Spivak's Calculus Question

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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
$|(x-x_0)+(y-y_0|\le |(x-x_0)|+|(y-y_0|<\frac{e}{2}+\frac{e}{2}=e.$