Contradiction Problem 1


Suppose not. That is, suppose \(\log_3 2\) were rational. Then we would have \(\log_3 2=\frac{a}{b}\) for some integers a and b, with b non-zero.

Since \(\log_3 2=\frac{a}{b}\), we must have \(3^{\log_3 2}=3^{\frac{a}{b}}\). Simplifying the lefthand side gives us \(2=3^\frac{a}{b}\). So \(2^b= (3^\frac{a}{b})^b = 3^a\). But 2 and 3 have no positive factors other than 1, so we can have \(2^b=3^a\) only when \(a=b=0\). This contradicts our assumption that b is non-zero.

Since our assumption that \(\log_3 2\) was rational led to a contradiction, it must be the case that \(\log_3 2\) is irrational. QED.