**Introduction**

**Irrational Numbers – **The numbers that cannot be written in the form of p/q, are called **Irrational Numbers**. Where p and q are integers and q ≠ 0. Irrational numbers are denoted by T. **Examples –** √2, √3, √5, π, etc.

In this section, we shall study the proof of the irrationality of an irrational number with the help of The Fundamental Theorem of Arithmetic. The proof of irrationality is based on a technique which is called **proof by contradiction**. This is the technique, in which we prove the irrationality by the wrong assumption.

Before we prove irrationality, there is a theorem to be studied, and whose proof is based on the fundamental theorem of arithmetic.

**Theorems Based on Irrationality**

**Theorem 1)** If **p** divides **a ^{2}, **then

**p**also divides

**a.**where

**p**is a prime number and

**a**is a positive integer.

**Proof – **Since **a** is a positive integer. Let the prime factorization of **a** be as follows

a = p_{1}×p_{2}×p_{3}…………….p_{n}

Where p_{1}, p_{2}, p_{3}, …… p_{n} are prime numbers (may be different or same)

Therefore, a^{2} = (p_{1}×p_{2}×p_{3}…………….p_{n})(p_{1}×p_{2}×p_{3}…………….p_{n}) = p_{1}^{2}×p_{2}^{2}×p_{3}^{2}…………….p_{n}^{2}

Now, it is given that **p** divides **a**^{2}. So, by the fundamental theorem of arithmetic, **p** will be one of the prime factors of **a ^{2}**.

But prime factors of **a**^{2} are p_{1}, p_{2}, p_{3}, …… p_{n} so **p** is one of the prime factors p_{1}, p_{2}, p_{3}, …… p_{n}.

Since **a** = p_{1}×p_{2}×p_{3}…………….p_{n}

Therefore, **p** divides **a**. **Hence Proved.**

**Theorem 2) **Prove that √2 is an irrational number.

**Proof – **Using the contradiction, let √2 be a rational number.

So, for integers a and b, we can write

√2 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√2b = a

By squaring both sides,

(√2b)^{2} = a^{2}

2b^{2} = a^{2}

Or b^{2} = a^{2}/2, it means 2 divides a^{2} therefore, 2 also divides **a** [by theorem 1] ………………..(1)

So, we can write for any integer c, a = 2c …………………….(2)

Putting the value of **a** from equation (2) to (1)

We get, ^{ }b^{2} = (2c)^{2}/2

b^{2} = 4c^{2}/2

b^{2} = 2c^{2}

Or b^{2}/2 = c^{2} it means 2 divides b^{2} therefore, 2 also divides **b** [by theorem 1] ………………(3)

From equations (1) and (3), 2 divides **a**, and 2 divides **b** this means **a** and **b** have at least 2 as a common factor.

But this contradicts (opposes) the fact that **a** and **b** have no common factor other than 1.

It shows that our assumption that √2 is a rational number is wrong.

So, we conclude that **√2 is an irrational number**. **Hence Proved.**

Let’s have some examples based on this theorem.

**Some Examples**

**Example 1) **Prove that √3 is an irrational number.

**Solution –** Let √3 be a rational number.

So, for integers a and b, we can write,

√3 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√3b = a

By squaring both sides

(√3b)^{2} = a^{2}

3b^{2} = a^{2}

Or b^{2} = a^{2}/3 it means 3 divides a^{2} therefore, 3 also divides **a** [by theorem 1] ………………..(1)

So, we can write for any integer c, a =3c …………………….(2)

Putting the value of **a** from equation (2) to (1)

We get, ^{ }b^{2} = (3c)^{2}/3

b^{2} = 9c^{2}/3

b^{2} = 3c^{2}

Or b^{2}/3 = c^{2} it means 3 divides b^{2} therefore, 3 also divides **b** [by theorem 1] ………………(3)

From equations (1) and (3), 3 divides **a**, and 3 divides **b** this means **a** and **b** have at least 3 as a common factor.

But this contradicts (opposes) the fact that **a** and **b** have no common factor other than 1.

It shows that our assumption that √3 is a rational number is wrong.

So, we conclude that **√3 is an irrational number**. **Hence Proved.**

**Example 2) **Prove that √5 is an irrational number.

**Solution –** Let √5 be a rational number.

So, for integers a and b, we can write,

√5 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√5b = a

By squaring both sides

(√5b)^{2} = a^{2}

5b^{2} = a^{2}

Or b^{2} = a^{2}/5 it means 5 divides a^{2} therefore, 5 also divides **a** [by theorem 1] ………………..(1)

So, we can write for any integer c, a = 5c …………………….(2)

Putting the value of **a** from equation (2) to (1)

We get, ^{ }b^{2} = (5c)^{2}/5

b^{2} = 25c^{2}/5

b^{2} = 5c^{2}

Or b^{2}/5 = c^{2} it means 5 divides b^{2} therefore, 5 also divides **b** [by theorem 1] ………………(3)

From equations (1) and (3), 5 divides **a,** and 5 divides **b** this means **a** and **b** have at least 5 as a common factor.

But this contradicts (opposes) the fact that **a** and **b** have no common factor other than 1.

It shows that our assumption that √5 is a rational number is wrong.

So, we conclude that **√5 is an irrational number**. **Hence Proved.**

**Note – **We know that the addition and subtraction of rational numbers and irrational numbers are also irrational numbers and multiplication and Division of non-zero rational numbers and irrational numbers are also irrational numbers.

**Example 3) **Prove that 5 + √3 is irrational.

**Solution – **Let 5 + √3 be a rational number.

So, for integers a and b,

5 + √3 = a/b where b ≠ 0 and a, b = coprime numbers [not any common factor other than 1]

√3 = a/b – 5

√3 = (a – 5b)/b …………………….(1)

Since a, b, and 5 are integers, so (a – 5b)/b is a rational number.

So, from equation (1), √3 will be a rational number.

But this contradicts the fact that √3 is an irrational number.

It shows that our assumption that 5 + √3 is a rational number is wrong.

So, we conclude that **5 + √3 is irrational**. **Hence Proved.**

**Example 4) **Prove that 3√5 is irrational.

**Solution – **Let 3√5 be a rational number.

So, for integers a and b,

3√5 = a/b where b ≠ 0 and a, b = coprime numbers

√5 = a/3b …………………….(1)

Since a, b, and 3 are integers, so a/3b is a rational number.

So, from equation (1), √5 will be a rational number.

But this contradicts the fact that √5 is an irrational number.

It shows that our assumption that 3√5 is a rational number is wrong.

So, we conclude that **3√5 is irrational**. **Hence Proved.**

**Example 5) **Show that √3 + √5 is irrational.

**Solution – **Let √3 + √5 be rational.

So, √3 + √5 = a/b where b ≠ 0 and a, b = coprime numbers

√5 = a/b – √3

Squaring both sides, (√5)^{2} = (a/b – √3)^{2}

5 = a^{2}/b^{2} – 2√3a/b + 3 [∵ (a – b)^{2} = a^{2} – 2ab + b^{2}]

2√3a/b = a^{2}/b^{2} + 3 – 5

2√3a/b = a^{2}/b^{2} – 2

2√3a/b = (a^{2} – 2b^{2})/b^{2}

√3 = (a^{2} – 2b^{2})/b^{2 }× b/2a

√3 = (a^{2} – 2b^{2})/2ab …………………….(1)

Since a, b, and 2 are integers, so (a^{2} – 2b^{2})/2ab will be rational.

So, from equation (1), √3 will be a rational number.

But this contradicts the fact that √3 is an irrational number.

It shows that our assumption that √3 + √5 is rational is wrong.

Therefore, **√3 + √5 is irrational**. **Hence Proved.**