Question

Show that${\log }_{2}3$ is an irrational number. Recall that an irrational number is a real number $x$that cannot be written as the ratio of two integers.

Short answer

${\log }_{2}3$ is an irrational number.

Step-by-step solution

Step 1

DEFINITIONS

An integer $p$ larger than $x$ is called prime if the only positive factors of are 1 and $p$.

An integer that is not prime is called composite.

Step 2

To prove: ${\log }_{2}3$ is an irrational number

PROOF BY CONTRADICTION

Let us assume that ${\log }_{2}3$ is a rational number, then there exist integers $p$and $q$with $q\ne 0$ such that:

${\log }_{2}3=\frac{p}{q}$

Let us take the exponential with base 2 of each side of the previous equation:

$3=2^{\raisebox{1ex}{$p$}\!\left/ \!\raisebox{-1ex}{$q$}\right.}$

Let us take the $qth$ power of each side of the previous equation:

$3^p=2^p$

As 2 and 3 are prime numbers, $3^q$and $2^p$ are prime factorization of the corresponding numbers. It is then not possible for $p$ and $q$ to be nonzero, since the prime factorization is unique for each number and thus $p=q=0$

However, we knew $q\ne 0$ and thus we obtained a contradiction. Our assumption that *${\log }_{2}3$ is a rational number* is then false and thus ${\log }_{2}3$ is an irrational number.