TAOCP 1.2.2 Exercise 10

Suppose, for the sake of contradiction, that $\log_{10} 2$ is rational.

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Section 1.2.2: Numbers, Powers, and Logarithms

Exercise 10. [18] Prove that $\log_{10} 2$ is not a rational number.

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Solution

Suppose, for the sake of contradiction, that $\log_{10} 2$ is rational. Then we can write

$\log_{10} 2 = \frac{p}{q},$

where $p$ and $q$ are integers with $q > 0$ and $\gcd(p,q) = 1$. By the definition of logarithm, Eq. (9), this implies

$10^{p/q} = 2. \eqno(1)$

Raising both sides of Eq. (1) to the $q$-th power gives

$(10^{p/q})^q = 10^p = 2^q. \eqno(2)$

Equation (2) asserts that $10^p$ equals $2^q$. Factorizing the left-hand side, we have

$10^p = (2 \cdot 5)^p = 2^p \cdot 5^p. \eqno(3)$

Substituting Eq. (3) into Eq. (2) yields

$2^p \cdot 5^p = 2^q. \eqno(4)$

Dividing both sides by $2^p$ gives

$5^p = 2^{q-p}. \eqno(5)$

Equation (5) asserts that a power of 5 equals a power of 2. Since $2$ and $5$ are distinct primes, the fundamental theorem of arithmetic implies that no nonzero powers of 2 and 5 can be equal. Therefore Eq. (5) holds if and only if

$p = 0 \quad \text{and} \quad q - p = 0.$

However, $p = 0$ contradicts the assumption that $\log_{10} 2 = p/q > 0$, since $2 > 1$. Hence no such integers $p$ and $q$ exist.

We conclude that $\log_{10} 2$ cannot be rational. This completes the proof.

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