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 Rational & Irrational Numbers « PreviousNext » Algebra 
Author: S.LAL
Created: 25 Aug, 2010; Last Modified: 29 Apr, 2018

Number Theory - 02

Surds

The root of a quantity, represented by a radical sign , is called a radical.

Irrational radicals of positive integers of the form a q (the "qth root of a"), such as 2 , 3 and 2 3 , as well as expressions comprising them, like 2 + 11 and 1 + 2 7 3 + 3 3 , are called surds.

q is referred to as the index of the roots. Note that in the case of a square root, where q = 2, the symbol 2 is written simply as .

The order of a surd is the index of the root. Thus, the surd a 3 is a third order surd.

A mixed surd is the product of a rational factor and a surd factor, where the rational factor is called the coefficient of the surd. Thus, 2 x 7 3 is a mixed surd with 2x as the coefficient.

Similar surds are surds which have the same irrational factors. 2 x 7 3 and 5 7 3 are similar surds.

Some identities related to square roots are helpful in solving problems related to surds. These are enumerated in Table . You may recognise that

Table 1: Some identities related to square roots.
Let a and b be positive real numbers. Then the following identities are applicable:
(1) a b = a b (2) a b = a b
(3) ( a + b ) ( a − b ) = a − b (4) ( a + b ) ( a − b ) = a 2 − b
(5) ( a + b ) ( c + d ) = a c + a d + b c + b d
(6) ( a + b ) 2 = a + 2 a b + b
  • identities (3) and (4) are derived from the expansion of ( x + y ) ( x − y ) , which is x 2 − y 2 , and
  • identity (6) is derived from the expansion of the expression ( x + y ) 2 , which is x 2 + 2 x y + y 2 .

Simplifying surd expressions

Simplifying a surd expression involves reducing it to a sum of rational(s) and irrational(s).

Simplify the expressions: (NCERT, Mathematics Class IX,, p. 22 Example 16)

(i) ( 5 + 7 ) ( 2 + 5 ) (ii) ( 5 + 5 ) ( 5 − 5 )
(iii) ( 3 + 7 ) 2 (iv) ( 11 − 7 ) ( 11 + 7 )

(i) ( 5 + 7 ) ( 2 + 5 ) = 5 + 5 5 + 2 7 + 35
(ii) ( 5 + 5 ) ( 5 − 5 ) = 5 2 − 5 = 25 − 5 = 20
(iii) ( 3 + 7 ) 2 = ( 3 ) 2 + 2 3 7 + ( 7 ) 2 = 3 + 2 21 + 7 = 10 + 2 21
(iv) ( 11 − 7 ) ( 11 + 7 ) = ( 11 ) 2 − ( 7 ) 2 = 11 − 7 = 4

Rationalisation

Rationalisation is the process of multiplying two surds together so that their product is without a surd. The two surds which are multiplied are called the rationalising factors of each other.

Find the least rationalising factor of (i) 27   (ii) 2 125 . (Bansal, Mathematics Class IX, p. 9 Ex. 4)

(i) 27 can be expressed as

27 = 3 × 3 × 3 = 3 2 × 3 = 3 2 × 3 = 3 3

Now, to rationalise 3 3 , all that needs to be done is to multiply it with 3 , which yields 3 3 × 3 = 9 , a rational number. Thus, 3 is the least rationalising factor.


(ii) 2 125 can be expressed as

2 125 = 2 5 2 × 5 = 2 × 5 5 = 10 5

Apparently, 3 is the least rationalising factor, multiplication with which gives the rational number 50.

Rationalising the denominator

When the denominator of a fractional expression contains a surd, the process of converting the expression into an equivalent one with a rational denominator is called rationalising the denominator.

For the simpler case of a fraction with only a single surd c in the denominator, such as a c , the solution is to multiply and divide the expression with the least rationalising factor, and simplifying if needed.

Rationalise the denominator of (i) 2 27  (ii) 7 2 3 5 .

(i) For the denominator we have

27 = 3 × 3 × 3 = 3 2 × 3 = 3 2 × 3 = 3 3 Thus, 3 is the least rationalising factor. Multiplying and dividing the expression with the factor gives;

2 27 = 2 3 3 × 3 3 = 2 3 9

(ii) For the denominator, the least rationalising denominator is 5 . Thus

7 2 3 5 = 7 2 3 5 × 5 5 = 7 10 15

For an expression of the form a b + c which has a surd expression b + c in the denominator, the denominator can be rationalised by multiplying the expression with b − c b − c    ( = 1 ) . Thus,

a b + c = a b + c × ( b − c b − c ) = a ( b − c ) b 2 − c

Since the value of the second fraction is in effect unity, the value of the original expression remains unchanged.

Rationalise the denominators of: (NCERT, Mathematics Class IX,, p. 23 Example 18-9)
       (i) 1 2 + 3  (ii) 5 3 − 5

(i) 1 2 + 3 = 1 2 + 3 × ( 2 − 3 2 − 3 ) = 2 − 3 2 2 − 3 = 2 − 3

(ii) 5 3 − 5 = 5 3 − 5 × ( 3 + 5 3 + 5 ) = 5 ( 3 + 5 ) 3 − 5 = − 5 2 ( 3 + 5 )

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List of References

Bansal, RK, Concise Mathematics I.C.S.E., Part I – Class IX, New Delhi: Selina Publishers, 2005.
NCERT, Mathematics – Textbook for Class IX, viewed 25 February, 2008, <http://www.ncert.nic.in/textbooks/testing/index.htm>, (n.d.).

Bibliography

Barry, S & Davis, S, Essential Mathematical Skills – for engineering, science and applied mathematics, Australia: UNSW Press, 2002.
NCERT, Mathematics – Textbook for Class IX, viewed 25 February, 2008, <http://www.ncert.nic.in/textbooks/testing/index.htm>, (n.d.).
Schultze, A, Elements of Algebra, NY, USA: The Macmillan Company, 1910.

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