# 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.

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.

Last updated date: 30th Sep 2023

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