Logarithm Base 10 of 2 is Irrational/Proof 1
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Theorem
- $\log_{10} 2 \approx 0.30102 \, 99956 \, 63981 \, 19521 \, 37389 \ldots$
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.
$\blacksquare$