Question

Prove that $\sqrt{3}$ is an irrational number. Hence, show that $7+2\sqrt{3}$ is also an irrational number.

Answer 1

Step 1: Proving $\sqrt{3}$ is an irrational number by contradiction.

Assume that $\sqrt{3}$ is a rational number then it can be written in the form of $\frac{p}{q}andq\ne 0$.

Let $\sqrt{3}=\frac{p}{q}$, here $p\mathrm{and}q$ are integers with $q\ne 0\mathrm{and}\mathrm{HCF}\left(p,q\right)=1$.

Square on both sides of the equation:

$$\begin{gathered} 3=\frac{p^2}{q^2} \\ \Rightarrow \frac{p^2}{3}=q^2...\left(1\right) \end{gathered}$$

$\because p^2$ is divisible by 3.

$\therefore p$ is also divisible by 3. …(2)

So, it can be assumed that $p=3m$ for some natural number $m$. Substitute the value of $p$ in equation (1):

$$\begin{gathered} {\left(3m\right)}^2=3q^2 \\ \Rightarrow 9m^2=3q^2 \\ \Rightarrow 3m^2=q^2 \\ \Rightarrow \frac{q^2}{3}=m^2 \end{gathered}$$

$\because q^2$ is divisible by 3.

$\therefore q$ is divisible by 3. …(3)

From the statement (2) and (3), 3 is a factor of $p\mathrm{and}q$, which is impossible as we had assumed $\mathrm{HCF}\left(p,q\right)=1$. Thus, our assumption leads us to a contradictory conclusion; therefore, it must be wrong. Hence, $\sqrt{3}$ is an irrational number.

Step 2: Proving $7+2\sqrt{3}$ is an irrational number

It is proved that $\sqrt{3}$ is an irrational number. On contrary, assume that $x=7+2\sqrt{3}$ is a rational number.

Isolate $\sqrt{3}$ on the left side of the equation:

$$\begin{gathered} x-7=2\sqrt{3} \\ \Rightarrow \sqrt{3}=\frac{x-7}{2}...\left(4\right) \end{gathered}$$

Observe that $\frac{x-7}{2}$ is also a rational number as the subtraction and the division of two rational numbers gives a rational number by the properties of rational numbers. Thus, according to equation (4), $\sqrt{3}$ must be a rational number.

Since it is already proved that $\sqrt{3}$ is an irrational number, it can be concluded that our assumption leads us to a contradictory conclusion. Hence, it must be wrong.

Final Answer:

Therefore, it is proved that $\sqrt{3}$ is an irrational number. Hence, $7+2\sqrt{3}$ is an irrational number.