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{\Large \textbf{The number} {\mathversion{bold}$\log_2 3$} \textbf{is irrational}}

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\textbf{Proof by contradiction:} Assume $\log_2 3$ is rational. Choose $a, b \in N^{+}$, $\log_2 3 = \frac{a}{b}$, such that \frac{a}{b} is in lowest terms.

\[\log_2 3 = \frac{a}{b}\]
\[3 = 2^\frac{a}{b}\]
\[3^b = 2^a\]

Since $3^b$ is always an odd number, and $2^a$ is always an even number, we have that an odd number is equal to an even number. This is a contradiction.
\\Thus, the hypothesis is true: $\log_2 3$ is irrational. 


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