How to do $\lvert x - 1 \rvert < \lvert x-3 \rvert$ 
Solve for $x$ such that $$\lvert x - 1 \rvert < \lvert x-3 \rvert$$

I understand that the question is essentially saying what are the possible values for $x$, centered at $1$ with a distance of $x-3$ on either side of $x$. But how does that work? 
 A: hint: Square both sides ! and solve $(x-1)^2 < (x-3)^2$ . Can you continue ?
A: The intuition here is that $x$ is closer to $1$ than to $3$, and this is clearly whenever $x$ is less than $2$.
This intuition can also help to see what is going on in other situations - for example in two or three dimensions when similar expressions are used.

You need to take care, though, that the signs are right. $|x-a|\lt|x-b|$ means that $x$ is closer to $a$ than $b$. If it were $|x+a|$ instead you'd be looking at the distance from $-a$. This is one of the easier places to trip up in a hurry when you are learning.
A: The question is asking for you to find those values of $x$ that make $|x-1| < |x-3|$ true. For example, $x=0$ makes it work. To make things easier, we can rewrite our inequality in the equivalent inequality
$$ 0 \leq |x-3| - |x-1| $$
And now this translates to find value of $x$ for which $f(x) = |x-3|-|x-1|$ is positive. can you find them? Here is a graph:

A: Several ways.
There are $4$ possibilities $x -1$ can be $<$ or $\ge 0$ and $x-3$ can be $<$ or $\ge 0$.
Or in other words $x$ can be $<$ or $\ge 1$ or $x$ can be $< $ or $\ge 3$. These are four possibilities but some may be contradictory (We can't have have $x < 1$ and $x \ge 3$ both) or redundant (if $x\ge 3$ then $x > 1$ is obviously also true.
So this breaks down into three cases.  $x < 1; 1 \le x < 3; x\ge 3$.
Case 1:  $x < 1$.  Then $|x-1| = 1-x$ and $|x-3| = 3-x$.
So $1 -x < 3-x\implies 1 < 3$ ... which is always true.  This means if $x< 1$ then $x$ an be any value less than $1$.
Case 2:  $1 \le x < 3$ so $|x-1| = x-1$ and $|x-3| = 3-x$.
So $x-1 < 3-x \implies 2x < 4\implies x < 2$.  This means if $1 \le x < 3$ then we know further than $x < 2$ and that $2\le x < 3$ is impossible.
Case 3:  $x \ge 3$ then $|x-1| = x-1$ and $|x-3| = x-3$ and we can $x-1 < x-3\implies -1 < -3$.  That is simply impossible.  So we know that $x \ge 3$ is not possible.
So we know 1) $x$ could be any value $< 1$.  2) $x$ could be any value so that $\le x < 2$ but $x$ can not be $2 \le x < 3$.  And 3) $x \ge 3$ is impossible.
Combining those three results we get .... $x < 2$.  And any value$x < 2$ will work.
====
Just for kicks and giggles:
$|m| < k$ means $-k < m < k$.
So we have $-|x-3| < x-1 < |x-3|$.  
If $x \ge 3$ then that is $-x + 3 < x - 1 < x-3$ so $4 < 2x < 2x - 2$  and $-2x + 4 < 0 < -2$ which is impossible.
So $x < 3$ and $x -3 < x - 1 < -x + 3$ so $2x -3 < 2x - 1 <  3$ and so $2x < 4$ and $x < 2$.
So $x < 2$.
====
And cute....
$\sqrt {k^2} = |k|$ and if $0 \le a$ and $0 \le b$ then $a^2 < b^2 \iff a < b$.
So $\sqrt {(x-1)^2} = |x-1| < |x - 3| = \sqrt{(x-2)^2} \iff $
$(x-1)^2 < (x-3)^2 \iff$
$x^2 -2x + 1 < x^2 - 6x + 9 \iff$
$4x < 8 \iff$
$x < 2$.
A: $y:=x-1$;
Let
A) $y \not =0$.
$|y| \lt |y-2|$;
$1\lt |1-2/y|;$
1) $1\lt 1-2/y$
$2/y \lt 0$, or 
$y <0$.
2)$1 \lt 2/y -1$,
$1\lt 1/y.$
$0<y <1.$
B) Now check for  $y=0$.
Combining 
1) $y <0$ and 2) $0<y<1$, with 
B) $y=0$:
we have $y <1$, or $x <2$.
(Recall $x= y+1$.)
A: Hint:
Let $x=a+ib$ where $a,b$ are real
$|x-1|=\sqrt{(a-1)^2+b^2}>0$
Now if $p>q, $ $$p^2-q ^2=(p-q)(p+q)>0$$  for $p+q>0$
