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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!

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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:

$$x\gt 2$$

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  • $\begingroup$ @BetterWorld, along these lines, what are the solutions of $$\frac{|y|}{y} > 0$$ They are $y > 0$. Hence avid19's answer. $\endgroup$ – Simon S Jun 28 '15 at 19:07
  • $\begingroup$ 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? $\endgroup$ – Ishan Jun 28 '15 at 19:08
  • $\begingroup$ @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. $\endgroup$ – user223391 Jun 28 '15 at 19:12
  • $\begingroup$ @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?. $\endgroup$ – Ishan Jun 28 '15 at 19:23
  • $\begingroup$ 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. $\endgroup$ – Ishan Jun 28 '15 at 19:24
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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 $\text {sgn}(x-2)$.

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