MathematicalPhysicist
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one (and perhaps the only) example is:2^4=4^2
what more examples are there?
what more examples are there?
2! does not = 4!!!!!Originally posted by loop quantum gravity
one (and perhaps the only) example is:2^4=4^2
what more examples are there?
ah, this is from c++ language which means does not equal to (m doesnt equal to n).Originally posted by StephenPrivitera
I think he means to say that 4!=24[x=] 2
Your example 2^4=4^2 doesn't satisfy the condition m!=n (m=4, n=2, 4![x=]2)
Prove it.Well...it is quite simple to prove that m must be equal to k*n...where k is an integer...
Yes, I was to tired to find the [x= symol. 2![x=]4 and 4![x=]2, so how does this work? When you plug the numbers back in it dosn't work.Silly me, I was also thinkg of inregers. Back to learning!Originally posted by bogdan
Well...it is quite simple to prove that m must be equal to k*n...where k is an integer...
n^(k*n)=(n*k)^n
(n^k)^n=(n*k)^n
n^k=n*k...
n^(k-1)=k....
if n=1...k=1...wrong...m=n...
if n=2...2^(k-1)=k...if k>2 then it's wrong...so k=2...m=4...
if n>2...n^(k-1)>2^(k-1)>k...(for k>1...)
So...n=2...m=4...it's the only solution...
can someone explain this? and what the hell is superfactorials used for?
Well, even if they were integers, it is not always possible to express one integer as k*n.Ooops...sorry...I thought m and n were integers...sorry again...stupid me
For those of you who don't care to read below, I put these approximate solutions first (generated by a program I built using a method described below). Because the program I built stores the numbers in a relatively small memory space the answers aren't as good as they could be.Originally posted by loop quantum gravity
so what other solutions are there?
No, other than 2 and 4, they don't exist because one of the natural numbers would have to be between 1 and e (2.718). The only natural number that exists to satisfy this condition is 2.Originally posted by loop quantum gravity
do you have examples of whole numbers?