Question

Prove that if\[|x - x_0| < \frac{\epsilon}{2}\]and \[|y - y_0| < \frac{\epsilon}{2}\] then \[|(x+y)-(x_0 + y_0)| < \epsilon\]\[|(x-y)-(x_0 - y_0)| < \epsilon\]

Answer 1

Can we assume the triangle inequality here?

Answer 2

If so, then
\[ |(x+y) - (x_0 + y_0)| = |(x-x_0) + (y-y_0)| < |x-x_0| + |y-y_0| = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\]
and
\[ |(x-y) - (x_0- y_0)| = |(x-x_0) - (y-y_0)| < |x-x_0| + |y-y_0| = \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon\]

Answer 3

Ah crap. Sorry, those lines are too long....the end like they look like they end :)

Answer 4

And also those equals signs before the epsilon / 2 's should be less than signs... o.0

Answer 5

what's the triangle inequality?

Answer 6

It has two parts...
\[|a+b| < |a| + |b|\]
and
\[|a+b| > |a| - |b| \text{ and } |b| - |a|\]

Answer 7

It's called the triangle inequality because if you imagine a and b to be vectors, then a, b, and a+b form a triangle. It is clear that a+b, the resultant, cannot exceed the algebraic sum of the lengths of a and b, nor can it be less than the difference.

Answer 8

Strictly speaking those should be less than or equal to signs, too.