<s>[INST] A problem regarding inequality There is a question in an algebra textbook that asks to prove $|x^2-1|\lt\frac{21}{100}$, given that $|x-1|\lt\frac{1}{10}$.
Is the question correct(solvable)? If it is, how?
 [/INST] Suppose $|x - 1| < \frac{1}{10}$. Then
$$|x^2 - 1| = |(x-1)(x+1)| = |x-1|\cdot|x+1| < \frac{1}{10} |x+1|$$
We now need to show $|x+1| < \frac{21}{10}$ in order to show $|x^2 - 1| < \frac{21}{100}$. 
Since $|x-1| < \frac{1}{10}$, this means
$$ -\frac{1}{10} < x-1 < \frac{1}{10}.$$
Adding 2 to each side gives us:
$$ -\frac{1}{10} + 2 < x-1+2 < \frac{1}{10} + 2.$$
We can do this without any issues since 2 is positive. Now, add everything together to get:
$$ \frac{19}{10} < x + 1 < \frac{21}{10}.$$
Since $-\frac{21}{10} < \frac{19}{10}$ and $\frac{19}{10} < x+1$, we have by the transitive property that ....
Can you take it from here?
</s>