Irrational Numbers

The word irrational means not rational. So, Irrational numbers are defined as the real numbers which cannot be written in the simple fraction form of  \[\frac{p}{q}\], where ‘p’ and ‘q’ are integers and q \[ \ne \] 0. 


Whereas any number which can be written in the form of  \[\frac{p}{q}\] , where ‘p’ and ‘q’ are integers and q\[ \ne \]0 is known as a rational number.

Examples of irrational numbers are ‘π’, \[\sqrt 2 ,\sqrt 3 \], etc.

How to Check if a Given Real Number is Irrational or Not.

The given real number is irrational or not can be checked by its decimal expansion.

The decimal expansion of an irrational number is neither terminating nor repeating or recurring.

Let us see the decimal expansion of some real numbers to understand the above statement in better way.

  1. 13 = 0.33333…., The decimal expansion of \[\frac{1}{3}\] repeats with the digit ‘3’ after decimal. So, \[\frac{1}{3}\]is not an irrational number, it is a rational number.


  1. \[\sqrt 2 \]= 1.414213562…, The decimal expansion of \[\sqrt 2 \]goes on expanding without any repetition of digits after the decimal. So, \[\sqrt 2 \]is an irrational number.


  1. \[\frac{7}{8}\]= 0.875 The decimal expansion of  \[\frac{7}{8}\]terminates after decimal. So, \[\frac{7}{8}\]  is not an irrational number, it is a rational number.


  1. \[\pi \left( {pi} \right)\] = 3.14159265358…, The decimal expansion of ‘π goes on expanding without any repetition of digits after the decimal. So, ‘π’ is an irrational number.


  1. \[\frac{{22}}{7}\]= 3.142857142…, The decimal expansion of \[\frac{{22}}{7}\] repeats with the digits ‘142857’ after decimal. So, \[\frac{{22}}{7}\]is not an irrational number, it is a rational number.

Note: 

1. We usually see that π = \[\frac{{22}}{7}\] , this does not mean that ‘π’ is a rational number. It is because we approximate the decimal values of ‘π’ with\[\frac{{22}}{7}\]  to simplify our calculations.

2. The square root of natural numbers other than perfect squares are irrational numbers.

3. If ‘p’ is a prime number, then \[\sqrt p \]  is an irrational number.   

Symbolic Representation of Irrational Numbers

Generally, the symbol used to represent the irrational numbers is ‘P’ or Q . This is most likely because the irrationals are defined negatively as the set of real numbers which are not rational.

The most common designation for the irrationals is R∖Q (or R−Q): the difference of the set of real numbers (R) minus the set of rational numbers (Q).


Flow-chart of Real numbers

Representation of an Irrational Number on Number Line

We know that a real number is either rational or irrational. So, we can say that every real number is represented by a unique point on the number line. Also, every point on the number line represents a unique real number. 

We can locate some of the irrational number of the form\[\sqrt n \], where n is a positive integer on the number line by using following steps:

Step 1:  Write the given number (without root) as the sum of the squares of two natural numbers (say ‘a’ and ‘b’, where a > b).

Step 2: Take the distance equal to these two natural numbers on the number line (‘a’ on number line and ‘b’ vertically) starting from 0 (say OA and AB) in such a way that one is perpendicular to other (say AB ⊥ OA). 

Step 3: Use Pythagoras theorem to find the distance OB.

Step 4: Take O as centre and OB as radius, draw an arc, which cut the number line at C (say).
Thus, the point C will represent the location of\[\sqrt n \]on the number line.

How to Find Irrational Numbers Between Two Real Numbers.

Let us find an irrational number between 2 and 3.

We know, square root of 4 is 2; \[\sqrt 4 \] =2
          and square root of 9 is 3; \[\sqrt 9 \] = 3
Therefore, the number of irrational numbers between 2 and 3 are \[\sqrt 5 ,\sqrt 6 ,\sqrt 7 {\text{ and }}\sqrt 8 \], as these are not perfect squares and cannot be simplified further.


Proof of Irrational Number

Irrational numbers like: \[\sqrt 2 ,\sqrt 3 ,\sqrt 5 ,\sqrt 7 \] and in general, if ‘p’ is a prime number then, \[\sqrt p \] is an irrational number. The above statement can be proved using the following theorem.

Theorem: Let p be a prime number. If p divides a2 , then p divides a, where a is a positive integer.

Proof: Let the prime factorisation of a be as follows :

a = p1 × p2 × p3…… × pn;             (1)

where  p1, p2 ,...... pn are primes, not necessarily distinct.                   

Squaring both the sides of equation (1),

Therefore, a= ( p1 p2…... pn )( p1 p2…... pn ) 

a2 = (p1)2( p2)2…... (pn)2 .

Now, we are given that p divides a2 .

Therefore, from the Fundamental Theorem of Arithmetic: (the prime factorization of a natural number is unique, except for the order of its factors), it follows that p is one of the prime factors of a2.

The only prime factors of a2 are p1, p2, p3…… pn. 

If p is a prime number and a factor of a2, then p is one of the  p1, p2,  p3…….. pn. So, p will also be a factor of a. Hence, if a2  is divisible by p, then p also divides a.


Properties of Irrational Numbers

The following are the properties of rational numbers:

  1. The sum or difference of a rational and an irrational number is always an irrational number.

Example: 5 +\[\sqrt 2 ,\sqrt 7 \]– 2.


  1. The product of a non-zero rational number and an irrational number is always an irrational number.

Example: \[3 \times \sqrt 2  = 3\sqrt 2 \]


  1. The sum or difference of two irrational numbers is not always an irrational number.

Example: sum- (5 +\[\sqrt 2 \]) + (5 -\[\sqrt 2 \]) =10; the resulting number 10 is a rational number.

      Difference- (5 +\[\sqrt 2 \]) - (5 -\[\sqrt 2 \]) = 2\[\sqrt 2 \], the resulting number 2\[\sqrt 2 \] is an irrational number.


  1. The product of two irrational numbers is not always an irrational number.

Example: (2 +\[\sqrt 3 \]) \[ \times \] (2 -\[\sqrt 3 \]) = 4 -3 = 1, which is a rational number.


  1. The L.C.M. of two irrational numbers may or may not exist.


Solved Examples:

Q.1. Check Which of the Following are Rational or Irrational Number.

         (a) 4.13468968968….

         (b) (7 +\[\sqrt 3 \]) \[ \times \] (\[\sqrt 3 \])

         (c) 0.098009800098….

Ans. (a) 4.13468968968…. is a rational number. Because, the digits ‘689’ in the expansion repeats after the decimal.

         (b) (7 + \[\sqrt 3 \]) \[ \times \](\[\sqrt 3 \])= 3 +7\[\sqrt 3 \] ,which is an irrational number.

         (c)  0.098009800098…. is an irrational number. Because, the decimal expansion is neither terminating nor repeating.

Q.2. Prove that \[\sqrt 2 \] is an Irrational Number.

Ans. Let us assume \[\sqrt 2 \] is a rational number. Then, by the definition of rational numbers, it can be written that,

\[\sqrt 2 \]=\[\frac{p}{q}\]  ……. (1)

Where p and q are co-prime integers and q \[ \ne \] 0 (Co-prime numbers are those numbers whose common factor is 1).

Squaring both the sides of equation (1), we have

2 = \[\frac{{p2}}{{q2}}\]

p2 = 2 q 2    ……. (2)

From the theorem stated above, if 2 is a prime factor of p2, then 2 is also a prime factor of p.

So, = 2 × c, where c is any integer.

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

(2c)2 = 2 q2

⇒ q2 = 2c2 

This implies that 2 is a prime factor of q2 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 arised due to the incorrect assumption that \[\sqrt 2 \]  is rational.

So, \[\sqrt 2 \]  is irrational.