$|x-1| \lt \frac{\textrm{1}}{\textrm{100}}$

Question

$$|x-1| \lt \frac{\textrm{1}}{\textrm{100}}$$

$$x\pm1 \lt \frac{\textrm{1}}{\textrm{100}}$$

$$\frac{\textrm{1}}{\textrm{100}}-1 \lt x \lt \frac{\textrm{1}}{\textrm{100}}+1$$

$$\frac{\textrm{-99}}{\textrm{100}} \lt x \lt \frac{\textrm{101}}{\textrm{100}}$$

but the answer is

$$\frac{\textrm{99}}{\textrm{100}} \lt x \lt \frac{\textrm{101}}{\textrm{100}}$$

i.e. it is +99 instead of -99 on left side of inquality

Where I am wrong?

Answer 1

As an alternative way, by the statement

$$|x-1| \lt \frac1{100}$$

we require that $x$ is far from $1$ less than $\frac1{100}$ that is

$$x_{inf}=1-\frac1{100} \quad x_{sup}=1+\frac1{100}$$

or

$$1-\frac1{100}<x<1+\frac1{100}$$

More in general with $\varepsilon > 0$

$$|x-a|<\varepsilon \iff -\varepsilon<x-a<\varepsilon \iff a-\varepsilon<x<a+\varepsilon$$

Answer 2

Hint :$$|x-1| \lt \frac{\textrm{1}}{\textrm{100}} \Longleftrightarrow -\frac{\textrm{1}}{\textrm{100}}\lt x-1 \lt \frac{\textrm{1}}{\textrm{100}}$$

Answer 3

$$|x-1| \lt \frac{\textrm{1}}{\textrm{100}}$$ is not equivalent to

$$x\pm1 \lt \frac{\textrm{1}}{\textrm{100}}$$ but to

$$-\frac{\textrm{1}}{\textrm{100}} < x- 1 <\frac{\textrm{1}}{\textrm{100}}$$