Let a = 2 and b = log_2(pi), so that pi=2^b. If b were rational, then pi would be algebraic. Thus, b is irrational.

## pi = 2^b, b irrational

Let a = 2 and b = log_2(pi), so that pi=2^b. If b were rational, then pi would be algebraic. Thus, b is irrational.