## Introduction of Irrational Numbers

An irrational number is a real number that cannot be expressed as a ratio of integers. example, √ 2. The real numbers which are not rational i.e. which cannot be expressed in the form of p/q, 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 √ 2, √ 3, √ pi, etc.

The above numbers are not a rational number as any number which can be represented in the form of p/q, such that, p and q are integers and q ≠ 0 is known as a rational number.

For Example, 2, 5/11, -5.12, 0.31 are all examples of rational numbers because the decimal expansion of an irrational number either terminates or repeats.

From the study of irrational numbers, if *p* is a prime number, then **√ **p is irrational. From this, it can be said that **√** 2, √ 3, √ pi, etc. are all irrational numbers.

## Proof that root 2 is irrational.

Let us assume that **√** 2 is a rational number.

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

**√** 2 =p/q …….(1)

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 (1), we have

2 = p^{2}/q^{2}

⇒ p^{2} = 2 q ^{2} ………. (2)

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} = 2 q ^{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 the incorrect assumption that **√** 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 **√** p is an irrational number. Similarly, it can be proved that for any prime number *p*,**√** p is irrational.

### Irrational Number theorem

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 **√** 2 is irrational.

### Solved examples of Irrational Numbers

**Question**: Which of the following are Rational Numbers or Irrational Numbers?

2, -.45678…, 6.5, **√** 3, **√** 2

**Solution**: Rational Numbers – 2, 6.5 as these are terminating numbers.

Irrational Numbers – -.45678…, **√** 3, **√** 2 as these are non-terminating numbers.

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