Answers

2016-04-26T13:08:29+05:30
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.
1 3 1