**Irrational numbers** are the numbers that cannot be represented as a simple fraction. It is a contradiction of rational numbers but is a type of real numbers. Hence, we can represent it as R\Q, where the backward slash symbol denotes ‘set minus’ or it can also be denoted as R – Q, which means set of real numbers minus set of rational numbers. The calculations based on these numbers are a bit complicated. For example,Â âˆš5,Â âˆš11,Â âˆš21, etc are irrational. If such numbers are used in arithmetic operations, then first we need to evaluate the values under root. These values could be sometimes recurring also. Now let us find out its definition, lists of irrational numbers, how to find them, etc., in this article.

**Table of Contents:**

## What are 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?Â 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

## Irrational Number symbol

Generally, the symbol used to represent the irrational symbol is “P”.Â Since the irrational numbers are defined negatively, the set of real numbers (R) that are not the rational number (Q), is called an irrational number. The symbol P is often used because of the association with the real and rational number. (i.e) because of the alphabetic sequence P, Q, R. But mostly, it is represented using the set difference of the real minus rationals, in a way R- Q or R\Q.

## Irrational Number Properties

The following are the properties of rational numbers:

- The addition of an irrational number and a rational number gives an irrational number.Â For example, let us assume that xÂ is an irrational number, yÂ is a rational number, and the addition of both the numbers x +y gives a rational number z.
- While Multiplying any irrational number with any nonzero rational number results in an irrational number. Let us assume that if xy=z is rational, then x =z/y is rational, contradicting the assumption that x is irrational. Thus, the product xy must be irrational.
- The least common multipleÂ (LCM) of any two irrational numbers may or may not exist.
- The addition or the multiplication of two irrational numbers may be rational; for example,Â âˆš2.Â âˆš2 = 2. Here,Â âˆš2 is an irrational number. If it is multiplied twice, then the final product obtained is a rational number. (i.e) 2.
- The set of irrational numbers is not closed under multiplication process, unlike the set of rational numbers.

## List ofÂ Irrational Numbers

The famous irrational numbers consist of Pi, Euler’s number, Golden ratio. Many square roots and cube roots numbers are also irrational, but not all of them. For example,Â âˆš3 is an irrational number butÂ âˆš4 is is a rational number. Because 4 is a perfect square, such as 4 = 2 x 2 andÂ âˆš4 = 2, which is a rational number.

Pi, Ï€Â | 3.14159265358979… |

Eulerâ€™s Number, e | 2.71828182845904… |

Golden ratio, Ï† | 1.61803398874989…. |

## Irrational Number Proof

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 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}, 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}, 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.

## How to Find an Irrational Number?

Let us assume a case ofÂ **âˆš**2. Now, how can we find ifÂ **âˆš**2 is an irrational number.

Suppose,Â **âˆš**2 is a rational number. Then,Â 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 Numbers Examples

**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.

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