Prove Inequality: |x-x_0|, |y-y_0| & xy-x_0y_0<\epsilon

[objectivesea]

Homework Statement

Prove that if |x-x_0| &lt; \textrm{min} \bigg ( \frac{\epsilon}{2|y_0|+1},1 \bigg ) and |y-y_0| &lt; \frac{\epsilon}{2|x_0|+1} then xy-x_0y_0&lt;\epsilon

Homework 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&gt;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 aand b, one and only one of the following holds:
(i) a=b
(ii) a &lt; b
(iii)a &gt; b

For any numbers a, b, and c, if a&lt;b and b&lt;c, then a&lt;c.

For any numbers a, b, and c, if a&lt;b, then a+c&lt;b+c.

For any numbers a, b, and c, if a&lt;b and 0&lt;c, then ac&lt;bc.

The Attempt at a Solution

I'm not even sure where to start.

 

[praharmitra]

Are you sure it's not
<br /> |y-y_0| &lt; \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )<br />?

 

[objectivesea]

Are you sure it's not
<br /> |y-y_0| &lt; \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )<br />?

I just double checked. I'm sure.

 

[objectivesea]

I just double checked. I'm sure.

But if |y-y_0|&lt;\textrm{min} \bigg ( \frac {\epsilon}{2|x_0|+1},1 \bigg )
then |y-y_0|&lt; \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|&lt;\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|&lt;1 \leq \frac {\epsilon}{2|x_0|+1}\

either way |y-y_0|&lt; \frac {\epsilon}{2|x_0|+1}

 

[praharmitra]

No, I actually want the fact that |y-y_0|&lt;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} &lt;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.