Chapter 6.3, Problem 1E

To determine

To show that $\sqrt{3}$ is irrational.

Explanation of Solution

Given information: The given radical is $\sqrt{3}$ .

Proof: In order to prove that $\sqrt{3}$ is irrational, assume that $\sqrt{3}$ is rational. If $\sqrt{3}$ is rational then it can be written in the form $\frac{p}{q}$ where $p,q$ belongs to integers and $q$ is non-zero.

  $\begin{array}{c}\therefore \frac{p}{q}=\sqrt{3}\mathrm{......}\left(\begin{array}{l}\text{Assumption that }\\ \sqrt{3}\text{}\text{is rational}\end{array}\right)\\ p=\sqrt{3}q\\ p^2=3q^2\mathrm{......}\left(1\right)\\ \text{or, }{q}^2=\frac{p^2}{3}\mathrm{......}\left(2\right)\end{array}$

From (2) $\frac{p^2}{3}$ implies that $3$ divides $p^2$ , therefore $3$ will also divide $p$ . Now the prime number 3 divides $p$ which means,

  $\begin{array}{c}\frac{p}{3}=r\mathrm{.....}\left(\begin{array}{l}\text{where }\left(r\right)\text{is }\\ \text{another integer}\end{array}\right)\\ p=3r\end{array}$

Plugging for $p=3r$ in (1) gives,

  $\begin{array}{c}{\left(3r\right)}^2=3q^2\\ 9r^2=3q^2\\ r^2=\frac{q^2}{3}\mathrm{....}\left(3\right)\end{array}$

Again from (3) $\frac{q^2}{3}$ implies that $\left(q^2\right)$ is divisible by $3$ . Hence $\left(q\right)$ is also divisible by $3$ .

Hence, summarized it can be said that both $p\text{ and }q$ are divisible by $3$ which means the common factor of $p\text{ and }q$ is $3$ . Therefore $p\text{ and }q$ are not co-prime numbers because they have a common factor $3$ which means our assumption is wrong.

  $\therefore \sqrt{3}$ is irrational by contradiction.

Hence proved.