Proof that log 2 is an irrational number

January 6, 2012 GB High School Mathematics, High School Number Theory

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

Proof:

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.

One comment irrational numbers, log 2, proof by contradiction, rational numbers

Proof that log 2 is an irrational number

Leave a ReplyCancel reply