Logarithm Base 10 of 2 is Irrational/Proof 1

Theorem

The common logarithm of $2$:

$\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$

is irrational.

Proof

Aiming for a contradiction, suppose $\log_{10} 2$ is rational.

Then:

\(\ds \log_{10} 2\)

\(=\)

\(\ds \frac p q\)

for some $p, q \in \Z_{\ne 0}$

\(\ds \leadsto \ \ \)

\(\ds 2\)

\(=\)

\(\ds 10^{p / q}\)

Definition of General Logarithm

\(\ds \leadsto \ \ \)

\(\ds 2^q\)

\(=\)

\(\ds 10^p\)

raising both sides to the power of $q$

Both $10^p$ and $2^q$ are integers, by construction.

But $10^p$ is divisible by $5$, while $2^p$, which has only $2$ as a prime factor, is not.

So $10^p \ne 2^q$.

So, by Proof by Contradiction, it follows that $\log_{10} 2$ is irrational.

$\blacksquare$