Can you help me with this inequality? $|x-1| < |x-3|$ I found this problem from Precalculus 7th ed, James Stewart, chapter 1.8, problem 128.
"Recall that $ |a-b| $ is the distance between a and b on the number line. For any number x, what do $|x-1|$ and $|x-3|$ represent? Use this interpretatin to solve the inequality  $|x-1| < |x-3|$ geometrically. In general, if a < b, what is solution of the inequality $|x-a| < |x-b|$ ? "
*I know that $|x-1|$ is the distance between x and 1, and geometrically, if $x =  2$, the distance between 1 and x is equal to the distance between 3 and x, at this point we can see that for $ x < 2 $ the distance between x and 3 is bigger than the distance between x and 1. I wanted to check this solution using the formal way to solve this inequality so I wrote the next solution, but I wasn't able to check it because, apparently, I did something wrong and I can't figure out what was wrong.
$|x-1| < |x-3|$
$-|x-3| < x-1 < |x-3|$
$-|x-3| < x-1  \land x-1 < |x-3|$
$(-x+1 < x-3 < x-1) \land (x-1 < x-3  \lor  x-1 > -x+3)$
$(-x+1 < x-3 \land x-3 < x-1) \land (x-1 < x-3  \lor  x-1 > -x+3)$
$(4 < 2x \land -3 <-1) \land (-1 < -3  \lor  2x > 4)$
$(2 < x \land -3 <-1) \land (-1 < -3  \lor  x > 2)$
My question is what I did wrong? and how to improve my formal solution, Thank you very much in advance for taking the time to read this.
EDIT: First of all, thank you everyone for all your answers now I can see various ways to solve this problem. Regarding my "cumbersome" solution Mr @AnuragA pointed out my mistake and now I came with this new solution, I hope now It is ok, Thank you again, and if you can point out any error in this new solution I will appreciate it.
$|x-1| < |x-3|$
$-|x-3| < x-1 < |x-3|$
$[|x-3| > -x+1]  \land [x-1 < |x-3|]$
$[x-3 > -x + 1  \lor -x + 1 < -(x-3)]\land[x-3>x-1 \lor x-1<-(x-3)] $
$[2x > 4  \lor 1 < 3]\land[-3>-1 \lor 2x<4] $
$[x >2   \lor \{ x  \in \Bbb R \Bbb \}]\land[\{ x \in \emptyset \} \lor x<2] $
$[ x  \in \Bbb R \Bbb ]\land[ x<2] $
$x<2$
 A: We can visually see the answer to this question and use the knowledge of geometry.

A: The inequality represents the set of all points $x$ on the real axis whose several distances from $1$ is less than the corresponding distances from $3.$
Square both sides, and transpose, to obtain $$(x-1)^2-(x-3)^2<0.$$ The LHS factors as $(x-1-x+3)(x-1+x-3)=2(2x-4).$ Thus, the inequality reduces to $$x-2<0,$$ which tells us that the set is the set of points to the left of $2$ on the real axis as usually oriented.
A: Hint
The locus of the points equidistant from $A$ and $B$ is given by the perpendicular bisector of $A$ and $B$. Given that the equidistant point has to lie on the real line, what can you conclude?
A: Set: $y:=x-1$;
Then: $|y| <|y-2|$;
Real number line:
Distance of of a point $y$ to the origin is less than distance of $y$ to $2$.
(Note: $y=1$ is equidistant from $0$ and $2$)
Hence : 
$y < 1$, or $ x-1<1$, and finally $x<2$.
A: If you know how to draw absolute value functions, then solve this kind of problems is very easy:

If you want to solve $|x-1| < |x-3|$ just you have to say in which interval the blue line is above the green line.
So you can say the answer is $\langle-\infty;2\rangle$
going further:
The other kind of equation is $|x-1| \ge |x-3|$, but you can solve this just by taking complement:
If $S$ is the set of solutions of $|x-1| \ge |x-3|$, then $S^c$ is the set of solutions of $|x-1| < |x-3|$ (and this one is the easiest problem)
A: Since we want to show an inequality that is $\alpha<\beta$, where $\alpha:=|x-1|>0)$ and $\beta:=|x-3|>0)$, we can do the following simple trick:
$$\alpha^2<\beta^2\Rightarrow x^2-2x+1<x^2-6x+9\Rightarrow x<2$$
Some other way to solve the given inequality is by comparing the graph of the absolute value of $x$, with the sifted graphs of the absolute values of $x-1$ and $x-3$ respectively, which is a little bit geometrically!
