Prove that $ |x-a|<b \iff x∈(a-b,a+b)$

Question

I am not sure if I am doing this correctly,

my forward proof is:

Since $|x-a|$ is an absolute value function, it can be defined as $|x-a|:= \max\{(x-a),-(x-a)\}$ .

Let $S={(x-a) ∈ ℝ}$ and $b ∈ S, b>0,$

then $ x-a>b

     x>b+a $

 2.-(x-a)>b

   -x+a>b

   -x>b-a

    x<a-b

which prove that x ∈ (a − b, a + b)

Am I doing this correctly, and how can I prove the backward?

Edit: What I am thinking about the backward equation:

a-b < x , then a-b-x<0, next -(a-b-x)>0, so -(a-b-x)>(a-b-x) -->max = a-b-x

a+b>x , then a+b-x<0, next -(a+b-x)>0, so -(a+b-x)>(a-b-x) --> max= a+b-x

then I am stuck.

Answer 1

We need to have $b > 0$ or the interval $(a - b, a + b)$ would not be expressed sensibly.

Recall that $|x-a| < b$ means that $-b < x-a < b$, after which we can add $a$ to obtain the chain of inequalities $a - b < x < a + b$, which says that $x \in (a-b, a+b)$ when rewritten using interval notation.

Answer 2

HINT

Assuming that $b > 0$, we have:

\begin{align*} |x-a| < b \Longleftrightarrow |x-a|^{2} < b^{2} \Longleftrightarrow (x-a)^{2} - b^{2} < 0 \Longleftrightarrow (x - a - b)(x - a + b) < 0 \end{align*}

Answer 3

Take for granted* that $$|t|<c\iff-c<t<c$$

and plug

$$t=x-a, c=b.$$

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*If you don't want to,

• $t\ge0\implies |t|=t\implies t<c$;

• $t\le0\implies |t|=-t\implies -t<c\implies -c<t$.