Proof that ln(r) is irrational for rational r ≠ 1

1. Clarifying the statement

The claim “the natural log is irrational” needs clarification. We prove the precise statement:

Theorem: If r is a positive rational number and r ≠ 1, then ln(r) is irrational.

2. Restating the theorem

Let r = a/b, where a and b are positive integers, gcd(a,b) = 1, and a ≠ b. We want to show that ln(a/b) is irrational.

Equivalently: ln(a) – ln(b) is irrational for integers a ≠ b.

3. Proof using the Lindemann–Weierstrass theorem

A special case of the Lindemann–Weierstrass theorem states:

If α is a nonzero algebraic number, then e^α is transcendental.

We apply this as follows:

Let α = ln(r), where r is rational, r > 0, and r ≠ 1.

Then e^α = r. Since r is rational, it is algebraic.

If α were algebraic and nonzero, the theorem would imply that e^α = r is transcendental—a contradiction.

Therefore, α cannot be algebraic and nonzero. Since r ≠ 1, α = ln(r) ≠ 0, so α must be transcendental.

All transcendental numbers are irrational, so ln(r) is irrational.

4. Elementary considerations for specific cases

For specific values like ln(2), more elementary proofs (using series expansions and divisibility arguments) can be constructed, but they are intricate. The transcendence-based proof above is both general and elegant.

5. Conclusion

We have proven:

For every rational number r > 0 with r ≠ 1, the natural logarithm ln(r) is irrational.

This follows directly from the Lindemann–Weierstrass theorem: if ln(r) were algebraic and nonzero, then r would be transcendental, contradicting the algebraicity of rational numbers.

Final result: ln(r) is irrational for all rational r > 0, r ≠ 1.