# Not understanding a cancellation step in an inequality proof from Spivak's Calculus.

have just been reading through and doing the questions in Spivak's Calculus, but am not entirely sure I am understanding a step in one of the proofs given in chapter $5$ (but was given as an exercise in the first chapter).

I had found a picture of the question online, but apparently I can't post it here. The proof is on page $89$, chapter $5$, and its for the $2nd$ lemma.

Lemma: If $|x-x_0|<\min\left(1,\frac{\varepsilon}{2(|y_0|+1)}\right)$ and $|y-y_0|<\frac\varepsilon{2(|x_0|+1)}$, then $|xy-x_0y_0|<\varepsilon$

Proof: $$|xy-x_0y_0|=|x(y-y_0)+y_0(x-x_0)|\le|x||y-y_0|+|y_0||x-x_0|$$ then \begin{align}&|x||y-y_0|+|y_0||x-x_0|<(1+|x_0|)\frac\varepsilon{2(|x_0|+1)}+|y_0|\frac{\varepsilon}{2(|y_0|+1)}\\ &=\frac\varepsilon2+\frac\varepsilon2=\varepsilon\end{align}

In the second to last line of the proof, I don't understand how the $\dfrac{\varepsilon \lvert y_0\rvert}{2(\lvert y_0\rvert+1)}$ is meant to cancel to $\dfrac\varepsilon2$.

Did they use $\dfrac{\lvert y_0\rvert}{\lvert y_0\rvert+1}<1$ so $\dfrac{\varepsilon \lvert y_0\rvert}{2(\lvert y_0\rvert+1)}<\dfrac\varepsilon2$? But if they did something like this, shouldn't that last line then have a "$<$" instead of a "$=$"?

• Please edit your question to make it self-contained - not all of us have a copy of Spivak handy (nor the time to root through it to find what proof you're talking about).
– user296602
Commented Jan 31, 2016 at 4:47
• Sorry, I had a picture of the proof, to save me having to go through and write it all out, especially as Im not familiar with latex, or any similar math writing styles. However apparently Im not allowed to attach pictures, so I've just now written out the question and the given proof best I can. Commented Jan 31, 2016 at 4:54
• The expression is $(1+|x_0|)(e/2(|x_0|+1) = \dfrac{(1+|x_0|)e}{2(|x_0|+1)} = \dfrac e2$ Commented Jan 31, 2016 at 5:00

You are right, it should be $<\frac{\varepsilon}{2} + \frac{\varepsilon}{2}$ rather than $=\frac{\varepsilon}{2} + \frac{\varepsilon}{2}$.