Proof that log 2 is an irrational number

Before doing the proof, let us recall two things: (1) rational numbers are numbers that can be expressed as \frac{a}{b} where a and b are integers, and b not equal to 0; and (2) for any positive real number y, its logarithm to base 10 is defined to be a number x such that 10^x = y. In proving the statement, we use proof by contradiction.

Theorem: log 2 is irrational


Assuming that log 2 is a rational number. Then it can be expressed as \frac{a}{b} with a and b are positive integers (Why?).  Then, the equation is equivalent to 2 = 10^{\frac{a}{b}}. Raising both sides of the equation to b, we have 2^b = 10^a. This implies that 2^b = 2^a5^a.  Notice that this equation cannot hold (by the Fundamental Theorem of Arithmetic) because 2^b is an integer that is not divisible by 5 for any b, while 2^a5^a is divisible by 5. This means that log 2 cannot be expressed as \frac{a}{b} and is therefore irrational which is what we want to show.