Logarithm Base 10 of 2 is Irrational
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Theorem
- $\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
is irrational.
Proof 1
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.
$\blacksquare$
Proof 2
Because $5$ is a divisor of $10$ but not $2$, it cannot be the case that $2^a = 10^b$ for $a, b \in \Z_{>0}$.
Hence this is a special case of Irrationality of Logarithm.
$\blacksquare$
Sources
- 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $10$