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Which number can be $x$?

$$\vert 1-\vert x-1\vert\vert\lt1$$

I got:

$$1-x+1=0 \Longrightarrow \boldsymbol{x_1} = 0$$ $$-1+x-1=0\Longrightarrow \boldsymbol{x_2} = 2$$

What is the method of calculation?


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1 Answer 1

up vote 3 down vote accepted

We have $|1-|x-1||\lt 1$ if and only if $$-1\lt 1-|x-1| \lt 1.$$ Subtract $1$ from each expression. So our inequalities are equivalent to $$-2 \lt -|x-1| \lt 0,$$ or equivalently $$0\lt |x-1| \lt 2.$$ The part $0\lt |x-1|$ is no problem, it holds everywhere except at $x=1$. So you are down to a familiar inequality $|x-1|\lt 2$, with the proviso that $x\ne 1$.

Another way: Our inequality holds if and only if $$(1-|x-1|)^2\lt 1.$$ Expand the square, and do some algebraic simplification. We arrive at the inequality $$|x-1|^2 \lt |x-1|.$$ Remembering that we cannot have $x=1$, we can cancel and arrive at $|x-1|\lt 2$, again, familiar territory. This says that the distance of $x$ from $1$ is $\lt 2$. So $-1\lt x\lt 3$, with $x\ne 1$. Alternately, we can give the answer by saying that $x$ satisfies the inequality if $x$ is in the interval $(-1,1)$ or the interval $(1,2)$.

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