Can you find a rational number which when multiplied by itself (square) gives 4 ?

You know the answer is 2. i.e. √4 = 2

What is the rational number whose square is 9⁄25 ?

As you know (3⁄5) x (3⁄5) = 9⁄25, the answer is 3⁄5. i.e. √(9⁄25) = 3⁄5.

Example 1

Now I ask you another question, Can you find a rational number whose square is 2 ?

In other words, Is √2 a rational number ?

If √2 is rational, let it be equal to a⁄b where a and b(≠0) are integers. Let us say a⁄b reduced to the simplest form be p⁄q, which means p and q are integers which do not have common factors other than 1.

√2 = p⁄q⇒ (√2)^{2} = (p⁄q)^{2}⇒ 2 = p^{2}⁄q^{2}
⇒p^{2} = 2q^{2}⇒ p^{2} is divisible by 2. ⇒ p is divisible by 2.-----(i)

Since p is divisible by 2, we can take p = 2r where r is an integer.

Substituting this in p^{2} = 2q^{2}, we get (2r)^{2} = 2q^{2}⇒ 4r^{2} = 2q^{2}⇒q^{2} = 2r^{2}⇒ q^{2} is divisible by 2. ⇒ q is divisible by 2.-----(ii)

From (i) and (ii), p and q have common factor 2. But p and q do not have any common factor other than 1.

This contradiction has arisen because of our assumption that √2 is a rational number.

So, we conclude that √2 is not a rational number.

Thus there is a need for extending the rational numbers.

Definition of Irrational Numbers

A number which can not be expressed in the form a⁄b, where a and b are integers, b≠0 is called an Irrational Number.

We have seen every terminating or repeating decimal is a rational number.

This gives us another way of defining Irrational Number.

Every number which when expressed in decimal form is expressible as a non-terminating and non-repeating decimal, is called an Irrational Number.

When a positive rational number is not a power of n, then nth root of the number is irrational and is called a surd.or radical.

We first take the statement that √3 is rational and prove that the statement is wrong.

Let √3 be rational, Let it be equal to a⁄b where a and b(≠0) are integers. Let us say a⁄b reduced to the simplest form be p⁄q, which means p and q are integers which do not have common factors other than 1.

√3 = p⁄q⇒ (√3)^{2} = (p⁄q)^{2}⇒ 3 = p^{2}⁄q^{2} ⇒p^{2} = 3q^{2}⇒ p^{2} is divisible by 3. ⇒ p is divisible by 3.-----(i)

Since p is divisible by 3, we can take p = 3r where r is an integer.

Substituting this in p^{2} = 3q^{2}, we get (3r)^{2} = 3q^{2}⇒ 9r^{2} = 3q^{2}⇒q^{2} = 3r^{2}⇒ q^{2} is divisible by 3. ⇒ q is divisible by 3.-----(ii)

From (i) and (ii), p and q have common factor 3. But p and q do not have any common factor other than 1.

This contradiction has arisen because of our assumption that √3 is a rational number.

Examples of irrational numbers which are not surds

(i) Example of irrational number which is not a surd is π, which is defined as the ratio of circumference and diameter of a circle.

The value of π = 3.1416.... which is a non-terminating and non-repeating decimal.

We some times take π = 22⁄7 which is only an approximate value of π.

(ii) Another Example of irrational number which is not a surd is e, the base of natural logarithm.

e is defined as the sum of the infinite series e = 1 + 1⁄1! + 1⁄2! + 1⁄3! + 1⁄4! + ..... where, ! stands for factorial (n! = product of natural numbers from 1 upto n)

The value of e = 2.71828.... which is a non-terminating and non-repeating decimal.

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If the product of two irrational numbers is a rational number, then each of the irrationals is called a rationalising factor of the other.

Examples : (i) √5 x √5= 5, so √5is the rationalising factor of √5.

(ii) (√3+ √2)x (√3- √2)= [(√3)^{2}- (√2)^{2}]= 3 - 2 = 1 which is rational. So, (√3+ √2)and (√3- √2)are rationalising factors of each other.

(iii) (6 + √5)x (6 - √5)= 6^{2} - (√5)^{2}= 36 - 5 = 31 which is rational. So, (6 + √5)and (6 - √5)are rationalising factors of each other.

Rationalisation of Denominator of a Fraction

Multiplying the numerator and denominator of the fraction by the rationalising factor of the denominator and then simplifying is called Rationalisation of the Denominator of a Fraction.

Example 3 of Irrational Numbers

Rationalise the denominators of each of the following.

(i) 2⁄√6

(ii) 3⁄(3 + √7)

(iii) 16⁄(√11- √3)

Solution: (i) 2⁄√6= (2 x √6)⁄(√6 x √6) = (2 x √6)⁄6= √6⁄3. Ans.

(ii) 3⁄(8 + √7)= [3 x (3 - √7)]⁄[(3 + √7)x (3 - √7)] = [3 x (3 - √7)]⁄[(3^{2} - √7)^{2}] = [3 x (3 - √7)]⁄[(9 - 7)]= 3(3 - √7)⁄2. Ans.

Rationalise the denominators of each of the following.

7⁄√14

8⁄(5 - √23)

27⁄(√10+ √7)

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