Question

Consider the Riemann zeta function, defined as ζ(s) = 1^(-s) + 2^(-s) + 3^(-s) + ..., where s is a complex number with real part greater than 1. Prove that the Riemann zeta function has no zeros in the critical strip 0 < Re(s) < 1.

Answer 1

To prove that the Riemann zeta function has no zeros in the critical strip 0 < Re(s) < 1, we can use the Euler product formula for the Riemann zeta function.

The Euler product formula states that for any complex number s with real part greater than 1, we have:

ζ(s) = ∏(p prime) (1 - p^(-s))^(-1)

where the product is taken over all prime numbers p.

Now, let's assume that there exists a complex number s = a + bi in the critical strip 0 < Re(s) < 1 such that ζ(s) = 0. This means that the Euler product formula must also be equal to zero.

Since the Euler product formula is a product of terms, in order for the product to be zero, at least one of the terms must be zero. Let's consider a prime number p. If p^(-s) = 1, then the term (1 - p^(-s))^(-1) is equal to 1, and it does not contribute to the product being zero.

On the other hand, if p^(-s) ≠ 1, then (1 - p^(-s))^(-1) is a non-zero complex number. This means that for each prime number p, the term (1 - p^(-s))^(-1) is non-zero.

Since the product of non-zero complex numbers is also non-zero, we can conclude that the Euler product formula is non-zero for any complex number s = a + bi in the critical strip 0 < Re(s) < 1.

Therefore, the Riemann zeta function ζ(s) cannot have any zeros in the critical strip 0 < Re(s) < 1.