Absolute value less than some value

Question

This is a noob question. If, $$\biggm| \frac{1}{2} - e \biggm | \le n$$ Then how do I get the following? $$e \le \frac{1}{2} + n$$

Answer 1

Having $\lvert a - b \rvert \leq c$ is basically the same as $\lvert b - a \rvert \leq c$.

You can also think of distances. The distance between $a$ and $b$ is less than $c$. If you imagine dots on the number line and keep the distance thingy in mind it should be very easy to see.

Answer 2

Observe that if $\dfrac{1}{2} - n \leq e$, the first equation holds.

Plot the graph of the absolut value function. All values of $x$ such that $|x|\leq n$ lie in the interval $[-n,n]$, i.e. $$x\in[-n,n] \ \iff \ |x|\leq n.$$

Substitute the right hand side of the equation where $x$ is to get $$| e + \dfrac{1}{2}| \leq n \ \iff \ e + \dfrac{1}{2} \in [-n,n] \ \iff -n\leq e + \dfrac{1}{2} \leq n.$$ For solving it, you need to solve both inequalities: $$-n\leq e + \dfrac{1}{2}$$ $$e + \dfrac{1}{2} \leq n.$$

Answer 3

You have to know that $$[a-b|\le c\iff b-c\le a\le b+c,$$ and remener tha $|a-b|=|b-a|.