Problem understanding a proof in Spivak Vol. 4

[SheldonG]

Homework Statement

This is from Spivak, Vol. 4 Page 102-103

Given |x-x_0| < 1, |x-x0| < Epsilon/(2(|y_0|+1))

Also given |y-y_0| < Epsilon/(2(|x_0| + 1))

Prove |xy-x_0y_0| < Epsilon

Homework Equations

See above

The Attempt at a Solution

The proof proceeds clearly enough. Using |x-x_0| < 1, he shows that |x| < |x_0| + 1.

Then

|xy-x_0y_0| = |x(y-y_0) + y_0(x-x_0)|

< |x(y-y_0)| + |y_0(x-x_0)|

< (1+|x_x0|)*Epsilon/(2(|x0|+1)) + |y_0|*Epsilon/(2(|y_0|+1))

= Epsilon/2 + Epsilon/2 = Epsilon

So.. Q.E.D., but I do not understand the second term...

How is |y_0|*Epsilon/(2(|y_0| + 1)) = Epsilon/2 ??

Any help would be most appreciated. This is for self-study, so I am without a teacher.

Thanks,
Shelly

 

[stringy]

That term, in fact, is not equal to [itex]\epsilon/2[/itex], it's merely less than it.

Try verifying the inequality

[tex] |y_0| \cdot \frac{ \epsilon}{2(|y_0|+1)} < \frac{\epsilon}{2}.[/tex]

Spivak's book was my first introduction to rigorous calculus too. In these proofs I remember trying to equate everything (and having it never work). It's quite frustrating at first! Just remember analysis is all about inequalities.

EDIT: Oh! Did you notice that that equal sign in the fourth line of your proof is supposed to be a less-than sign? Maybe that's where you got mixed up...?

 

[SheldonG]

Stringy - thank you *so* much. That is a tremendous help.

Much appreciated!

Shelly