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Proof of Inequality

  1. May 7, 2011 #1
    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
     
  2. jcsd
  3. May 8, 2011 #2
    Are you sure it's not
    [tex]
    |y-y_0| < \textrm{min} \bigg (\frac{\epsilon}{2|x_0|+1},1 \bigg )
    [/tex]?
     
  4. May 8, 2011 #3
    I just double checked. I'm sure.
     
  5. May 8, 2011 #4
    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]
     
  6. May 8, 2011 #5
    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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