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anonymous

  • one year ago

please help,

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  1. anonymous
    • one year ago
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    let x be an arbitary non-empty set and p: X*X---> R be discrete metric on x, defined by p(x,y)= k^2 - 4 if x \[\neq y\] and 0 if x=y ... obtain the possible values of k

  2. anonymous
    • one year ago
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    @zzr0ck3r , @Kainui , @oldrin.bataku

  3. anonymous
    • one year ago
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    @dan815

  4. anonymous
    • one year ago
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    @Loser66

  5. anonymous
    • one year ago
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    @zzr0ck3r , @Kainui , @oldrin.bataku

  6. anonymous
    • one year ago
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    @Loser6

  7. anonymous
    • one year ago
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    @dan815

  8. zzr0ck3r
    • one year ago
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    what is the first rule of being a metric?

  9. Loser66
    • one year ago
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    I think if \(x\neq y\) then the distance between them \(\neq 0\) and = k^2 -4 Moreover, the distance is never negative, hence k^2 -4 >0 and k<-2 or k >2

  10. zzr0ck3r
    • one year ago
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    The discrete metric just puts the same distance between all points (think integers). 1st rule of metric club is that we only obtain non negative values. i.e. \[K^2-4\ge 0\]

  11. Loser66
    • one year ago
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    @zzr0ck3r cannot be =0, right?

  12. Loser66
    • one year ago
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    but at that time, x =y while we are considering \(x\neq y\)

  13. Loser66
    • one year ago
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    Because it is a piece wise function. |dw:1438908728253:dw|

  14. Loser66
    • one year ago
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    when k =2, p(x,y) =0 we must use the second piece of the function to work out. Like if it is |dw:1438908882357:dw| 0 stays at only 1 part of the function. Can't stay in both.

  15. zzr0ck3r
    • one year ago
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    So we have \(K^2-4>0\).

  16. Loser66
    • one year ago
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    :)

  17. zzr0ck3r
    • one year ago
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    I deleted my stuff as not to confuse @GIL.ojei

  18. anonymous
    • one year ago
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    ok sir, thanks . are the values K<-2 or k>2????

  19. anonymous
    • one year ago
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    please is that correct sirs?

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