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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?

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    $\begingroup$ $x-1= \frac{1}{10 }$;$ x+1=\frac{21}{10 } $;$x^2-1= (x-1)(x+1)$ $\endgroup$ – N.S.JOHN Oct 11 '16 at 17:09
  • $\begingroup$ Hint: Write $x=1+\epsilon$ so that $|\epsilon|<\frac 1{10}$. Then remark that $x^2=1+2\epsilon+\epsilon^2$. $\endgroup$ – lulu Oct 11 '16 at 17:13
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Make the change of variables $y=x-1$. We have

$$|y|<\frac{1}{10}$$

and we want to prove that

$$|y||y+2| = |y(y+2)|<\frac{21}{100}$$

Can you take it from here?

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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?

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  • $\begingroup$ Are 19/10<x+1<21/10 and -21/10<x+1<21/10 the same? Don't they represent different sizes of interval? $\endgroup$ – Bernhard Listing Oct 14 '16 at 19:20
  • $\begingroup$ They aren't necessarily the same, but being greater than $\frac{19}{10}$ also means you are greater than $-\frac{21}{10}$, which will lead you to $x+1$ being greater than $-\frac{21}{10}$ as well. Yes they represent different sizes of intervals, but that doesn't matter in this context. $\endgroup$ – josh Oct 15 '16 at 1:16

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