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There's no shortage of questions Famine could ask.
Depends on the value of n.
If n>0 || n<=-2, then n^n>n.
If n>-2 && n<0 && n!==-1, then n^n==Oh_crap
If n==0 || n==-1, then n^n==n.
But what if n<=-2, then you can get something imaginary, so is an imaginary number bigger than a real number?
(e.g. n = -3.5)
*Edit - I though it was <><> for infinity, not n
Though they have the potential to be the same, which is greater? n or n^n?
Depends on the value of n.
If n>0 || n<=-2, then n^n>n.
If n>-2 && n<0 && n!==-1, then n^n==Oh_crap
If n==0 || n==-1, then n^n==n.
but n can be nything. That is the mystery...
In that case, n is neither less, equal to, or more than n^n.
But what if n<=-2, then you can get something imaginary, so is an imaginary number bigger than a real number?
(e.g. n = -3.5)
Oh, you’re right. Let’s revise the formulae:But what if n<=-2, then you can get something imaginary, so is an imaginary number bigger than a real number?
....
I do believe that only a select few of you know exactly what the hell you are doing. Do you realize that if "n>0 && n!==1, then n^n>n" is not a valid equation? And do you realize that if "n==0 || n==1, then n^n==n", then the very existence of this universe is impossible?![]()