# Spivak's Calculus Question

• July 11th 2011, 07:35 AM
RogueDemon
Spivak's Calculus Question
Prove that if $|x - x_0| < e/2$ and $|y - y_0| < e/2$, then

$|(x + y) - (x_0 + y_0)| < e$,
$|(x - y) - (x_0 - y_0)| < e$.

I'm not looking for a complete solution, just a hint as to where to begin.

Thanks.
• July 11th 2011, 08:07 AM
Plato
Re: Spivak's Calculus Question
Quote:

Originally Posted by RogueDemon
Prove that if $|x - x_0| < e/2$ and $|y - y_0| < e/2$, then
$|(x + y) - (x_0 + y_0)| < e$,
$|(x - y) - (x_0 - y_0)| < e$.

just a hint as to where to begin.

$(x+y)-(x_0+y_0)=(x-x_0)+(y-y_0)$
• July 11th 2011, 08:48 AM
RogueDemon
Re: Spivak's Calculus Question
Will the following identity be of any use in completing the proof?

$|(x - x_0) + (y - y_0)| <= |x - x_0| + |y - y_0|$
• July 11th 2011, 08:51 AM
Plato
Re: Spivak's Calculus Question
Quote:

Originally Posted by RogueDemon
Will the following identity be of any use in completing the proof?

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

That is the only to do it. That is the triangle inequality.
• July 11th 2011, 09:57 AM
RogueDemon
Re: Spivak's Calculus Question
This is what I've got so far:

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

$|x + y - x_0 - y_0| < e$

$|x - x_0 + y - y_0| < e$

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

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

$|x - x_0| < e/2$

$2|x - x_0| < e$

$|x - x_0| + |x - x_0| < e$

$|x - x_0| < e - |x - x_0|$

$|x - x_0| + |y - y_0| < e - |x - x_0| + |y - y_0|$

$|(x - x_0) + (y - y_0)| <= |x - x_0| + |y - y_0| < e - |x - x_0| + |y - y_0|$

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

Any idea as to how to turn the $e - |x - x_0| + |y - y_0|$ into a simple $e$? Am I even on the right track?
• July 11th 2011, 10:13 AM
Plato
Re: Spivak's Calculus Question
Given that $|x-x_0|<\frac{e}{2}~\&~|y-y_0|<\frac{e}{2}$ then
$|(x-x_0)+(y-y_0|\le |(x-x_0)|+|(y-y_0|<\frac{e}{2}+\frac{e}{2}=e.$