Inequalities and Absolue Values: Problem Solving Approach

[Saladsamurai]

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)):

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

$$\begin{align}<br /> |x| =<br /> \begin{cases}<br /> x, & \text{if }x\ge0 \\<br /> -x, & \text{if }x\le0<br /> \end{cases}<br /> \end{align}$$

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 1 for 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, which is the same as specifying 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?

 

[mathman]

It is simpler just to look at the domains for x.
There are 3 cases, x > 2, 2 > x > 1, 1 > x. These correspond to your first 3 cases (not in the same order).
Your case 4 is non-existent, x > 2 and x < 1 is impossible.

 

[Saladsamurai]

It is simpler just to look at the domains for x.
There are 3 cases, x > 2, 2 > x > 1, 1 > x. These correspond to your first 3 cases (not in the same order).
Your case 4 is non-existent, x > 2 and x < 1 is impossible.

Hi mathman Yes, we have not gotten to the other 3 cases yet. I am focusing on the simple one: Case 1. I am afraid people might not be understanding question, but I am not sure how else to word it.

 

[micromass]

The case $x>1~\wedge~x>2$ is indeed equivalent to x>2.

If you solve the equation under the premisse that x>2, then every solution must satisfy that. If you find that the equation is true for all x>-2, then only the x>2 will count.

For example, if you solve $|x-1|>-5$ (I know you can easily see that all x will be a solution, but I'm just setting an example).

You can split up

1) $x\geq 1$, in that case |x-1|=x-1. So the equation is x-1>-5. This is true for x>-4. However, you originally set $x>1$, so in this case the only solutions are all $x>1$ (and not all x>-4).

2) x<1, in that case |x-1|=1-x. The equation becomes 1-x>5, or x<6. In this case the solutions are all x<1 (and not x<6).

Adding (1) and (2) yields that all real numbers are a solution.

 

[LCKurtz]

This thread is a duplicate of:
https://www.physicsforums.com/showthread.php?t=577008

 

[micromass]

Salad convinved me that a separate thread does have some merit. This thread is not the same as the HW question, but is rather more general. So I'll allow this thread to stay open.