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Tchebysheff proof, help understanding transition to last step

  1. Jun 4, 2009 #1
    proof attached as pdf in link provided
     

    Attached Files:

  2. jcsd
  3. Jun 4, 2009 #2

    EnumaElish

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    Using A for |A| and kk for k squared:

    1 > Akk/n
    1 < n/(Akk)
    A < n/kk
    A/n < 1/kk
    1 - A/n > 1 - 1/kk

    So the stated result actually holds with strict equality.
     
  4. Jun 4, 2009 #3
    I actually got some help with the algebraic manipulation, which was clouding my conceptual understanding, after looking at this more I realized what it all meant and why the use of >= instead of > at the end


    Divide by k^2 results in 1/(k^2)>|A|/n

    multiply both sides by -1 (fips inequality) and add 1 to both sides results in 1- 1/k^2 < 1-|A|/n

    change from strict inequality to weak inequality to account for proportion of all elements "~A" INTERSECTION "B" results in
    1- 1/k^2 <= 1-|A|/n
     
    Last edited: Jun 5, 2009
  5. Jun 4, 2009 #4
    But the only way 1 - A/n > 1 - 1/kk describes the proportion of measurements in B, i.e. ~A, is to switch to weak inequalitysince strict inquality leaves out part of B, t/f weak inequality is required.

    What do you think?
     
    Last edited: Jun 5, 2009
  6. Jun 4, 2009 #5

    EnumaElish

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    |A| is a real number, like 2.

    If 2/n < 1/kk then 1 - 2/n > 1 - 1/kk. Not so?
     
  7. Jun 4, 2009 #6
    the cardinality of A, i.e. |A|, is a natural number, an integer in the set {0,1,2,3,...}

    if |A|=1, and k =1 and n = 1 then the equations are equal. thats assuming that there is no restriction on n being equal to 1 and card(A) being equal to 1.
     
    Last edited: Jun 4, 2009
  8. Jun 4, 2009 #7

    EnumaElish

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    Of course, if the premise were 2/n < 1/kk then 1 - 2/n > 1 - 1/kk would be the case. But the premise is stated with strict inequality.
     
  9. Jun 4, 2009 #8
    yes you're right. So the use of weak inequality may have something to do with describing the complement of A?
     
    Last edited: Jun 5, 2009
  10. Jun 4, 2009 #9

    EnumaElish

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    Or what they meant was 2/(n-1) < 1/kk, then they replaced n-1 with n but forgot to change < to <.
     
  11. Jun 5, 2009 #10
    Perhaps that's the case, because the strict inequality was used to derive the last line. Moreover, perhaps my set element argument was more of a way to try and rationalize the use of the weak inequality. I'm going to try and get a response from the author of this proof, and see what he says. Thanks for your responses EnumaElish.
     
  12. Jun 6, 2009 #11
    I corresponded with a professor from a previous class about this and he reminded me about the use of logic in proofs

    If you assume p is true then you may also conclude that ( p OR q) is true, no matter what q is! q doesn't ever have to be true since we've assumed p is. It's a fact in logic referred to as (addition or generalization).

    p is LHS > RHS, and q is LHS = RHS.
     
    Last edited: Jun 6, 2009
  13. Jun 6, 2009 #12

    EnumaElish

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    I am not saying the proof is wrong; I am saying it can be more definitive.
     
    Last edited: Jun 6, 2009
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