Authors: Gang tae geuk
Riemann hypothesis means that satisfying ζ(s)=0(ζ(s) means Riemann Zeta function) unselfevidenceable root's part of true numbers are 1/2. Dennis Hejhal, and John Dubisher explained this hypothesis to : "Choosed Any natural numbers(exclude 1 and constructed with two or higher powered prime numbers) then the probability of numbers that choosed number's forming prime factor become an even number is 1/2." I'll prove this explain to prove Riemann hypothesis indirectly. In binomial coefficient, C(n,0)+C(n+1)+...+C(n,n)=2^n. And C(n,1)+C(n,3)+C(n,5)+...+C(n,n) and C(n,0)+C(n,2)+C(n,4)+...+C(n,n) is 2^(n-1). If you pick up 8 prime numbers, then you can make numbers that exclude 1 and constructed with two or higher powered prime numbers, and the total amount of numbers that you made is 2^8. Same principle, if you pick the numbers in k times(k is a variable), the total amount of numbers you made is C(8,k). If k is an even number, the total amount of numbers you can make is C(8,0)+C(8,2)+...+C(8,8)-1(because we must exclude 1,same for C(8,0)), and as what i said, it equals to 2^(8-1)-1. So, the probability of the numbers that forming prime factor's numbers is an even number is 2^(8-1)-1/2^8 If there are amount of prime numbers exist, and we say that amount to n(n is a variable, as the k so), and sequence of upper works sameas we did, so the probability is 2^(n-1)/2^n. If you limits n to inf, then probability convergents to 1/2. This answer coincident with the explain above, so explain is established, same as the Riemann hypothesis is.
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