Number System

NUMBER SYSTEM CONTENTS  Number system  Decimal representation of Rational numbers  Conversion of decimal numbers into rational numbers  Representing irrational numbers on the numbers line   Surds or Radicals   Some Rules For Exponents  Natural Numbers : The simplest numbers are 1, 2, 3, 4....... the numbers being used in counting. These are called natural numbers.  Whole numbers : The natural numbers along with the zero form the set of whole numbers i.e. numbers 0, 1, 2, 3, 4 are whole numbers. W = {0, 1, 2, 3, 4....}  Integers : The natural numbers, their negatives and zero make up the integers. Z = {....–4, –3, –2, –1, 0, 1, 2, 3, 4,....} The set of integers contains positive numbers, negative numbers and zero.  Rational Number : (i) A rational number is a number which can be p put in the form , where p and q are both q integers and q  0. (ii) A rational number is either a terminating or (iii) A rational number may be positive, negative or zero.  Complex numbers : Complex numbers are imaginary numbers of the form a + ib, where a and b are real numbers and i= – 1 , which is an imaginary number.  Factors : A number is a factor of another, if the former exactly divides the latter without leaving a remainder (remainder is zero) 3 and 5 are factors of 12 and 25 respectively.  Multiples : A multiple is a number which is exactly divisible by another, 36 is a multiple of 2, 3, 4, 9 and 12.  Even Numbers : All integers which are multiples of 2 are even number (i.e.) 2,4, 6, 8............... are even numbers.  Odd numbers : All integers which are not multiples of 2 are odd numbers.  Prime and composite Numbers : All natural numbers which cannot be divided by any number other than 1 and itself is called a prime number. By convention, 1 is not a prime number. 2, 3, 5, 7, 11, 13, 17 ............. are prime numbers. Numbers which are not prime are called composite numbers.  The Absolute Value (or modulus) of a real Number : If a is a real number, modulus a is written as |a| ; |a| is always positive or zero.It means positive value of ‘a’ whether a is positive or negative |3| = 3 and |0| = 0, Hence |a| = a ; if a = 0 or a > 0 (i.e.) a  0 |–3| = 3 = – (–3) . Hence |a| = – a when a < 0 Hence, |a| = a, if a > 0 ; |a| = – a, if a < 0  Irrational number :  EXAMPLES  (i) All real numbers are irrational if and only if there decimal representation is non-terminating and non-repeating. e.g. 2 , Ex.1 3 , ............ etc. (ii) Rational number and irrational number taken together form the set of real numbers. Sol. (iii) If a and b are two real numbers, then either (i) a > b or (ii) a = b or (iii) a < b (iv) Negative of an irrational number is an irrational number. Ex.2 (v) The sum of a rational number with an irrational number is always irrational. Sol. Is zero a rational number? can you write it in p the form , where p and q are integers q and q  0? Yes, zero is a rational number. It can be 0 0 0 written as = = etc. where denominator 1 2 3 q  0, it can be negative also. 3 4 and . 5 5 rs A rational number between r and s is . 2 A rational number between Find five rational numbers between 4 1 3 4 3 7 and =    = . 5 5 2  5 5  10 (vi) The product of a non-zero rational number with an irrational number is always an irrational number. And a rational number between (vii) The sum of two irrational numbers is not always an irrational number. 1 3 7 and = 5 10 2 (viii) The product of two irrational numbers is not always an irrational number. 5 27 31 3 4 , , are between and . 8 40 40 5 5 So, five rational number between Similarly;  Rational Numbers : 3, 4, 7 5 3 , ,  , 2.7, 3.923, 1.42 7 , 1.2343434, 3 2 7 etc. 2 , 3 , 5 , 6 , , 1.327185………… Ex.3 Find six rational numbers between 3 and 4. Sol. We can solve this problem in two ways. A rational number between r and s is rs . 2 Therefore, a rational number between 3 and 1 7 4= (3 + 4) = 2 2  Imaginary Numbers :  3  , ……….. 8  Note:-  = 3.14159265358979…………. while 22 = 3.1428571428…………. 7    4 3 5 13 7 31 27 and are , , , , 5 5 8 20 10 40 40 Method 1 :  Irrational Numbers : 5  2 ,  49 , 3i,   7  3 7  13   =  5 10  20 22 22 but for calculation we can take   . 7 7 A rational number between 3 and 7 1 67 13 = = . We can accordingly 2 2 2 4 proceed in this manner to find three more rational numbers between 3 and 4. Hence, six rational numbers between 3 and 4 15 13 27 7 29 15 are , , , , , . 8 4 8 2 8 4 Method 2 : Since, we want six numbers, we write 3 and 4 as rational numbers with denominator 6 + 1, 21 28 i.e., 3 = and 4 = . Then we can check 7 7 22 23 24 25 26 27 that , , , , , and are all 7 7 7 7 7 7 between 3 and 4. Ex.7 Sol. 7 in the decimal form by long 8 division method. We have, Express Hence, the six numbers between 3 and 4 are 22 23 24 25 26 27 , , , , , and 7 7 7 7 7 7 Ex.4 Are the following statement true or false? Give reasons for your answer.  (i) Every natural number is a whole number. (ii) Every integer is a whole number. (iii) Every rational number is a whole number. Sol. (i) True, because natural number starts from 1 to  and whole number starts from 0 to . Ex.8 Sol.  7 = 0.875 8 35 into decimal form by long 16 division method. We have, Convert (ii) False, because negative integers are not whole number. (iii) False, because rational number such that 1 is not whole number. 2 Ex.5 Find 3 irrational numbers between 3 & 5. Sol.  3 and 5 both are rational  35 = 2.1875 16 The irrational are 3.127190385…………… 3.212325272930……… Ex.9 Express 2157 in the decimal form. 3.969129852937………… Sol. We have, Ex.6 Find two rational & two irrational numbers between 4 and 5. Sol. Rational numbers & 625 625 2154.0000 3.4512 1875 2820 2500 3200 3125 750 625 1250 1250 0 45  4.5 Ans. 2 4.5  4 8.5   4.25 2 2 Ans. Irrational numbers 4.12316908……… Ans. 4.562381032…….. Ans.  2157 = 3.4512 625 –17 in decimal form by long 8 division method. Ex.10 Express Ex.12 Express 2 as a decimal fraction. 11 By long division, we have Sol. –17 In order to convert in the decimal form, 8 17 we first express in the decimal form and 8 –17 will be negative of the decimal form of 8 17 the decimal form of 8 Sol. we have,    –17  = – 2.125 8   2 = 0.181818 ..... = 0.18 11 Ex.13 Find the decimal representation of By long division, we have Sol. 8 Ex.11 Find the decimal representation of . 3 By long division, we have Sol.     8 = 2.6666 ... = 2. 6 3  16 = 0.3555 .... = 0.3 5 45 Hence, 16 = – 0.3 5 45 16 45 Ex.14 Find the decimal representation of Sol. 22 . 7 (B) Non terminating recurring (repeating) By long division, we have (remainder  0, but equal to devidend) Ex. 10 = 3.333 ……….. or 3. 3 3 1 = 0.1428514285……….or 0.142857 7 2320 = 23.434343……..or 23.43 99 These expansion are not finished but digits are continusely repeated so we use a line on those digits, called bar ( a ). So we can say that rational numbers are of the form either terminating, non repeating or non terminating repeating (recurring). m n Case I : When the decimal number is of terminating nature. Algorithm : Step-1 : Obtain the rational number. 22  = 3.142857142857 ..... = 3. .142857 7 So division of rational number gives decimal expansion. This expansion represents two types (A) Terminating (remainder = 0) 6 8 7 , , ,………are equal to 1.2, 1.6, 5 5 4 1.75 respectively, so these are terminating and non repeating (recurring) Ex. Step-2 : Determine the number of digits in its decimal part Step-3 : Remove decimal point from the numerator. Write 1 in the denominator and put as many zeros on the right side of 1 as the number of digits in the decimal part of the given rational number. Step-4 : Find a common divisor of the numerator and denominator and express the rational number to lowest terms by dividing its numerator and denominator by the common divisor. 5 Ex.15 Express each of the following numbers in the p form . q (i) 0.15 Sol. (ii) 0.675 (i) 0.15 = (iii) –25.6875 Step-2 : Write the number in decimal form by removing bar from the top of repeating digits and listing repeating digits at least twice. For sample, Dividingnumeratorand denominator  by the commondivisor5 of numerator and denominator  3 20 (ii) 0.675 = = = write x = 0.8 as x = 0.888.... and x = 0.14 as x = 0.141414...... Step-3 : Determine the number of digits having bar on their heads. Step-4 : If the repeating decimal has 1 place repetition, multiply by 10; a two place repetition, multiply by 100; a three place repetition, multiply by 1000 and so on. 675 1000 675  25 27 = 1000  25 40 (iii) – 25.6875 = Algorithm : Step-1 : Obtain the repeating decimal and pur it equal to x (say) 15 100 15  5 = 100  5 =  Conversion of a pure recurring decimal to the p form q 256875 10000 256875 625 411 = 10000 625 16 Case II : When decimal representation is of nonterminating repeating nature. In a non terminating repeating decimal, there are two types of decimal representations (i) A decimal in which all the digit after the decimal point are repeated. These type of decimals are known as pure recurring decimals. For example: 0.6, 0.16, 0.123 are pure recurring decimals. (ii) A decimal in which at least one of the digits after the decimal point is not repeated and then some digit or digits are repeated. This type of decimals are known as mixed recurring decimals. For example, 2.16, 0.35, 0.785 are mixed recurring decimals. Step-5 : Subtract the number in step 2 from the number obtained in step 4 Step-6 : Divide both sides of the equation by the coefficient of x. Step-7 : Write the rational number in its simplest form.  EXAMPLES  Ex.16 Express each of the following decimals in the p form : q (i) 0.6 (ii) 0.35 (iii) 0.585 Sol. (i) Let x = 0.6 then, x = 0.666....... ....(i) Here, we have only one repeating digit, So, we multiply both sides of (i) by 10 to get 10 x = 6.66.... ....(ii) Subtracting (i) from (ii), we get 10 x – x = (6.66 .....) – (0.66.....) 6  9x = 6  x= 9 2 2 x = Hence 0.6 = 3 3 (ii) Let x = 0.35      x = 0.353535.... ....(i) Here, we have two repeating digits after the decimal point. So, we multiply sides of (i) by 102 = 100 to get 100x = 35.3535...... ....(ii) Subtracting (i) from (ii), we get 100 x – x = (35.3535....) – (0.3535....)  99 x = 35 35  x= 99 Hence, 0.35  (iii) Let x = 0.585  x = 0.585585585... ....(i) Here, we have three repeating digits after the decimal point. so, we multiple both sides of (i) by 103 = 1000 to get 1000 x = 585.585585......... ....(ii) Subtracting (i) from (ii), we get 1000x – x = (585.585585...) – (0.585585585...) 1000x – x = 585  999x = 585 585 195 65   x= = = 999 333 111 The above example suggests us the following rule to convert a pure recurring decimal into a p rational number in the form . q Ex.17 Convert the following decimal numbers in the p form : q (i) 5.2 (ii) 23.43 Sol. (i) Let x = 5.2  x = 5.2222 ....(i) Multiplying both sides of (i) by 10, we get 10 x = 52.222 ...... ....(ii) Subtracting (i) from (ii), we get 10x–x = (52.222...) – (5.222....)  9x = 47 47   x= 9   (ii) Let x = 23.43  x = 23.434343..... Multiplying both sides of (i) by 100, we get 100 x = 2343.4343....... ....(ii) Subtracting (i) from (ii), we get 100x – x = (2343.4343...) – (23.4343....)   99 x = 2320 2320   x= 99  Conversion of a mixed recurring decimal to the p Form :q Algorithm : Step-1 : Obtain the mixed recurring decimal and write it equal to x (say) Step-2 : Determine the number of digits after the decimal point which do not have bar on them. Let there be n digits without bar just after the decimal point Step-3 : Multiply both sides of x by 10n so that only the repeating decimal is on the right side of the decimal point. Step-4 : Use the method of converting pure p and obtain the recurring decimal to the form q value of x  EXAMPLES  Ex.18 Express the following decimals in the form (i) 0.32 (ii) 0.123 Sol. (i) Let x = 0.32 Clearly, there is just one digit on the right side of the decimal point which is without bar. So, we multiply both sides of x by 10 so that only the repeating decimal is left on the right side of the decimal point.  10x = 3.2  10x = 3 + 0.2 2 9 9 3  2 29 =  10x = 9 9  10x = 3 +  10x    x= 2   0.2  9    29 90    (ii) Let x = 0.123 Clearly, there are two digits on the right side of the decimal point which are without bar. So, we multiply both sides of x by 102 = 100 so that only the repeating decimal is left on on the right side of the decimal point.  100x = 12. 3  100x = 12 + 0. 3 3   100x = 12 + 9 12  9  3  100x = 9 108  3  100x = 9 111   100x = 9 111 37   x= = 900 300 Ex.19 Express each of the following p recurring decimals in the form ; q (i) 4.32 Sol. (i) Let x       Ex.20 Represent 3.765 on the number line. Sol. This number lies between 3 and 4. The distance 3 and 4 is divided into 10 equal parts. Then the first mark to the right of 3 will represent 3.1 and second 3.2 and so on. Now, 3.765 lies between 3.7 and 3.8. We divide the distance between 3.7 and 3.8 into 10 equal parts 3.76 will be on the right of 3.7 at the sixth mark, and 3.77 will be on the right of 3.7 at the 7th mark and 3.765 will lie between 3.76 and 3.77 and soon. mixed (ii) 15.712 = 4.32  10x = 43.2 [Multiplying both sides of x by 10] 2  10x = 43 + 0.2   10x = 43 + 9 43  9  2 387  2  10x =   10x = 9 9 389 389  10x =    x = 9 90 (ii) Let x = 15.712 . Then, 10x = 157.12  10x = 157 + 0.12 12  10x = 157 + 99 4  10x = 157 + 33 157  33  4  10x = 33 5181  4  10x = 33 5185 5185 1037  10x =  x = = 33 330 66 Ex.21 Visualize 4.26 on the number line, upto 4 decimal places. Sol. We have, 4.26 = 4.2626 This number lies between 4 and 5. The distance between 4 and 5 is divided into 10 equal parts. Then the first mark to the right of 4 will represent 4.1 and second 4.2 and soon. Now, 4.2626 lies between 4.2 and 4.3. We divide the distance between 4.2 and 4.3 into 10 equal parts 4.2626 lies between 4.26 and 4.27. Again we divide the distance between 4.26 and 4.27 into 10 equal parts. The number 4.2626 lies between 4.262 and 4.263. The distance between 4.262 and 4.263 is again divided into 10 equal parts. Sixth mark from right to the 4.262 is 4.2626. Ex.22 Express the decimal 0.00352 in the form p q Let x = 0.00352 Clearly, there is three digit on the right side of the decimal point which is without bar. So, we multiply both sides of x by 103 = 1000 so that only the repeating decimal is left on the right side of the decimal point.   1000x = 3.52   1000x = 3 + 0.52 52   1000x = 3 + 99 3  99  52 297  52   1000x = 1000x = 99 99 349 349   1000x = x = 99 99000 Ex.23 Give an example of two irrational numbers, the product of which is (i) a rational number (ii) an irrational number Sol. Sol. (i) The product of 27 and which is a rational number. 3 is 23 = 6 We know that, if a and b are two distinct positive irrational numbers, then ab is an irrational number lying between a and b.  Irrational number between Irrational number between 2 6 3 is 6 = 61/4 2 3 = 1/ 4 2 and 1/4 =2 1/4 2 and 6 is 1/8 ×6 . Hence required irrational number are 61/4 and 21/4 × 61/8 Ex.27 Find two irrational numbers between 0.12 and 0.13. Sol. Let a = 0.12 and b = 0.13. Clearly, a and b are rational numbers such that a < b. We observe that the number a and b have a 1 in the first place of decimal. But in the second place of decimal a has a 2 and b has 3. So, we consider the numbers c = 0.1201001000100001 ...... and, d = 0.12101001000100001....... Clearly, c and d are irrational numbers such that a < c < d < b. Ex.28 Find two rational numbers 0.232332333233332.... and between 0.252552555255552...... Sol. Let a = 0.232332333233332.... and b = 0.252552555255552..... Ex.25 Find two irrational numbers between 2 and 2.5. Sol. If a and b are two distinct positive rational numbers such that ab is not a perfect square of a rational number, then ab is an irrational number lying between a and b.   Irrational number between 2 and 2.5 is 2  2.5 = 5 Similarly, irrational number between 2 and 2 5 So, required numbers are Sol. 81 = 9, (ii) The product of 2 and 3 is 6 , which is an irrational number. Ex.24 Insert a rational and an irrational number between 2 and 3. Sol. If a and b are two positive rational numbers such that ab is not a perfect square of a rational number, then ab is an irrational number lying between a and b. Also, if a,b are ab rational numbers, then is a rational 2 number between them.   A rational number between 2 and 3 is 23 = 2.5 2 An irrational number between 2 and 3 is 5 is Ex.26 Find two irrational numbers lying between 2 and 3 . The numbers c = 0.25 and d = 0.2525 Clearly, c and d both are rational numbers such that a < c < d < b. Ex.29 Find a rational number and also an irrational number between the numbers a and b given below: a = 0.101001000100001...., 5 and 2 5 . b = 0.1001000100001... Sol. Since the decimal representations of a and b are non-terminating and non-repeating. So, a and b are irrational numbers. We observed that in the first two places of decimal a and b have the same digits. But in the third place of decimal a has a 1 whereas b has zero. Ex.30 Find one irrational number between the number a and b given below : a = 0.1111..... Clearly, a and b are rational numbers, since a has a repeating decimal and b has a terminating decimal. We observe that in the third place of decimal a has a 1, while b has a zero. Sol.  a>b Construction of a rational number between a and b : As mentioned above, first two digits after the decimal point of a and b are the same. But in the third place a has a 1 and b has a zero. So, if we consider the number c given by  a>b Consider the number c given by c = 0.111101001000100001..... Clearly, c is an irrational number as it has non-repeating and non-terminating decimal representation. c = 0.101 Then, c is a rational number as it has a terminating decimal representation. We observe that in the first two places of their decimal representations b and c have the same digits. But in the third place b has a zero whereas c has a 1. Since b has a zero in the third place of decimal and c has a 1.  bb Also, comparing a and d, we obtain a > d Thus, d is an irrational number such that b < d < a.  Represent 2 & 3 on the number line : Greeks discovered this method. Consider a unit square OABC, with each side 1 unit in lenght. Then by using pythagoras theorem 1 C 1 1 1 O OB = 1 1 = B A 2 10 Now, transfer this square onto the number line making sure that the vertex O coincides with zero  Real Number Line : 2 With O as centre & OB as radius, draw an arc, meeting OX at P. Then OB = OP = 2 units Then, the point represents 2 on the number line Now draw, BD  OB such that BD = 1 unit join OD. Then OD = ( 2 ) 2  (1) 2 = 3 3 units Irrational numbers like 2 , 3 , 5 etc. can be represented by points on the number line. Since all rational numbers and irrational numbers can be represented on the number line, we call the number line as real number line.  Surds : 2 , 3 , 5 , 21 ,................ are irrational numbers, These are square roots (second roots), of some rational numbers, which can not be written as squares of any rational number. (i) If a is rational number and n is a positive integer such that the nth root of a is an irrational number, then a1/n is called a surd or radical. 2 Wit h O as centre & OC as radius, draw an arc, meeting OX at Q. Then OQ = OD = 3 on the real line Remark : In the same way, we can locate n for any positive integer n, after n  1 has been located.  Existence of The value of 5, 2, 3 etc. (ii) If is a surd then ‘n’ is known as order of surd and ‘a’ is known as radicand. 3 units Then, the point Q represents e.g. n for a positive real number : 4.3 geometrically : - (iii) Every surd is an irrational number but every irrational number is not a surd.  Quadratic Surd: A surd of order 2 is called a quadratic surd. 3 = 31/2 is a quadratic surd but Ex.31 not a quadratic surd, because Draw a line segment AB = 4.3 units and extend it to C such that BC = 1 unit. Find the midpoint O of AC. With O as centre and OA a radius, draw a semicircle. a rational number. So, 9 = 91/2 is 9 = 91/2 = 3 is 9 is not a surd.  Cubic surd : A surd of order 3 is called a cubic surd. Ex.32 The real number real number surd. 3 3 4 is a cubic surd but the 8 is not a cubic surd as it not a  Biquadratic surd : Now, draw BD  AC, intersecting the semicircle at D. Then, BD = 4.3 units. With B as centre and BD as radius, draw an arc, meeting AC produced at E. Then, BE = BD = 4.3 units A surd of order 4 is called a biquadratic surd. A biquadratic surd is also called a quadratic surd. Ex.33 4 5 is a biquadratic surd but 4 81 is not a biquadratic surd as it is not a surd.  Laws of radicals : For any positive integer ‘n’ and a positive rational number ‘a’. n (a) (b) n a n  EXAMPLES  Ex.34. State with reasons which of the following are surds and which are not : a n (i) a  b = ab [one of either a or b should be non-negative integer] na a = n (c) n b b n m a (d) m n a = mn a = (e) (f) p p an m = p an  am = (iii) (ii) 20  45 45 (iv) 8 10  4 15 (v) 3 12  6 27 (vi) 3 5  3 25 64 = 8 Sol. (i) 8 is a rational number, hence 64 is not a surd. a n. m p (ii) 45 = 9  5 = 3 5 a n m Thus (iv) A surd which has unity only as rational factor is called a pure surd. (v) A surd which has a rational factor other than unity is called a mixed surd. (vi) Surds having same irrational factors are called similar or like surds. (iii) We have = Surds of same order can be multiplied or divided and 3  2 or 2 + 5 and 2 – 5 are conjugate surds. 30 8 10 ÷ 4 15 = 2 = 30 8 10 4 15 = (xi) A surd consisting of one term only is called a monomial surd. (xiii)The binomial surds which differ only in sign (+ or –) between the terms connecting them, are called conjugate surds e.g. 3  2 30  30 = 900 (iv) we have (x) If the product of two surds is a rational number, then each one of them is called the rationalising factor of the other. 2  5 , 3  2, 2  3 etc. are binomial surds. 20  45 = Which is a rational number and therefore 20  45 is not a surd. (ix) If the surds to be multiplied or to be divided are not of the same order, we first reduce them to the same order and then multiply or divide. (xii) An expression consisting of the sum or difference of two monomial surds or the sum or difference of a monomial surds and a rational number is called binomial surd. e.g. 45 is an irrational number. Because the rational number 45 is not the square of any rational number, hence 45 is a surd. (vii) Only similar surds can be added or subtracted by adding or subtracting their rational parts (viii) 64 n 8 4 =    2 2 10  15 8  8  10 4  4  15 8  8  10 640  4  4  15 240 8 = 3 8 3 Which is an irrational number. 8 is not the 3 square of any rational number, hence the given expression is a surd. Because the rational number (v) 3 12 ÷ 6 27 = 3 12 6 27 6 108   972   2 2 Ex.37 Simplify each of the following: 12 27 Sol. 3  3  12 3  3  12   6  6  27 = 3 = 6  6  27 (i) 3 3· 3 4 (ii) 3 128 (i) 3 3· 3 4 = 3 (ii) 3 128 = 3 64  2 = 3 64 1 1 = 9 3  = 1 is a rational number, therefore 3 3 12 ÷ 6 27 is not a surd. Since (vi)   3 5 × = 3 3 25 = 3  (i) 3 Hence, Sol. (i) 3 4 8 27 53 = 5 (i) ( 3 5 )3 (ii) 3 (i) ( 5 ) = 5 3 64 = 3 (ii) 3 8 = 27 = 3 64 (i) Sol. (i) 3 4 (ii) (Using Ist Law) 43 = 4 3 4x  7 – 5 = 0 (ii) 4 23 3 43 = 4] 4 4 = 3888 4 48 [Using IIIrd Law] = 33 3888 = 48 2 3 [Using Ist Law] 3888 48 81 = 4 [Using IIIrd Law] 34 = 3 [Using Ist Law] Ex.39 Siplify each of the following 3x  1 = 2 (i) 4 3 3 (i) 4 3 3 = (ii) 2 3 5 4x  7 – 5 = 0 3 Sol.     ( 3 4x  7 )3 = 53   4x – 7 = 125   4x = 132   x = 33 (ii) 4   ( 4 3x  1 )4 = 24   3x + 1 = 16   3x = 15   x=5 12 3 [Using IVth law] 6 5 [Using IVth law] 4x  7 = 5 (ii) 2 3 5 = [( n a ) n = a] Ex.40 Simplify :  3 8 27 4 (Using Ist Law) Ex.36 Find the value of x in each of the following: 3 2 Ex.38 Simplify each of the following: Ex.35 Simplify the following : (ii) 3 = 4 2 [Using I, 3 Sol. 2 [Using IInd Law] 43 · 5  25 Which is a rational number. 3 5 × 3 25 is not a surd. 3 3 3 3 555 3 3 4 = 3 12 [Using IInd Law] Sol. ( 23 ) 4 Using the above property, we have 5 4 3x  1 = 2 5 4 (23 )4 = 5 23 = 5 8.  Pure And Mixed Surds : [( n a ) n = a] (i) Pure Surd : A surd which has unity only as rational factor, the other factor being irrational, is called a pure surd. Ex.41 3 , 5 2 , 4 3 are pure surds. Ex.42 6 , 3 12 are pure surds. Ex.45 Expressed each of the following as pure surds- (ii) Mixed Surd : Sol. (i) A surd which has a rational factor other than unity, the other factor being irrational, is called a mixed surd. Ex.43 2 3 . 108 3 (i) EXAMPLES 3 1/ 3 1/ 3  3   3 4 3 (ii) . 2 = (8×4)1/3 = (32)1/3 = (iv) 3 4 1/ 4 8 3 1/ 4  32  ×   243  1/ 4  32  ×   243  1/ 4  2 =    3 = 4 1/ 4  81 32  =    16 243  2 3 Ex.46 Express each of the following as pure surd : = (4×3)1/2 = 121/2 = 12 (ii) 2. 32 3  32  32 × =  2  243  243  81  =    16  Sol. (i) 2 3 = 2 ×31/2 =(22)1/2 × 31/2 =41/2 ×31/2 3 1/ 4 4   3 4  =     2     (ii) 2. 3 4 32 × (108)1/3 1/3  8  =   108   27  Ex.44 Express each of the following as a pure surd. (iii) 32 243  8  × (108)1/3 =    27  1/ 4 (i) 2 3 4 2 × (108)1/ 3 3 108 = 3  =  2   2 3 , 5 3 12 , 2 4 5 are mixed surds. TypeI : On expressing of mixed surds into pure surds 2 3 3 2 (ii) (ii) a 3 b 2 (i) a a  b (iii) 2ab 3 ab 4 = 2×41/3 = (23)1/3 ×41/3 = 81/3 ×41/3 Sol. (i) a a  b = a × (a + b)1/2 1/3 1/ 3 = (8  4) = (32) = 3 32 = (a2)1/2 × (a + b) 1/2 = [a2× (a + b)]1/2 (iii) 3 4 2 32 =  3    32 =  4 2  3    32  4 = (a3 + a2b)1/2 = a3  a 2b 3 (ii) a b 2 = (a3) 1/3 × (b2)1/3 = (a3×b2)1/3 = 9  32 = 18 16 = (iii) (iv) 3 4 2 8 = 3    8 = 4 2ab. 3 3 a 3b 2 ab = 2ab 3 1/ 3  (ab)1/ 3 2 3   8 4 = (8 a3b3.ab)1/3 = (8 a4b4)1/3 = = 9 8 = 16 9 2 3 8a 4 b 4 TypeII : On expressing given surds as mixed surds in the simplest form. EXAMPLES Ex.49 Ex.47 Express each of the following as mixed surd in its simplest form: (i) 80 (iii) 5 (v) Sol. (i) (ii) (iii) 5 288 5 3 Ex.50 2 2 16  5 = 4  5 = 4 × 5 = 4 5 72 = 3 8  9 = 3 23  9 = 3 23 × 288 = 5 32 9 = 5 25  9 = 5 5 320 = 5 32  10 = 5 3 9 = 23 9 (iii) Ex.51 25 × 5 9 = 2 5 9 25  10 = (121) 3  33 = (112)3 × 11 = 116 × 111 3 = 117 Ans. (vi) 5. 3 135 (iv) 1350 = 225 6 = 152  6 = 15 2 × 6 =15 6 (v) = 73 × 7 = 73+1 = 74 Ans. (ii) (ap)q = apq 72 (iv) 1350 320 80 = 3 (ii) ( 7) 3  21 =? 3 5 a p q   a q 1 / a q  p ap (169) 2 132 Ex.52 3 3 25 × 5 10 3 [(13) 2 ]2 132  134 132 53  2 4  49 625  32  7 = 2 5 10 3 if q  p = 134–2 = 132 = 169 Ans. = 3  if p  q 53  2 4  7 2 5 4  25  7 = = 3 (vi) 5. 135 = 5 27  5 = 5 3  5 = 5 3 × 5 1 5 4 3  1 2 5 4  7 2 1 7 7  Ans. 5  2 10 (iv) (ab)p = apbp 3 =5×3× Ex.48 Express form : Sol. 4 = 4 1280 = 4 3 5 = 15 5 . 1280 as mixed surd in its simplest 4 256  5 = 4 256  4 5 44  4 5 = 4 4 5 Ex.53 ( 63) 4  144 ? 132  9 (9  7) 4  (12) 2 (11 12)  32 = (32 ) 4  7 4  32  (2 2 ) 2 38  7 4  3 2  2 4 = 11  2 2  31 2 11  2 2  33 = 2 4  38  2  7 4 2 2  33  11 = 2 2  37  7 4 Ans. 11 Let a, b > 0 be real numbers & p, q are rational numbes then (i) ap. ap = ap+q i.e. if base is same with different or same powers are multiply each other then powers are add. (32  7) 4  (3  2 2 ) 2 11  2 2  3  32 = (v) a0 = 1 = = 2 4 2  310 3  7 4 11 Ex.54 ( 2 0  30 ) 5 2 25 = = Ex.56 (1  1) .52 52 2 5  2 2 (vi) a+p = 2 2  2 50 1 1 a or a–p = p Ans. 1 ap 2 ? 4 Ex.55 = = ( 2)1/ 2 22 1 1 2 ( 2) 2  1 ( 2) 3 / 2 = 2–3/2 Ans. 11 / 3 ? (11 / 3) 7 = 1 (11 / 3) 7 1 = 1 (11 / 3) 6 6 3 =   Ans.  11 