- #1
Seydlitz
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- 4
Homework Statement
Prove that if
##\left |x-x_{0} \right | < \frac{\varepsilon }{2}## and ##\left |y-y_{0} \right | < \frac{\varepsilon }{2}##
then
##|(x+y)-(x_0+y_0)| < \varepsilon ## and ##|(x-y)-(x_0-y_0)| < \varepsilon ##
Homework Equations
Postulate and proof with real numbers as well as inequalities.
The Attempt at a Solution
First we understand that the first two statements imply the existence of ##\varepsilon ## such that
##\frac{\varepsilon }{2} - |x-x_0| > 0##
##\frac{\varepsilon }{2} - |y-y_0| > 0##
Hence we can add the two statements.
##\varepsilon - |x-x_0| - |y-y_0| > 0##
##|x-x_0| + |y-y_0| < \varepsilon ##
Considering Triangle Inequality
##|a+b| \leq |a|+|b|##
##|x-x_0+y-y_0| \leq |x-x_0| + |y-y_0| < \varepsilon##
Rearranging
##|(x+y)-(x_0+y_0)| \leq |x-x_0| + |y-y_0| < \varepsilon##
Considering the inequality we can therefore say
##|(x+y)-(x_0+y_0)| < \varepsilon##
The similar reasoning can be done to get
##|(x-y)-(x_0-y_0)| < \varepsilon##
The proof is done as required. ##\blacksquare##
I have not read the chapter about limit yet, since this is taken from Spivak's Calculus problem in prologue chapter. So I might miss some important fact or theorem.
I also hope you guys can give me advice on how to make the above proof smoother or up to the rigorous standard, and additionally to write with LaTex properly. I've spend almost 1 hour just to transfer what I've done in paper to computer. I know I can use my drawing stylus, but I decided to try Latex in spite of that.
Thank You