anonymous
  • anonymous
please help,
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
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
anonymous
  • anonymous
@zzr0ck3r , @Kainui , @oldrin.bataku
anonymous
  • anonymous
@dan815

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anonymous
  • anonymous
@Loser66
anonymous
  • anonymous
@zzr0ck3r , @Kainui , @oldrin.bataku
anonymous
  • anonymous
@Loser6
anonymous
  • anonymous
@dan815
zzr0ck3r
  • zzr0ck3r
what is the first rule of being a metric?
Loser66
  • Loser66
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
zzr0ck3r
  • zzr0ck3r
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. \[K^2-4\ge 0\]
Loser66
  • Loser66
@zzr0ck3r cannot be =0, right?
Loser66
  • Loser66
but at that time, x =y while we are considering \(x\neq y\)
Loser66
  • Loser66
Because it is a piece wise function. |dw:1438908728253:dw|
Loser66
  • Loser66
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.
zzr0ck3r
  • zzr0ck3r
So we have \(K^2-4>0\).
Loser66
  • Loser66
:)
zzr0ck3r
  • zzr0ck3r
I deleted my stuff as not to confuse @GIL.ojei
anonymous
  • anonymous
ok sir, thanks . are the values K<-2 or k>2????
anonymous
  • anonymous
please is that correct sirs?

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