## Introduction

An irrational number is a real number that cannot be expressed as a ratio of integers. Any number which can be represented in form of \( \frac pq \) , such that, *p* and *q* are integers and *q* ≠ 0 is known as a rational number.

For Example: 2, \( \frac {5}{11} \) , -5.12 , 0.31 are all examples of rational numbers. The decimal expansion of an irrational number either terminates or repeats.

The real numbers which are not rational i.e. which cannot be expressed in the form of \( \frac pq \) , where *p* and *q* are integers and *q* ≠ 0 are known as** irrational numbers**. The decimal expansion of an irrational number is *neither terminating nor recurring*.

**For Example:** \( \sqrt{2} \) ,\( \sqrt{3} \), \( \pi \) , etc.

From the study of irrational numbers, if *p* is a prime number, then \( \sqrt{p}\) is irrational. From this, it can be said that \( \sqrt{2} \) , \( \sqrt{3} \) , \( \sqrt{5} \) etc. are all irrational numbers.

The following theorem is used to prove the above statement

**Theorem**: Given *p* is a prime number and \(a^2\) is divisible by *p *(where *a* is any positive integer), then it can be concluded that p also divides *a*.

**Proof:** Using the Fundamental Theorem of Arithmetic, the positive integer a can be expressed in the form of the product of its primes as:

\( a \) = \( p_1~×~p_2~×~p_3~×……….×~p_n\) ……..(1)

Where, \( p_1,p_2,p_3……….,p_n\) represent all the prime factors of *a*.

Squaring both the sides of equation (1),

\( a^2 \) = \( (p_1~×~p_2~×~p_3~×……….×~p_n)(p_1~×~p_2~×~p_3~×……….×~p_n)\)

⇒\(a^2 \) = \( p_1^2~×~p_2^2~×~p_3^2~×……….×~p_n^2\)

According to the Fundamental Theorem of Arithmetic, the prime factorization of a natural number is unique, except for the order of its factors.

The only prime factors of \(a^2\) are \(p_1,p_2,p_3……….,p_n\). If *p* is a prime number and a factor of \(a^2\), then p is one of \( p_1,p_2,p_3……….,p_n\). So, *p* will also be a factor of *a*. Hence, if \(a^2\) is divisible by *p*, then *p* also divides *a*.

Now, using this theorem, we can prove that \( \sqrt{2} \) is irrational.

**Proof that, root 2 is irrational.**

Let us assume that \( \sqrt{2} \) is a rational number.

Going by the definition of rational numbers, it can be written that,

\( \sqrt{2} \) = \( \frac pq \) …….(2)

Where *p* and *q* are co-prime integers and *q* ≠ 0 (Co-prime numbers are those numbers whose common factor is 1).

Squaring both the sides of equation (2), we have

2 = \(\frac {p^2}{q^2}\)

⇒ \( p^2 \) = \( 2q^2 \) ………. (3)

From the theorem stated above, if 2 is a prime factor of \(p^2\), then 2 is also a prime factor of *p*.

So, *p *= 2 × *c*, where *c* is any integer.

Substituting this value of *p* in equation (3), we have

\( (2c)^2 \) = \( 2q^2 \)

⇒\(q^2\) = \( 2c^2 \)

This implies that 2 is a prime factor of \(q^2\) also. Again from the theorem, it can be said that 2 is also a prime factor of *q*.

Since according to initial assumption, *p* and *q* are co-primes but the result obtained above contradicts this assumption as *p* and *q* have 2 as a common prime factor other than 1. This contradiction arose due to incorrect assumption that \( \sqrt{2} \) is rational.

**So, root 2 is irrational.**

Similarly, we can justify the statement discussed in the beginning that if *p* is a prime number, then \( \sqrt{p} \) is an irrational number. Similarly, it can be proved that for any prime number *p*, \( \sqrt{p} \) is irrational.

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