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Proof of Inequality

by objectivesea
Tags: inequalities, proof, real analysis
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objectivesea
#1
May7-11, 08:20 PM
P: 6
1. The problem statement, all variables and given/known data

Prove that if [tex] |x-x_0| < \textrm{min} \bigg ( \frac{\epsilon}{2|y_0|+1},1 \bigg )[/tex] and [tex]|y-y_0| < \frac{\epsilon}{2|x_0|+1}[/tex] then [tex]xy-x_0y_0<\epsilon[/tex]

2. Relevant equations
We can use basic algebra and the following axioms:
For any number [tex]a[/tex], one and only one of the following holds:
(i) [tex]a=0[/tex]
(ii) [tex]a[/tex] is in the collection [tex]P[/tex]
(iii) [tex]-a[/tex] is in the collection [tex]P[/tex]
Note: A number [tex]n[/tex] is the collection [tex]P[/tex] if and only if [tex]n>0[/tex]

If [tex]a[/tex] and [tex]b[/tex] are in [tex]P[/tex], then [tex]a+b[/tex] is in [tex]P[/tex].

If [tex]a[/tex] and [tex]b[/tex] are in [tex]P[/tex], then [tex] a \cdot b [/tex] is in P.

We may also use the following consequences of the above axioms:

For any numbers [tex]a[/tex]and [tex]b[/tex], one and only one of the following holds:
(i) [tex]a=b[/tex]
(ii) [tex]a < b[/tex]
(iii)[tex] a > b[/tex]

For any numbers [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], if [tex]a<b[/tex] and [tex]b<c[/tex], then [tex]a<c[/tex].

For any numbers [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], if [tex]a<b[/tex], then [tex]a+c<b+c[/tex].

For any numbers [tex]a[/tex], [tex]b[/tex], and [tex]c[/tex], if [tex]a<b[/tex] and [tex]0<c[/tex], then [tex]ac<bc[/tex].


3. The attempt at a solution

I'm not even sure where to start.
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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praharmitra
#2
May8-11, 03:02 AM
P: 311
Are you sure it's not
[tex]
|y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )
[/tex]?
objectivesea
#3
May8-11, 01:39 PM
P: 6
Quote Quote by praharmitra View Post
Are you sure it's not
[tex]
|y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )
[/tex]?
I just double checked. I'm sure.

objectivesea
#4
May8-11, 01:48 PM
P: 6
Proof of Inequality

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

either way [tex]|y-y_0|< \frac {\epsilon}{2|x_0|+1}[/tex]
praharmitra
#5
May8-11, 09:02 PM
P: 311
No, I actually want the fact that [tex]|y-y_0|<1[/tex] in addition to all your above expressions. Now I am wondering, given your first inequality, is it possible to prove that

[tex]\frac{\epsilon}{2|x_0|+1} <1[/tex]. Do you think you can try to prove that?

If you can do that, then afterwards use the following:

1. [itex] min(a,b) \leq a,~min(a,b)\leq b[/itex]

2. [itex]|x-x_0| \geq |x|-|x_0| [/itex]

and proceed from there. But first try the first thing. Something I am not able to do.


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