Prove $|a - b|< c$ if and only if $b - c < a < b + c$.

Question

Prove $|a - b|< c$ if and only if $b - c < a < b + c$.
It is a task from real analysis and I am failing the class I tried doing it on a quiz, but I got it incorrect.

Answer 1

($\to$): Suppose $|a-b|<c$. Then $c>a-b$ and $c>-(a-b)$ by the definition of absolute value. That is, $c>a-b \Longleftrightarrow b+c>a$ and $c>-(a-b) \Longleftrightarrow a>b-c$.

($\leftarrow$): Suppose $b-c<a<b+c$. Then $b-c<a$ and $a<b+c$. Thus, $$ b-c<a \Longleftrightarrow -(b-c)>-a \Longleftrightarrow c>b-a=-(a-b). $$ And we also have that $a<b+c \Longleftrightarrow c>a-b$.

Thus, the claim follows.

Answer 2

You should first try with $|x| < c$. Then plug in $x=a-b$.

Let us assume that $c \geq 0$.

For example, if $x \geq 0$ then $|x|=x$. In which case $|x| < c$ becomes $x < c$.

Answer 3

Hint: $|a-b| = \max\{a-b, b-a\}$. So $c > |a-b|$ means that $$c > a-b \quad \text{and} \quad c > b-a.$$ Can you go on?

Answer 4

Write out the definition of $|\cdot|$ and the task transforms into:

Prove $$\cases{a-b < c & if $a-b\ge 0$\\ b-a < c & if $a-b<0$} \Leftrightarrow b-c<a<b+c$$

Now since the LHS is all about $a-b$, you'll first want to get an $a-b$ into the RHS by subtracting $b$ from all inequalities: $$b-c<a<b+c \Leftrightarrow -c<a-b<c$$ can you do the rest?