# Proof of Inequality

by objectivesea
Tags: inequalities, proof, real analysis
 P: 6 1. The problem statement, all variables and given/known data Prove that if $$|x-x_0| < \textrm{min} \bigg ( \frac{\epsilon}{2|y_0|+1},1 \bigg )$$ and $$|y-y_0| < \frac{\epsilon}{2|x_0|+1}$$ then $$xy-x_0y_0<\epsilon$$ 2. Relevant equations We can use basic algebra and the following axioms: For any number $$a$$, one and only one of the following holds: (i) $$a=0$$ (ii) $$a$$ is in the collection $$P$$ (iii) $$-a$$ is in the collection $$P$$ Note: A number $$n$$ is the collection $$P$$ if and only if $$n>0$$ If $$a$$ and $$b$$ are in $$P$$, then $$a+b$$ is in $$P$$. If $$a$$ and $$b$$ are in $$P$$, then $$a \cdot b$$ is in P. We may also use the following consequences of the above axioms: For any numbers $$a$$and $$b$$, one and only one of the following holds: (i) $$a=b$$ (ii) $$a < b$$ (iii)$$a > b$$ For any numbers $$a$$, $$b$$, and $$c$$, if $$a  P: 311 Are you sure it's not [tex] |y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )$$?
P: 6
 Quote by praharmitra Are you sure it's not $$|y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )$$?
I just double checked. I'm sure.

P: 6
Proof of Inequality

 Quote by objectivesea I just double checked. I'm sure.
But if $$|y-y_0|<\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )$$
then $$|y-y_0|< \frac {\epsilon}{2|x_0|+1}$$
Since if $$\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )=\frac {\epsilon}{2|x_0|+1}$$ then $$|y-y_0|<\frac {\epsilon}{2|x_0|+1}\leq 1$$.
And if $$\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )=1$$ then $$|y-y_0|<1 \leq \frac {\epsilon}{2|x_0|+1}\$$

either way $$|y-y_0|< \frac {\epsilon}{2|x_0|+1}$$
 P: 311 No, I actually want the fact that $$|y-y_0|<1$$ in addition to all your above expressions. Now I am wondering, given your first inequality, is it possible to prove that $$\frac{\epsilon}{2|x_0|+1} <1$$. Do you think you can try to prove that? If you can do that, then afterwards use the following: 1. $min(a,b) \leq a,~min(a,b)\leq b$ 2. $|x-x_0| \geq |x|-|x_0|$ and proceed from there. But first try the first thing. Something I am not able to do.

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