## Introduction of Irrational Numbers

An irrational number is a real number that cannot be expressed as a ratio of integers, for example, âˆš 2 is an irrational number. Again, the decimal expansion of an irrational number is *neither terminating nor recurring*. **How do you know a number is irrational? **The real numbers which cannot be expressed in the form of p/q, where p and q are integers and q â‰ 0 are known as** irrational numbers**. For ExampleÂ âˆš 2 and âˆš 3 etc. Whereas any number which can be represented in the form of p/q, such that, p and q are integers and q â‰ 0 is known as a rational number.

## Proof that root 2 is irrational.

Let us assume that **âˆš** 2Â is a rational number.

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

**âˆš** 2 =p/qÂ Â …….(1)

Where *p* and *q* are co-prime integers and *q* â‰ 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}

â‡’ p^{2} = 2 qÂ ^{2}Â Â ………. (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, *pÂ *= 2 Ã— *c*, where *c* is any integer.

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

(2c)^{2}Â =Â 2 qÂ ^{2}

â‡’Â q^{2} = 2cÂ ^{2}Â

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 **âˆš** 2Â 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 **âˆš**Â pÂ is an irrational number. Similarly, it can be proved that for any prime number *p*,**âˆš**Â 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}Â is divisible by *p,Â *(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}Â Ã— p_{2Â }Ã— p_{3………..Â Â }Ã— p_{n …..(1)}

Where, p_{1,}Â p_{2}_{,Â }p_{3},_{ ……,Â }p_{n}Â represent all the prime factors of *a*.

Squaring both the sides of equation (1),

a^{2} = ( p_{1}Â Ã— p_{2Â }Ã— p_{3………..Â Â }Ã— p_{n) (} p_{1}Â Ã— p_{2}Â Ã— p_{3}………..Â Â Ã— p_{n})

â‡’a^{2} = (p_{1})^{2}Â Ã— (p_{2})^{2}_{Â }Ã— (^{p}_{3}^{Â })^{2}_{………..}Ã— (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}Â , p_{2 ,}Â p_{3………..,}Â p_{n}. If *p* is a prime number and a factor of a^{2}, then p is one ofÂ p_{1}Â , p_{2 ,}Â p_{3………..,}Â p_{n} . So, *p* will also be a factor of *a*. Hence, if a^{2} Â is divisible by *p*, then *p* also divides *a*.

Now, using this theorem, we can prove that **âˆš** 2Â is irrational.

### Solved Examples of Irrational Numbers

**Question 1**: Which of the following are Rational Numbers or Irrational Numbers?

2, -.45678…, 6.5,Â **âˆš**Â 3,Â **âˆš** 2

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

Irrational Numbers –Â -.45678…,Â **âˆš**Â 3,Â **âˆš** 2 as these are non-terminatingÂ numbers.

**Question 2:** Check if below numbers are rational or irrational.Â

2, 5/11, -5.12, 0.31Â

**Solution:** Since decimal expansion of an irrational number either terminates or repeats. So, 2, 5/11, -5.12, 0.31 are all rational numbers.

To know more about rational and irrational numbers, download BYJUâ€™S-The Learning App orÂ Register with us to watch interesting videos on irrational numbers.