Hello all! In my quest to re-teach myself the basics of mathematics in a more rigorous fashion, I have found out that inequalities and absolute values are a weak point if mine. So I am working to address that. I am getting much better at it (with help from PF), but I have recently encountered seemingly simple problem that turned out to be a little trickier than I thought. Though I can arrive at the correct answer, I am not sure that my procedure is sound. Hopefully you can offer some insight. Take the following problem from chapter 1 of Spivak's Calculus (Problem 11 (iv)):(adsbygoogle = window.adsbygoogle || []).push({});

Find all ##x## for which ##|x-1|+|x-2| > 1 \qquad(1)##.

My approach to these has been to use the fact that the definition of absolute value is

[tex]

\begin{align}

|x| =

\begin{cases}

x, & \text{if }x\ge0 \\

-x, & \text{if }x\le0

\end{cases}

\end{align}

[/tex]

so then for each quantity enclosed by absolute value signs, there are 2 cases that needed to be evaluated. Applying this to (1) we have

Case 1: ##(x-1)>0 \wedge (x-2)>0## then

##(x-1) + (x-2) > 1 \implies x > 2.##

Case 2: ##(x-1)<0 \wedge (x-2)<0## then

## (1-x) + (2-x) > 1 \implies x<2.##

Case 3: ##(x-1)>0 \wedge (x-2)<0## then

## (x-1)+(1-x) > 1 \implies 0 >1. ##

Case 4: ##(x-1)<0 \wedge (x-2)>0## then

## (1-x) + (x+2) > 1 \implies 3 > 1. ##

Let's just look at.Case 1for a moment

Assuming that ##(x-1)>0 \wedge (x-2)>0## is the same as assuming ## x > 1 \wedge x>2.## This is clearly only true for ##x>2##, so there is really no need to specify that ##x>1.## But when it comes time to solve the actual problem, I need to use the expression ##(x-1)## under the assumption that it is a positive quantity, whichis the same asspecifying that ##x>1##. The answer I got is ##x>2## and is valid, but I feel like I might miss something in future problems if I do not pay attention to this detail.

So my question is this: Do I simply solve the inequality as I have done and then restrict the solution to ##x>2## if I were to get something less than 2?

Does my question make sense?

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# Inequalities and Absolue Values: Problem Solving Approach

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