Question

Prove that $2+\sqrt{3}$ is an irrational number.

Answer 1

Step-1 Definition of rational number:

Let us assume $2+\sqrt{3}$ to be a rational number.

For a number to be consider a rational number, it must be able to be expressed in the $\frac{p}{q}$ form where,

1. $p,q$ are integers.
2. $q\ne 0$
3. $p,q$ are co-primes.(no other factor than $1$)

Expressing $2+\sqrt{3}$ in the $\frac{p}{q}$ form

$\Rightarrow$$2+\sqrt{3}=\frac{p}{q}$

$\Rightarrow$ $\sqrt{3}=\frac{p}{q}-2$

$\Rightarrow$ $\sqrt{3}=\frac{p-2q}{q}$

Step-2 : Prove of $\sqrt{3}$is an irrational number:

Let, $\sqrt{3}=\frac{r}{t},$where $r$and $t$ are intergers($t\ne 0$) and co-prime.

Square both sides

$$\begin{gathered} 3=\frac{r^2}{t^2} \\ \Rightarrow r^2=3t^2...\left(i\right) \end{gathered}$$

It means $r^2$ is divisible by $3$,so we can say $r=3k$

Now, Put value $r=3k$ in equation (i)

$$\begin{gathered} {\left(3k\right)}^2=3t^2 \\ \Rightarrow 9k^2=3t^2 \\ \Rightarrow t^2=3k^2 \end{gathered}$$

It means $t^2$ is divisible by $3$,so we can say $t=3m$

We get $r=3k$ and $t=3m$ which give both $r$ and $t$ are divisible by $3$

But our assumptions is that $r$ and $t$ are co-prime.

Which is contradiction of our statement.

Hence $\sqrt{3}$ is an irrational number.

But $\sqrt{3}$ is an irrational number and $\frac{p-2q}{q}$ is a rational number as $p,q$ are integers.

A rational number can not be equal to an irrational number.

Hence, this contradicts our assumption that $2+\sqrt{3}$ is a rational number.

Hence, it is proven that $2+\sqrt{3}$ is an irrational number.