[Guest]

prove that the log_e(x) for any integer x other than 1 is irrational.

[Guest]

If ln(x) is rational, then

ln(x) = a/b for integers a and b

b ln(x) = a

ln(x^b) = a

x^b = e^a

Let c = x^b

e^a - c = 0

It is well known that e is transcendental and therefore cannot be the solution to any non-constant polynomial equation with rational coefficients.

In this case, we consider an a'th order polynomial with leading coefficient 1, all other coefficients zero except the constant term which is -c.

Because x and b are integers c is rational.

Because e is transcendental it cannot be expressed by such a polynomial, which contradicts our assertion that ln(x) = a/b

Hence ln(x) is irrational for any positive integer x, except the trivial case where x=1. For that case, asserting x=1 immediately implies that a is 0 and our polynomial equation is a constant polynomial equation, which the transcendental property says nothing about.

Edited March 20, 2010 by mmiguel1