A non-equality and an inequality involving $y$ and $y_0$ from Spivak Calculus 4th ed. It's Problem 22. from Chapter 1.
I'm given:
$y_0 \neq 0$
$|y - y_0| < \frac{|y_0|}{2}$
$|y - y_0| < \frac{\epsilon|y_0|^2}{2}$
and I must use them to prove that:
$y \neq 0$
$|\frac{1}{y} - \frac{1}{y_0}| < \epsilon$
I haven't really ever done proof based problems before this book, so I'm having a little hard time.I can do the problems once I get the general direction, but I'm not sure how exactly to start.Can someone offer a hint where to start?I'm also confused by this:
$|y - y_0| < \frac{\epsilon|y_0|^2}{2} , |y - y_0| \geq 0 => \frac{\epsilon|y_0|^2}{2} > 0  => \epsilon > 0, y_0 > 0$
(you can just use the $\frac{|y_0|}{2}$ part, but the 2nd part also tells you about $\epsilon$)
However $y_0 \neq 0$ is already granted at the start, even tho it's obviously the first thing you notice, so does it mean the author is hinting at something with this?
 A: PROOF:
Step 1: $|y| > \frac{|y_{0}|}{2}$, since $|y|-|y_{0}| < |y-y_{0}| < \frac{|y_{0}|}{2}$.
Step 2: $|\frac{y - y_{0}}{y\cdot y_{0}}|<\frac{|y-y_{0}|}{(\frac{|y_0|^{2}}{2})} < \varepsilon\cdot \frac{(\frac{|y_{0}|^{2}}{2})}{(\frac{|y_{0}|^{2}}{2})} = \varepsilon$.
I hope this helps.
A: You are given that $\varepsilon > 0$ and $|y - y_0| < \min(|y_0|/2,\varepsilon|y_0|^2/2)$, and asked to prove that $\varepsilon$ is greater than $|1/y - 1/y_0|$. A useful first step would be to transform $|1/y - 1/y_0|$ into an expression that's easier to work with, given the assumptions. To this end, note that $$\left|\frac{1}{y} - \frac{1}{y_0}\right| = \left|\frac{y_0 - y}{y\cdot y_0}\right| = \frac{|y-y_0|}{|y|\cdot|y_0|} = \frac{1}{|y|} \cdot \frac{1}{|y_0|} \cdot |y - y_0|$$
We already have a bound on $|y - y_0|$ involving $\varepsilon$, namely $|y - y_0| < \varepsilon|y_0|^2/2$. Thus, it remains to find an appropriate bound on $1/|y|$. We note that if $|y - y_0| < \varepsilon|y_0|^2/2$, then in order to prove the desired result, we require that $1/|y| < 2/|y_0|$. I will leave this proof as an exercise, but note that you should use the fact that $|y - y_0| < |y_0|/2$ in conjunction with the reverse triangle inequality and the fact that $a > b$ implies $a^{-1} < b^{-1}$ for all positive real $a$ and $b$. I hope this answer was helpful.
