# Non-calc problem taken out from Spivak's Calculus, I reach a dead end

1. Nov 18, 2013

### Hivoyer

1. The problem statement, all variables and given/known data

The problem states that:

$$y_0 \neq 0$$
$$|y - y_0| < \frac{|y_0|}{2}$$
$$|y - y_0| < \frac{\epsilon|y_0|^2}{2}$$

And I am supposed to use these to prove that:

$$y \neq 0$$
$$|\frac{1}{y} - \frac{1}{y_0}| < \epsilon$$

2. Relevant equations

$$|a| - |b| \leq |a - b|$$
(Reverse Triangle Inequality - my professor hinted me to use it)

3. The attempt at a solution

Ok proving the first part was easy, I simply used the reverse triangle inequality like this:

1.I change $$|y - y_0|$$ to $$|y_0 - y|$$
2.I use the inequality: $$|y_0 - y| \geq |y_0| - |y| < \frac{|y_0|}{2} => -|y| < \frac{|y_0|}{2} - |y_0| => -|y| < -\frac{|y_0|}{2} => |y| > \frac{|y_0|}{2}$$ and since we know $$y_0$$ is greater than 0, it's square and it's fractions are also greater than 0, so this means $$y$$ is greater than 0.

Now for part 2 I played around with $$|\frac{1}{y} - \frac{1}{y_0}|$$ and even after transforming it by completing the fraction addition: $$|\frac{y_0 - y}{yy_0}|$$ I couldn't figure out how to proceed.Could someone help for this second part?

2. Nov 18, 2013

### Ray Vickson

For $v > 0$ the inequality $|u| < v$ is the same as $-v < u < v$. Now just consider the two cases $y_0 \geq 0$ and $y_0 < 0$.

3. Nov 18, 2013

### Hivoyer

I thought $$y_o$$ could not be equal to 0.You mean $$y_0 > 0$$ instead of $$y_0 \geq 0$$ right?

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