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Prove that the following number is an irrational number:
${{\log }_{2}}3$

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Hint: A rational number can always be expressed in the form $\dfrac{p}{q}$ where $p$ and $q$ are integers and $q\ne 0$.

We will use proof by contradiction to prove that ${{\log }_{2}}3$ is an irrational number.
First , we will consider ${{\log }_{2}}3$ to be a rational number.
We know , a rational number is a number which can be written in the form $\dfrac{p}{q}$ where $p$ and $q$ both are integers and $q\ne 0$.
Now , since we have considered ${{\log }_{2}}3$ to be a rational number , so , we can write ${{\log }_{2}}3$ in the form $\dfrac{p}{q}$ where $p$ and $q$ both are integers and $q\ne 0$.
Now , we know ${{\log }_{a}}b=\dfrac{\log b}{\log a}$
So , we can write ${{\log }_{2}}3$ as $\dfrac{\log 3}{\log 2}$.
Now , since we have considered ${{\log }_{2}}3$ to be a rational number, so, $\dfrac{\log 3}{\log 2}$ is also a rational number.
Hence, $\dfrac{\log 3}{\log 2}$ can be written in the form of $\dfrac{p}{q}$ where $p$ and $q$ both are integers and $q\ne 0$.
So, let $\dfrac{\log 3}{\log 2}=\dfrac{p}{q}$.
Now, we will cross multiply the numbers.
On cross multiplication, we get
$q\log 3=p\log 2........\left( i \right)$
Now, we know $\log {{x}^{a}}=a\log x$.
So, we can write $q\log 3$ as \[\log {{3}^{q}}\] and $p\log 2$ as \[\log {{2}^{p}}\].
So, equation $\left( i \right)$ becomes
\[\log {{3}^{q}}=\log {{2}^{p}}\]
\[\Rightarrow {{3}^{q}}={{2}^{p}}\]
Now,
We know, when an even number is raised to an integer power, it always gives an even number (except power $0$) and an odd number raised to an integer power always gives an odd number.
So, ${{3}^{q}}$ can never be equal to ${{2}^{p}}$
So, our assumption that ${{\log }_{2}}3$ is rational was wrong.
Hence, ${{\log }_{2}}3$ is irrational.

Note: Always remember that ${{\log }_{a}}b$ is defined only if \[a\] and \[b\] both are positive and \[a\] is not equal to \[1\]. Generally , this fact is forgotten by students and many questions are answered wrong.
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