Revisiting Irrational Numbers

Introduction

In Class 9, you learned about irrational numbers and some of their properties. You studied how rational and irrational numbers together make up the set of real numbers and learned how to locate irrationals on the number line. However, we didn’t formally prove their irrationality. In this section, we will use proof by contradiction to establish that numbers like $\boldsymbol{\sqrt{2}}$ , $\boldsymbol{\sqrt{3}}$ , and $\boldsymbol{\sqrt{5}}$ are irrational.

A number is considered irrational if it cannot be expressed in the form $\displaystyle\boldsymbol{\frac{p}{q}}$ , where $\boldsymbol{p}$ and $\boldsymbol{q}$ are integers and $\boldsymbol{q \neq 0}$ . Examples of irrational numbers include:

$\boldsymbol{\sqrt{2}, \sqrt{3}, \sqrt{15}}$
$\boldsymbol{\pi}$
Non-repeating, non-terminating decimals like $\boldsymbol{0.10110111011110 \ldots}$

Theorem for Proving Irrationality

To prove that numbers like $\boldsymbol{\sqrt{2}}$ and $\boldsymbol{\sqrt{3}}$ are irrational, we need an important theorem based on the Fundamental Theorem of Arithmetic.

Theorem 1: If $\boldsymbol{p}$ is a prime number and $\boldsymbol{p}$ divides $\boldsymbol{a^2}$ , then $\boldsymbol{p}$ must also divide $\boldsymbol{a}$ , where $\boldsymbol{a}$ is a positive integer.

This theorem is crucial for proving the irrationality of certain square roots.

Proof of Theorem 1:

Let the prime factorization of $\boldsymbol{a}$ be $\boldsymbol{a = p_1 p_2 \ldots p_n}$ , where $\boldsymbol{p_1, p_2, \ldots, p_n}$ are prime factors (not necessarily distinct).
Then, $\boldsymbol{a^2 = (p_1 p_2 \ldots p_n)^2 = p_1^2 p_2^2 \ldots p_n^2}$ .
If $\boldsymbol{p}$ divides $\boldsymbol{a^2}$ , then by the Fundamental Theorem of Arithmetic, $\boldsymbol{p}$ must be one of the prime factors of $\boldsymbol{a}$ .

This theorem is essential in proving that numbers like $\boldsymbol{\sqrt{2}}$ are irrational by contradiction.

Proving that $\boldsymbol{\sqrt{2}}$ is Irrational

To show that $\boldsymbol{\sqrt{2}}$ is irrational, we use proof by contradiction.

Proof:

Assume, to the contrary, that $\boldsymbol{\sqrt{2}}$ is rational.
Then we can write $\displaystyle\boldsymbol{\sqrt{2} = \frac{r}{s}}$ , where $\boldsymbol{r}$ and $\boldsymbol{s}$ are integers with no common factors other than 1 (i.e., they are coprime).
Squaring both sides, we get $\displaystyle\boldsymbol{2 = \frac{r^2}{s^2}}$ , which simplifies to $\boldsymbol{r^2 = 2s^2}$ .
Since $\boldsymbol{r^2}$ is divisible by 2, $\boldsymbol{r}$ must also be divisible by 2 (by Theorem 1). Let $\boldsymbol{r = 2c}$ for some integer $\boldsymbol{c}$ .
Substituting $\boldsymbol{r = 2c}$ into $\boldsymbol{r^2 = 2s^2}$ , we get $\boldsymbol{(2c)^2 = 2s^2}$ , which simplifies to $\boldsymbol{4c^2 = 2s^2}$ or $\boldsymbol{s^2 = 2c^2}$ .
This shows that $\boldsymbol{s^2}$ is divisible by 2, so $\boldsymbol{s}$ must also be divisible by 2.

Since both $\boldsymbol{r}$ and $\boldsymbol{s}$ are divisible by 2, they have a common factor of 2, which contradicts our assumption that $\boldsymbol{r}$ and $\boldsymbol{s}$ are coprime.

Conclusion: This contradiction proves that $\boldsymbol{\sqrt{2}}$ is irrational.

Proving that $\boldsymbol{\sqrt{3}}$ is Irrational

The proof for $\boldsymbol{\sqrt{3}}$ follows a similar pattern.

Proof:

Assume, to the contrary, that $\boldsymbol{\sqrt{3}}$ is rational, so $\displaystyle\boldsymbol{\sqrt{3} = \frac{a}{b}}$ , where $\boldsymbol{a}$ and $\boldsymbol{b}$ are coprime integers.
Squaring both sides, we get $\displaystyle\boldsymbol{3 = \frac{a^2}{b^2}}$ , which implies $\boldsymbol{a^2 = 3b^2}$ .
Since $\boldsymbol{a^2}$ is divisible by 3, $\boldsymbol{a}$ must also be divisible by 3 (using Theorem 1). Let $\boldsymbol{a = 3c}$ for some integer $\boldsymbol{c}$ .
Substituting $\boldsymbol{a = 3c}$ into $\boldsymbol{a^2 = 3b^2}$ , we get $\boldsymbol{(3c)^2 = 3b^2}$ , which simplifies to $\boldsymbol{9c^2 = 3b^2}$ , or $\boldsymbol{b^2 = 3c^2}$ .
This shows that $\boldsymbol{b^2}$ is divisible by 3, so $\boldsymbol{b}$ must also be divisible by 3.

Since both $\boldsymbol{a}$ and $\boldsymbol{b}$ are divisible by 3, they have a common factor of 3, which contradicts the assumption that $\boldsymbol{a}$ and $\boldsymbol{b}$ are coprime.

Conclusion: This contradiction proves that $\boldsymbol{\sqrt{3}}$ is irrational.

Properties of Irrational Numbers

Sum or Difference of a Rational and Irrational Number:
The sum or difference of a rational and an irrational number is always irrational. For example, $\boldsymbol{5 - \sqrt{3}}$ is irrational.
Product and Quotient of a Non-Zero Rational and an Irrational Number:
The product and quotient of a non-zero rational number and an irrational number are always irrational. For instance, $\boldsymbol{3 \cdot \sqrt{2}}$ and $\displaystyle\boldsymbol{\frac{\sqrt{5}}{2}}$ are irrational.

Example Problems

Example 1: Prove that $\boldsymbol{5 - \sqrt{3}}$ is irrational.

Solution:

Assume, to the contrary, that $\boldsymbol{5 - \sqrt{3}}$ is rational.
Then we can write $\displaystyle\boldsymbol{5 - \sqrt{3} = \frac{a}{b}}$ , where $\boldsymbol{a}$ and $\boldsymbol{b}$ are integers and $\boldsymbol{b \neq 0}$ .
Rearranging, we get $\displaystyle\boldsymbol{\sqrt{3} = 5 - \frac{a}{b}}$ .
Since $\displaystyle\boldsymbol{\frac{a}{b}}$ and 5 are rational, their difference $\boldsymbol{\sqrt{3}}$ would also be rational.

This contradicts the fact that $\boldsymbol{\sqrt{3}}$ is irrational. Thus, $\boldsymbol{5 - \sqrt{3}}$ is irrational.

Example 2: Prove that $\boldsymbol{\sqrt{3} \cdot \sqrt{2}}$ is irrational.

Solution:

Assume, to the contrary, that $\boldsymbol{\sqrt{3} \cdot \sqrt{2}}$ is rational.
Then $\displaystyle\boldsymbol{\sqrt{3} \cdot \sqrt{2} = \frac{a}{b}}$ , where $\boldsymbol{a}$ and $\boldsymbol{b}$ are coprime integers.
Rearranging, we get $\displaystyle\boldsymbol{\sqrt{2} = \frac{a}{b \cdot \sqrt{3}}}$ .
Since $\displaystyle\boldsymbol{\frac{a}{b \cdot \sqrt{3}}}$ is rational, this implies that $\boldsymbol{\sqrt{2}}$ is rational, contradicting its known irrationality.

Therefore, $\boldsymbol{\sqrt{3} \cdot \sqrt{2}}$ is irrational.