### 1.3.1 Introduction – Surds

You have studied about the irrational numbers like √3, ∛7, ∜5 and so on. You have here square root, cube rot and forth root of numbers. There are also irrational numbers which cannot be written in such a form; for example √(3+∛2), π. In the Surds we study more about numbers of the form is a natural number. *‘a*’ is a rational number. these are special class of irrational numbers.

## 1.3.2 Rational Exponent of a Number – Surds

Recall the statement from unit 2: Real Numbers given a natural number n, for every positive real number *a* there exists a unique positive real number b such that b^{n } = a. Here b is called the n-th root of *a* and we write

We can now define rational power of a positive number. let a be a positive real number. Let *r = ^{p}/_{q} *to be a rational number, where p is an integer and q is a natural number.

**Let a and b be positive real numbers. Let r _{1} and r_{2} be two rational numbers. Then we have:**

**Example 1: Simplify ^{2^1/2}/_{4^1/6}**

**Solution:**

4^{1/6} = (2^{2})^{1/6} = 2^{2x(1/6)} = 2^{1/3}

Hence,

^{2^1/2}/_{4^1/6} = ^{2^1/2}/_{2^1/3} = 2^{(1/2)-(1/3)} = 2^{1/6}

## Surds – Exercise 1.3.2

**Simplify the following using laws of indices:**

**(i) (16)**^{-0.75 }**x ****(64)**^{4/3 }

(16)^{-0.75 }x (64)^{4/3 }[16 = 2^{4} ]

= (24)^{-0.75 }x (24)^{4/3 }[64 = 2^{6} ]

= 2^{4} x ^{-3/4 }x 2^{6} x ^{4/3 }

= 2^{-3 }x 2^{8}

= 2^{5}

**= 32**

**(ii) (0.25)**^{0.5}**x ****(100)**^{-1/2}

(0.25)^{0.5} x (100)^{-1/2}

= (0.25)^{1/2} x (^{1}/_{100} )^{1/2}

= (0.25)^{1/2} x (^{1}/_{10}^{2} )^{1/2}

= (0.5) x ( ^{1}/_{10} )

= ( ^{5}/_{10} ) x ( ^{1}/_{10})

= ^{5}/_{100}

**= **^{𝟏}/_{𝟐𝟎}

**(iii) (6.25)**^{0.5}**x ****10**^{2}**x ****(100)**^{-1/2}**x ****(0.01)**^{-1}

= ( ^{625}/_{100} )^{1/2} x 10^{2} x (^{1}/_{10})^{1/2} x ( ^{1}/_{100} )^{-1}

= ( ^{25}/_{10} )^{1/2} x 10^{2} x (^{1}/_{10} )^{1/2} x (^{1}/_{100})^{-1}

= ( 25^{2}/_{10} )^{1/2} x 10^{2} x (^{1}/_{10}^{2} )^{1/2} x (100)^{1}

= ( ^{25}/_{10} ) x 100 x 1 10 x 100

**= 2500 **

**(iv) (3**^{-1/2}**x ****2**^{-1/3}**) ÷ (3**^{-3/4}**x ****2**^{-5/6}**) **

= 3^{−1/2}× 2^{−1/3} 3^{−3/4}× 2^{−5/6}

= 3^{-1/2+3/4} x 2^{-1/3+5/6 }

= 3^{1/4 }x 2^{3/6 }

= 3^{1/4 }x 2^{1/2 }

= 3^{1/4 }x 2^{1/4 }

=(3 × 2^{2})^{1/4}

= (3 × 4)^{1/4}

**= **𝟏𝟐^{1/4}

**Find the value of the expression.**

**[ 3 ^{1/3 }{5^{-1/2 }x 3^{-1/3 }x (225^{2})^{1/3}}^{1/2}]^{6}**

Solution:

= [ 3^{1/3 }{5^{-1/2}^{x}^{-1/2 }x 3^{-1/3}^{x}^{1/2 }x (225^{2})^{2/3}^{x}^{1/2}]^{6}

= [ 3^{1/3 }{5^{-1/4 }x 3^{1/6 }x (225^{2})^{-2/6}]^{6}

= [ 3^{1/3 }^{x }^{6 }{5^{1/4 }^{x }^{6 }x 3^{1/6 }^{x }^{6 }x (225^{2})^{-2/6 }^{x }^{6}]

= [ 32 x 5^{1/4 }^{x }^{6 }x 3^{1/6 }^{x }^{6 }x 225^{2-2/6 }^{x }^{6}]

= [32 x 5^{3/2} x 3^{1} x 225^{-2}]

= [32 x 5^{2/3} x 3^{1 }x 15^{-4}]

= [32 x 5^{-2/3 }x 31 x 3^{-4} x 5^{-4}]

= [3^{2 + 1 – 4 }x 5^{3/2} – 4]

= ^{1}/_{3}^{1} x ^{1}/_{5}^{5/2}

= ^{1}/_{3}^{1} x ^{1}/_{5}^{5/2}

= ^{1}/_{3}^{1} x ^{1}/_{√5}^{5}

= ^{1}/_{3} x ^{1}/_{√3125}

= ^{1}/_{3√3125}

**Simplify:**

**[{(3**^{5/2 }**x ****5**^{3/4}**) ÷ 2**^{-5/4}**} ÷ {16 / (5**^{2}**x ****2**^{1/4}**x ****3**^{1/2}**)}]**^{1/5}

Solution:

= [{(3^{5/2} × 5^{3/4}) ÷ 2^{−5/4}} ÷ {16÷ (5^{2}× 2^{1/4}× 3^{1/2})}] ^{1/5 }

= [{(3^{5/2} × 5^{3/4})/(_{2}^{−54}) ÷ ^{16}/(_{5}^{2}× _{2}^{1/4}× _{3}^{1/2})}] ^{1/5}

= [(3^{5/2} × 5^{3/4})/(_{2}^{−54}) ÷ (5^{2}× 2^{1/4}× 3^{1/2})/_{2}^{4}]^{1/5}

= [(3^{5/2} × 5^{3/4} × 5^{2/1} × 2^{1/4} × 3^{1/2})/(_{2}^{−5/4} × _{2}^{4 })]^{1/5}

= [2^{1/4} × 3^{5/2 + 1/2} × 5^{3/4 + 2/1}/_{2}^{−5/4 + 4/1}]^{1/5}

= 2^{1/4} × 3^{5/2 + 1/2} × 5^{11/4}/_{2}^{11/4}]^{1/5}

= [2^{1/4 }^{– }^{4/11} × 3^{3} × 5^{11/4 }] ^{1/5}

= [2^{10/4 × 1/5} × 3^{3 }^{× 1/5} × 5^{11/4 }^{+ }^{1/5}]

= 2^{−}^{1/2} × 3^{1/5} × 5^{11/20}

## 1.3.3 Surds and their properties – Surds:

Consider the following real numbers:

√17, 8 + ∛12 , ^{3}/_{5} + √(^{7}/_{11}) , ∛(3 + √5))

They are all irrational numbers. Nevertheless, you see that they are all different types.

A surd is real number of the form , where n is an integer larger than 1 and a is a rational number such that it is not an n-th power of any rational number. For example, ^{25}/_{36} is the square of ^{5}/_{6}. Thus, √(^{5}/_{6}) is not a surd. On the other hand √(^{24}/_{17}) is a surd.

### A simplest form of a surd:

If , where c does not contain any n-th power of a rational number, then it is the simplest form. Here b is the coefficient of the surd.

### Pure surd:

A surd which in its simplest form has 1 as coefficient.

### Mixed Surd:

A surd which has some rational number not equal to 1 as its coefficient written its simplest form.

### Similar surds:

Two surds are called similar surds or like surds if when written in simplest form they have the same order and the same radicand. Otherwise they are called unlike surds.

## Reduction of Surds of different orders to the same order:

Example: Reduce the surds √(24/98) and ∛16 to the surds of the same order

Solution:

Here the order are 2 and 3 respectively. Hence their LCM is 6. We write

## Surds – Exercise 1.3.3

**Write the following surds in their simplest form:**

**(i) √𝟕𝟔**

= √(19 × 4)

= √(19 ×2^{4})

**= **𝟐√𝟏𝟗

**(ii) **∛** (**𝟏𝟎𝟖)

= ∛(3^{3} × 2^{2})

= ∛2^{2}

**= **∛𝟒

**(iii) ****∜**𝟓𝟎𝟎𝟎

= **∜**5000

= **∜(**5^{11} × 8^{4})

**= **𝟓**∜**𝟖

**(iv) **∛^{𝟏𝟖𝟗}/_{𝟐𝟓}

= ∛[(3^{3} × 7)/_{5}^{2}]

**= 3**∛^{𝟕}/_{𝟐𝟓}

**(v) ****∜**^{𝟒𝟎𝟎}/_{𝟒𝟗}

= **∜(**2^{4} × 5^{2})/_{7}^{24}

= 2**∜**5^{2}/7^{2}

**= 2****∜**^{𝟐𝟓}/_{𝟒𝟗 }

**Classify the following in to like surds**

**(i) √**𝟐𝟒 **, √**𝟏𝟐𝟖 **, √**𝟕𝟓 **, √**𝟕𝟐 **, √**𝟓𝟒 **, √**𝟐𝟒

= **√**3^{5} , **√**2^{7} , **√(**5^{2}×3) , **√(**3^{2}×2^{3}) , **√(**2 × 3^{3}) , **√(**2^{3}×3)

= 3^{2}**√**3 , 2^{3}**√**2 , 5**√**3 , 3×2**√**2 , 3**√**(2×3), 2**√**(2×3)

= 9**√**3, 8**√**2, 5**√**3, 6**√**2, 3**√**6, 2**√**6

= {8**√**2, 6**√**2}, {9**√**3, 5**√**3}, {2**√**6, 3**√**6,}

**={ √**𝟏𝟐𝟖**, √**𝟕𝟐**}, { √**𝟐𝟒𝟑**, √**𝟕𝟓**}, { √**𝟓𝟒**, √**𝟐𝟒**,}**

**(ii) **∛𝟐𝟎𝟎𝟎**, **∛𝟔𝟖𝟔**, **∛𝟔𝟒𝟖**, **∛𝟑𝟕𝟓**, **∛𝟏𝟐𝟖**, **∛𝟐𝟒

= ∛(2×52), ∛(73×2), ∛(63× 3), ∛(5^{3}×3), ∛(4^{3}×2), ∛(2^{3}× 3)

= 10∛2, 7∛2, 6∛3, 5∛3, 4∛2, 2∛3

**= {4**∛𝟐**, 7**∛𝟐**, 10**∛𝟐**} & {2**∛𝟑**, 5**∛𝟑**, 6**∛𝟑**} **

**= { **∛𝟐𝟎𝟎𝟎**, **∛𝟔𝟖𝟔 **, **∛𝟏𝟐𝟖**} & { **∛𝟔𝟒𝟖**, **∛𝟑𝟕𝟓**, **∛𝟐𝟒**} **

**Which of the following are pure surds?**

(i) **√**296 = **√(**3^{3}×37) = 2**√(**37×2) = **2√**𝟕𝟒 – not pure surds

(ii) **√**729 = **√**3^{6} = 3^{3} = **27 – **not pure surds

(iii) ∛ 211 Cannot be reduced further hence it is a pure surd. **– Yes, a pure surd**

(iv) **∜** 75 is also a pure surd. **– Yes, **a pure surds

(v) ∛ 296 = ∛(2^{3}×37) = **2** ∛𝟑𝟕 – no, not a pure surd

(vi) **∜**296 = cannot be reduced further, hence it is a pure surd **– Yes **

**Write the following irrational numbers sure from**

**(i) √(**𝟏𝟓√(𝟐𝟕))

= **√**27 = **√**3^{3} = 3**√**3

= [15(27)^{1/2}]^{1/2}

= 15^{2/4} x 27^{1/4} [^{1}/_{2} = ^{4}/_{2}]

= ∜(15^{3} × 27)

= ∜(225 × 27)

**= **∜𝟔𝟎𝟕𝟓

** **

**(ii) √(**𝟒𝟎(**√**𝟏𝟐))

= [40 × (12)^{1/2}] ^{½}

= (40)^{1/2} × (12)^{1/4}

= ∜(40^{2} × (12)^{1})

= ∜(40^{2}× 121)

= ∜(1600×12)

= ∜19200

**(iii) √(**𝟓(**√**𝟒𝟖))

= [5^{1/2} ×(48)^{1/2}]^{½}

= [5^{1/2} × 48^{1/4}]

= ∜ (5^{2}× 48)

= ∜(25×48)

= ∜1200

**5. Reduce the following to surds of the same order. **

**(i) ****∛** **, √**𝟐**and **𝟓^{1/5}

The orders are 3, 2, and 5

LCM of 2, 3 and 5 is 30

∛2 = 2^{1/3 }= 2^{10/30} = (𝟐^{𝟏𝟎})^{ 1/3𝟎}

**√**2 = 2^{1/2 }= 2^{15/30} = (𝟐^{𝟏𝟓 })^{ 1/3𝟎}

5^{1/5} = 5^{1/5} = 5^{6/30} = (𝟓^{𝟔}) ^{1/3𝟎}

(1024)^{1/3𝟎} , (32768)^{ 1/3𝟎} , (15625)^{ 1/3𝟎}

They have the same order 30

**(ii) √**𝟓 **, (√**𝟏𝟓)^{1/4} **, and (√**𝟓𝟎)^{1/8}

Order is 2, 4, 8

Their LCM is 8

**√**5 = 5^{1/2} = 5^{4/8} = (5^{4 })^{1/8} = (125)^{1/8}

**(√**15)^{1/4} = (**√**15)^{1/4} = (15)^{2/8} = ((15)^{2})^{ 1/8} = (225)^{1/8}

(50)^{1/8} is in its simplest form

(50)^{1/8} **, **𝟐𝟐𝟓𝟖 **, **𝟏𝟐𝟓𝟖

∴ Thus they all are of the same order 8.

**(iii) √**𝟐 **, √**𝟕^{1/3} **, (√**𝟏𝟏)^{1/4} **and (√**𝟏𝟔𝟕𝟏)^{𝟏/1𝟐}

Order is 2, 3, 4, and 12

Their LCM is 12

**√**2 = 2^{1/2} = 2^{6/12} = (**√**2^{6})^{1/12} = (64)^{1/12}

**√**7^{1/3} = 7^{4/12} = (7^{4} )^{1/12} = (2401)^{1/12}

**(√**11)^{1/4} = (11)^{3/12} = ((11)^{3})^{1/12} = (1331)^{1/12}

167112 is in its simplest form

𝟔𝟒^{𝟏/1𝟐} **, **𝟐𝟒𝟎𝟏^{𝟏/1𝟐} **, **𝟏𝟑𝟑𝟏^{𝟏/1𝟐} **, **𝟏𝟔𝟕𝟏^{𝟏/1𝟐}

∴ Thus they all are of the same order 12

## 1.3.4 Comparing Surds and Some Irrational Numbers:

Example 7: Find which surd is larger

## Surds – Exercise 1.3.4

**Find which is larger:**

**(i) 3****∛**𝟑 **and 4****∜**𝟒

**∛(**3×3^{2}) 4×424

= **∛**27 x **∛**3 = **∜**256 x **∜**4

= **∛**81 = **∜**1024

= ((81)^{4})^{1/12} = ((1024)^{4})^{1/12}

= (43046721)^{1/12} = (1073741824)^{1/12}

**4****∜**𝟒 **is greater 3****∛**𝟑

**Compare the following and decide which is larger.**

**(i) (∜(30)) ^{1/7} and ∛(28^{1/10})**

(28^{1/10}) ^{1/3} = (28^{1/10})^{1/3} = 28^{1/30}

(30^{1/4}) ^{1/7} = (30^{1/4})^{1/7} = 30^{1/28}

LCM of 30 and 28 is 420

(28^{1/10})^{1/3} = 28^{1/30} = (28^{1/14})^{1/420}

(30^{1/4}) ^{1/7} = 30^{1/28} = (30^{15})^{1/420 }

28^{14} =2^{2} x 7^{14} = 2^{28} x 7^{14}

30^{15} = (5 x 6)^{15} = 5^{15} x 6^{15 }

Comparing the 2 numbers we conclude

30^{15} > 28^{14}

(∜(30))^{1/7} > ∛(28^{1/10})

**(ii) √(∜8) ** **and ∛(∛9)**

√(∜8) = (8^{1/4})^{1/2} = 8^{1/8} = 2^{3/8} [8 = 2^{3}]

∛(∛9) = (9^{1/3})^{1/3} = (3^{2/3})^{1/3} = 3^{2/9}

LCM of 8 and 9 is 72

√(∜8) = 2^{3/8} = (2^{3/8})^{9/9} = 2^{27/72} = (2^{27})^{1/72}

∛(∛9) = 3^{2/9} = (3^{2/9})^{8/8} = 3^{16/72} = (3^{16})^{1/72 }

By comparing we find that (2^{27}) is larger than 3^{16}.

**Hence **√(∜8) **> **∛(∛9)

**Write the following in ascending order:**

√𝟐 **, ****∛**𝟑𝟑 **, **𝟔^{1/6}

2^{1/2} , 3^{1/3} , 6^{1/6 }

LCM of 2, 3 and 6 is 6

2^{1/2} = 2^{3/6} = (2^{3})^{1/6} = 8^{1/6}

3^{1/3} = 3^{2/6} = (3^{3})^{1/6} = 9^{1/6}

6^{1/6} = 6^{3/6 }= (6)^{1/6} = 6^{1/6}

**Ascending order is **6^{1/6 } **, **2^{1/2}**, **3^{1/3}

**Write the following descending order:**

√(**∛**𝟔) **, **∛**(**∜𝟏𝟐) **, **√** (**∜𝟖)

(6^{1/3})^{1/2} , (12^{1/4})^{1/3} , (8^{1/4})^{1/2}

6^{1/6}, 12^{1/12} , 8^{1/8 }

LCM of 6, 12, 8 is 24

6^{4/24}, 12^{2/24}, 8^{3/24}

(6^{4})^{1/24}, (12^{2})^{1/24}, (8^{3})^{1/24}

(1296)^{1/24}, (144)^{1/24}, (512)^{1/24}

Descending order is √(**∛**𝟔) , √** (**∜𝟖) , ∛**(**∜𝟏𝟐)