inequality with absolute value on both sides

Question

determine the set $B={x∈R: |x-1|<|x|}$ we consider cases so that the absolute value can be removed. so we took the cases:
1) $x\geq1$
2) $0\leq x<1$
3) $x<0$

Answer 1

Solving the inequality by substituting the absolute values for square gives

$x>\frac{1}{2}$

Answer 2

Pedestrian approach :

Note $x\not = 0.$

Divide both sides of the inequality by $|x|.$

$|1-1/x|\lt 1, $

$-1 \lt 1-1/x \lt 1.$

1)$ 1-1/x \lt 1$ implies

$0<1/x $, hence $x>0.$

2)$ -1 \lt 1-1/x$ implies

$1/x \lt 2.$

$x \lt 0,$ which is ruled out by 1),

or for $x>0$: we get $x>1/2$.

Answer 3

Without cases it's $$x^2-2x+1<x^2$$ or $$x>\frac{1}{2}.$$