
• @avid19 Sorry for that. It's just that out of respect I call everybody Sir on this forum. I managed to work out a reason for applying the Wavy Curve in the first case. The critical point is evidently$0.$ However, if we look at $\dfrac{x-2}{x-2},$ it is actually $(x-2)^0.$ Hence, the function never crosses the $X-axis$ since the power to which the factor is raised is $0$ ie it is a double point. However, I cannot understand how to extend this argument to the second case without multiplying both sides by $-1.$ Could you please help me?. – Ishan Jun 28 '15 at 19:23
• Also, could you please explain why the second case is wrong?I'm not doubting it - I just would like to know if the reason for it being so could be extended to a general case. – Ishan Jun 28 '15 at 19:24

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 $\text {sgn}(x-2)$.