Chapter 2.1, Problem 11E

(a)

To determine

To prove: $\left\{x\in ℝ:\frac{3}{4}x-2>10\right\}=\left(16,\infty \right)$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:\frac{3}{4}x-2>10\right\}$ .

Proof:

Let the sets be named:

  $\begin{array}{l}A=\left\{x\in ℝ:\frac{3}{4}x-2>10\right\}\\ B=\left(16,\infty \right)\end{array}$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ \frac{3}{4}x-2>10\\ \Updownarrow \\ x>16\\ \Updownarrow \\ x\in B\end{array}$

Since x is arbitrary: A = B

(b)

To determine

To prove: $\left\{x\in ℝ:\left|x-4\right|=2\left|x\right|-2\right\}=\left\{-6,2\right\}$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:\left|x-4\right|=2\left|x\right|-2\right\}$ .

Proof:

Let the sets be named:

  $A=\left\{x\in ℝ:\left|x-4\right|=2\left|x\right|-2\right\}$

  $B=\left\{-6,2\right\}$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ \left|x-4\right|=2\left|x\right|-2\\ \Updownarrow \\ \text{Assume: }x\ge 4\Rightarrow x-4=2x-2\Rightarrow x=-2\text{ contradiction with }x\ge 4\\ \vee \\ \text{Assume: }0\le x<4\Rightarrow -x+4=2x-2\Rightarrow x=2\\ \vee \\ \text{Assume: }x<0\Rightarrow -x+4=-2x-2\Rightarrow x=-6\\ \Updownarrow \\ x\in B\end{array}$

Since x is arbitrary: A = B

(c)

To determine

To prove: $\left\{x\in ℝ:2\left|x+3\right|+x=0\right\}=\left\{-6,-2\right\}$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:2\left|x+3\right|+x=0\right\}$ .

Proof:

Let the sets be named:

  $A=\left\{x\in ℝ:2\left|x+3\right|+x=0\right\}$

  $B=\left\{-6,-2\right\}$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ 2\left|x+3\right|+x=0\\ \Updownarrow \\ \text{Assume: }x\ge -3\Rightarrow 2x+6+x=0\Rightarrow x=-2\\ \vee \\ \text{Assume: }x<-3\Rightarrow -2x-6+x=0\Rightarrow x=-6\\ \Updownarrow \\ x\in B\end{array}$

Since x is arbitrary: A = B

(d)

To determine

To prove: $\left\{x\in ℝ:\left|x\right|=6-\left|2x\right|\right\}=\left\{-2,2\right\}$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:\left|x\right|=6-\left|2x\right|\right\}$ .

Proof:

Let the sets be named:

  $A=\left\{x\in ℝ:\left|x\right|=6-\left|2x\right|\right\}$

  $B=\left\{-2,2\right\}$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ \left|x\right|=6-\left|2x\right|\\ \Updownarrow \\ \text{Assume: }x\ge 0\Rightarrow x=6-2x\Rightarrow x=2\\ \vee \\ \text{Assume: }x<0\Rightarrow -x=6+2x\Rightarrow x=-2\\ \Updownarrow \\ x\in B\end{array}$

Since x is arbitrary: A = B

(e)

To determine

To prove: $\left\{x\in ℝ:\left|x+3\right|\le -4x-2\right\}=(-\infty ,-1]$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:\left|x+3\right|\le -4x-2\right\}$ .

Proof:

Let the sets be named:

  $A=\left\{x\in ℝ:\left|x+3\right|\le -4x-2\right\}$

  $B=(-\infty ,-1]$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ \left|x+3\right|\le -4x-2\\ \Updownarrow \\ \text{Assume: }x\ge -3\Rightarrow x+3\le -4x-2\Rightarrow x\le -1\\ \vee \\ \text{Assume: }x<-3\Rightarrow -x-3\le -4x-2\Rightarrow x\le \frac{1}{3}\wedge x<-3\\ \Updownarrow \\ x\in B\end{array}$

Since x was arbitrary: A = B

(f)

To determine

To prove: $\left\{x\in ℝ:\left|x+3\right|=5-\left|x\right|\right\}=\left\{-4,1\right\}$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:\left|x+3\right|=5-\left|x\right|\right\}$ .

Proof:

Let the sets be named:

  $A=\left\{x\in ℝ:\left|x+3\right|=5-\left|x\right|\right\}$

  $B=\left\{-4,1\right\}$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ \left|x+3\right|=5-\left|x\right|\\ \Updownarrow \\ \text{Assume: }x\ge 0\Rightarrow x+3=5-x\Rightarrow x=1\\ \vee \\ \text{Assume: }-3\ge x<0\Rightarrow x+3=5+x\Rightarrow \text{ contradiction}\\ \Updownarrow \\ \text{Assume: }x<-3\Rightarrow -x-3=5+x\Rightarrow x=-4\\ \Updownarrow \\ x\in B\end{array}$

Since x was arbitrary: A = B

(g)

To determine

To prove: $\left\{x\in ℝ:\left|x-4\right|<x\right\}=\left(2,\infty \right)$ .

Explanation of Solution

Given information: $\left\{x\in ℝ:\left|x-4\right|<x\right\}$ .

Proof:

Let the sets be named:

  $A=\left\{x\in ℝ:\left|x-4\right|<x\right\}$

  $B=\left(2,\infty \right)$

  $\begin{array}{l}x\in A\\ \Updownarrow \\ \left|x-4\right|<x\\ \Updownarrow \\ x>0\\ \Updownarrow \\ \text{Assume: }4>x\ge 0\Rightarrow -x+4<x\Rightarrow 2<x<4\\ \vee \\ \text{Assume: 4}\le x\Rightarrow x-4<x\Rightarrow -4<0\Rightarrow 4\le x\\ \Updownarrow \\ x\in B\end{array}$

Since x was arbitrary: A = B

(h)

To determine

To prove: $\left\{3\right\}\text{ is a proper subset of }\left\{x\in ℝ:\left|x-2\right|=4-\left|x\right|\right\}$ .

Explanation of Solution

Given information: $\left\{3\right\}\text{ is a proper subset}$ .

Proof:

Let the sets be named:

First of all, let’s see if 3 is an element of $\left\{x\in ℝ:\left|x-2\right|=4-\left|x\right|\right\}$ .

  $\left|3-2\right|=1=4-3=4-\left|3\right|$

3 is in element of the set.

Now, prove that 3 is not only element of $\left\{x\in ℝ:\left|x-2\right|=4-\left|x\right|\right\},$

We will do th is by finding that element.

For $x=-1$ :

  $\left|-1-3\right|=\left|-3\right|=3=4-1=4-\left|-1\right|$

Therefore, $-1\in \left\{x\in ℝ:\left|x-2\right|=4-\left|x\right|\right\}$

  $\left\{3\right\}\cancel{\subseteq }\left\{x\in ℝ:\left|x-2\right|=4-\left|x\right|\right\}$