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NUMBER SYSTEM, , 1, CHAPTER, , 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, , NUMBER SYSTEM, , 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, non-terminating but recurring (repeating), decimal., , (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 :, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 1
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(i) All real numbers are irrational if and only if, there decimal representation is non-terminating, and non-repeating. e.g. 2 , 3 , ............ etc., (ii) Rational number and irrational number taken, together form the set of real numbers., , Sol., , 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., , Ex.2, , Find five rational numbers between, , (iv) Negative of an irrational number is an, irrational number., , 4, 3, and ., 5, 5, r+s, A rational number between r and s is, ., 2, A rational number between, , (v) The sum of a rational number with an, irrational number is always irrational., , 1 3 4, 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, 1 3 7 13, 3, 7, and, = + =, 10, 5, 2 5 10 , 20, , (vii) The sum of two irrational numbers is not, always an irrational number., , Similarly;, , (iii) If a and b are two real numbers, then either, (i) a > b or (ii) a = b or (iii) a < b, , Sol., , 4, 5 27 31, 3, ,, ,, are between, and ., 8 40 40, 5, 5, So, five rational number between, 4, 5 13 7 31 27, 3, and, are ,, ,, ,, ,, 8 20 10 40 40, 5, 5, , (viii) The product of two irrational numbers is not, always an irrational number., Rational Numbers :, 7 5, 3, 3, 4, ,, , − , 2.7, 3.923, 1.427 , 1.2343434,, 3 2, 7, etc., , Ex.3, , Find six rational numbers between 3 and 4., , Sol., , We can solve this problem in two ways., Method 1 :, , Irrational Numbers :, , A rational number between r and s is, , 2 , 3 , 5 , 6 , , 1.327185…………, , Therefore, a rational number between 3 and, 1, 7, 4=, (3 + 4) =, 2, 2, , Imaginary Numbers :, 5, − 2 , − 49 , 3i, +, 7, , , − 3 , , ……….., 8 , , A rational number between 3 and, 13, 1 6+7, 7, =, =, . We can accordingly, 2, 2 2, 4, proceed in this manner to find three more, rational numbers between 3 and 4., , Note:- = 3.14159265358979…………. while, 22, = 3.1428571428…………., 7, , , r+s, ., 2, , Hence, six rational numbers between 3 and 4, 15 13 27 7 29 15, are, ,, ,, , ,, ,, ., 8 4, 8 2 8 4, , 22, 22, but for calculation we can take , ., 7, 7, , Method 2 :, , ❖ EXAMPLES ❖, , 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, q 0?, , , , , , and, that, are all, 7 7 7 7 7, 7, between 3 and 4., CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, , Ex.1, , Is zero a rational number? can you write it in, p, the form , where p and q are integers and, q, , Number system, , 2
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Hence, the six numbers between 3 and 4 are, 22 23 24 25 26, 27, , , , , , and, 7 7 7 7 7, 7, Ex.4, , 8 7.000 0.875, 64, 60, 56, 40, 40, 0, 7, , = 0.875, 8, , 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., , (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, into decimal form by long, 16, division method., We have,, 16 35.0000 2.1875, 32, 30, 16, 140, 128, 120, 112, 80, 80, , Convert, , The irrational are 3.127190385……………, , 0, , 3.212325272930………, , , , 3.969129852937…………, Ex.6, , Sol., , Find two rational & two irrational numbers, between 4 and 5., 4+5, = 4.5 Ans., Rational numbers, 2, , &, , Ex.9, Sol., , 35, = 2.1875, 16, , 2157, in the decimal form., 625, We have,, , Express, , 625 2154.0000 3.4512, 1875, 2820, 2500, 3200, 3125, 750, 625, 1250, 1250, 0, , 4 .5 + 4 8 .5, =, = 4.25 Ans., 2, 2, , Irrational numbers 4.12316908……… Ans., 4.562381032…….. Ans., , ➢ DECIMAL REPRESENTATION OF, RATIONAL NUMBERS, , Ex.7, Sol., , 7, in the decimal form by long, 8, division method., We have,, , , , Express, , 2157, = 3.4512, 625, , –17, in decimal form by long, 8, division method., , Ex.10 Express, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 3
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Sol., , In order to convert, , –17, in the decimal form,, 8, , 17, in the decimal form and, 8, –17, the decimal form of, will be negative of, 8, 17, the decimal form of, 8, we have,, 8 17.000 2.125, 16, 10, 8, 20, 16, 40, 40, , 11 2.00 0.181818, 11, 90, 88, 20, 11, 90, 88, , we first express, , , , 0, –17, = – 2.125, 8, , , , 8, ., 3, , By long division, we have, , 3 8.0000 2.6666, 6, 20, 18, 20, 18, 20, 18, , Sol., , −16, 45, , By long division, we have, 45 160 0.3555, 135, 250, 225, 250, 225, 250, 225, 25, 16, , = 0.3555 .... = 0.3 5, 45, −16, Hence,, = – 0.3 5, 45, , Ex.14 Find the decimal representation of, , 20, 18, 2, , , 2, = 0.181818 ..... = 0. 18, 11, , Ex.13 Find the decimal representation of, Sol., , Ex.11 Find the decimal representation of, Sol., , 20, 11, 90, 88, 2, , 22, ., 7, , By long division, we have, , 8, = 2.6666 ... = 2. 6, 3, , 2, as a decimal fraction., 11, By long division, we have, , Ex.12 Express, Sol., , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 4
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7 22 3.142857142857, 21, 10, 7, 30, 28, 20, 14, 60, 56, , 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)., , ➢ CONVERSION OF DECIMAL NUMBERS INTO, m, RATIONAL NUMBERS OF THE FORM, n, , 40, 35, 50, 49, 10, 7, 30, 28, 20, 14, 60, 56, 40, 35, 50, 49, 1, , , Case I : When the decimal number is of, terminating nature., Algorithm :, Step-1 : Obtain the rational number., 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., , 22, = 3.142857142857 ..... = 3. .142857, 7, , So division of rational number gives decimal, expansion. This expansion represents two, types, , Ex.15 Express each of the following numbers in the, form, , (A) Terminating (remainder = 0), , p, ., q, , (i) 0.15, 6 8 7, Ex. , , ,………are equal to 1.2, 1.6,, 5 5 4, 1.75 respectively, so these are, terminating and non repeating (recurring), , Sol., , (B) Non terminating recurring (repeating), , (iii) –25.6875, , Dividing numerator and deno min ator , by the common divisor 5 of numerator, and deno min ator, , , (remainder 0, but equal to devidend), Ex., , (ii) 0.675, 15, (i) 0.15 =, 100, 15 5, =, 100 5, , 10, = 3.333 ……….. or 3. 3, 3, , =, , 1, = 0.1428514285……….or 0.142857, 7, , 3, 20, , (ii) 0.675 =, , 675, 1000, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 5
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675 25, 27, =, 40, 1000 25, −256875, (iii) – 25.6875 =, 10000, −256875 625, −411, =, =, 10000 625, 16, , Step-7 : Write the rational number in its simplest, form., , =, , ❖ EXAMPLES ❖, Ex.16 Express each of the following decimals in the, form, , Case II : When decimal representation is of nonterminating repeating nature., , (i) 0.6, (ii) 0.35, , 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., , p, :, q, , (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, , , For example, 2.16, 0.35, 0.785 are mixed, recurring decimals., , x=, , 2, 3, , Hence 0.6 =, , 2, 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, , Conversion of a pure recurring decimal to the, p, form, q, , Algorithm :, Step-1 : Obtain the repeating decimal and pur it, equal to x (say), 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,, write x = 0.8 as x = 0.888.... and x = 0.14 as, x = 0.141414......, , Hence, 0.35, , Step-3 : Determine the number of digits having, bar on their heads., , (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, , 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., 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., CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 6
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999x = 585, 65, 195, 585, , x=, =, =, 999, 111, 333, 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, , Step-4 : Use the method of converting pure, p, recurring decimal to the form, and obtain the, 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, 2, , 10x = 3 + 0.2, 0.2 = 9 , , , 2, 10x = 3 +, 9, 93+ 2, 29, 10x, =, 10x =, 9, 9, 29, x=, 90, , (ii) Let x = 23.43, , (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, , 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, , Ex.19 Express, , each, , of, , the following, p, recurring decimals in the form ;, q, , Step-3 : Multiply both sides of x by 10n so that, only the repeating decimal is on the right side of, (i) 4.32, (ii) 15.712, the decimal point., CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 7, , mixed
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Sol. (i) Let x, , 3, , = 4.32, , 3.1 3.2 3.3, , 10x = 43 .2 [Multiplying both sides of x by, 10], 10x = 43 + 0.2, 2, 10x = 43 +, 9, 43 9 + 2, 10x =, 9, 387 + 2, 10x =, 9, 389, 10x =, 9, 389, , x=, 90, , 3.76 3.77, , 3.80, 3.78 3.79, , To mark 3.765 we have to use magnifying glass, , Ex.21 Visualize 4.26 on the number line, upto 4, decimal places., 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., , Sol., , 12, 99, 4, 10x = 157 +, 33, 157 33 + 4, 10x =, 33, 5181 + 4, 10x =, 33, 5185, 5185 1037, 10x =, x=, =, 330, 66, 33, , 10x = 157 +, , , , 3.9, , 3.770, 3.765, 3.766, 3.767, 3.768, 3.769, 3.761 3.762 3.763 3.764, , 10x = 157 + 0.12, , , , 4, , 3.7 3.8, , 3.760, , 10x = 157 .12, , , , 3.6, , 3.70, 3.71 3.72 3.73 3.74 3.75, , (ii) Let x = 15.712 . Then,, , , , 3.4 3.5, , 4, , 4.2, 4.21 4.22 4.23 4.24 4.25, , 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., , 4.26, 4.261, , 4.262 4.263, , 5, , 4.7 4.8, , 4.1 4.2 4.3 4.4 4.5 4.6, , 4.26 4.27, , 4.9, , 4.3, 4.28 4.29, , 4.27, 4.264 4.265 4.266 4.267 4.268 4.269, , 4.262, , 4.263, 4.2621, 4.2622, , 4.2623, , 4.2625, , 4.2624, , 4.2627, , 4.2626, , 4.2629, , 4.2628, , Ex.22 Express the decimal 0.00352 in the form, Sol., , 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, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 8, , p, q
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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. (i) The product of 27 and 3 is 81 = 9,, which is a rational number., (ii) The product of 2 and, an irrational number., , 3 is, , 6 , which is, , 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, 23 = 6, 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, , 5 is, , 2 5, , So, required numbers are, , 5 and, , 2 5 ., , Ex.26 Find two irrational numbers lying between, 2 and 3 ., Sol., 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, 2 3 =, , 2 and, , 6 = 61/4, , Irrational number between, , 2 and 61/4 is, , 2 61 / 4 = 21/4 × 61/8., 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 between, 0.232332333233332.... and, 0.252552555255552......, Sol., Let a = 0.232332333233332...., and b = 0.252552555255552....., 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....,, 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., 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, c = 0.101, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 3 is, , 9
Page 10 : Then, c is a rational number as it has a, terminating decimal representation., Since b has a zero in the third place of, decimal and c has a 1., b<c, We also observe that c < a, because c has, zeros in all the places after the third place of, decimal whereas the decimal representation, of a has a 1 in the sixth place., Thus, c is a rational number such that, b < c < a., Hence , c is the required rational number, between a and b., Construction of an irrational number between, a and b : Consider the number d given by, d = 0.1002000100001......, Clearly, d is an irrational number as its, decimal representation is non-terminating and, non-repeating., We observe that in the first three places of, their decimal representation b and d have the, same digits but in the fourth place d and a 2, whereas b has only a 1., d >b, Also, comparing a and d, we obtain a > d, Thus, d is an irrational number such that, b < d < a., , , c<a, Hence, b < c < a, Thus, c is the required irrational number, between a and b., ➢, , 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, , B, 1, , 1, 1, , O, , OB = 1 + 1 =, , A, , 2, , Now, transfer this square onto the number line, making sure that the vertex O coincides with zero, C, , B, 2, , a = 0.1111..... = 0. 1 and b = 0.1101, 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., 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., 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., b<c, Also, c and a have the same digits in the first, four places of their decimal representations, but in the fifth place c has a zero and a has a 1., , 1, , 1, , X, , Ex.30 Find one irrational number between the, number a and b given below :, Sol., , REPRESENTING IRRATIONAL NUMBERS, ON THE NUMBER LINE, , –3, , –1, , –2, , O, , A, , P2, , 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 units, D, 1, 3, , X, , –3, , –2, , –1 O, , 2, , B, , A PQ 2, , 3, , With O as centre & OC as radius, draw an arc,, meeting OX at Q. Then, OQ = OD = 3 units, Then, the point Q represents, , 3 on the real line, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , X, , 3, , 10, , X
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Remark : In the same way, we can locate n for, n − 1 has been, any positive integer n, after, located., , Existence of, The value of, , 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, a rational number. So,, , n for a positive real number :, , Ex.32 The real number, real number, surd., , Find the midpoint O of AC., With O as centre and OA a radius, draw a, semicircle., D, , B, , C E, , X, , Now, draw BD ⊥ AC, intersecting the semicircle, at D. Then, BD = 4.3 units., , ➢, , (a), , Real Number Line :, , (e), (f), , Surds :, , 4, , 5 is a biquadratic surd but 4 81 is not a, biquadratic surd as it is not a surd., , ( a), n, , n, , =a, , n, , p, , (a ), , p, , an am =, , n m, , =, , p, , a n .m, p, , a n+m, , (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., , are, irrational, 2 , 3, 5 , 21 ,................, 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,, , 8 is not a cubic surd as it not a, , n m, (d) m n a = mn a =, a, , 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., , 5,, , 4 is a cubic surd but the, , a n b = n ab [one of either a or b, should be non-negative integer], na, a, (c), = n, n b, b, (b), , 4.3 units, , SURDS OR RADICALS, , e.g., , 3, , 3, , Laws of radicals :, For any positive integer ‘n’ and a positive rational, number ‘a’., , With B as centre and BD as radius, draw an arc,, meeting AC produced at E., Then, BE = BD =, , 9 is not a surd., , Biquadratic surd :, A surd of order 4 is called a biquadratic surd. A, biquadratic surd is also called a quadratic surd., Ex.33, , O, , 9 = 91/2 = 3 is, , Cubic surd :, A surd of order 3 is called a cubic surd., , 4.3 geometrically : -, , Draw a line segment AB = 4.3 units and extend it, to C such that BC = 1 unit., , A, , 9 = 91/2 is, , (vii) Only similar surds can be added or subtracted, by adding or subtracting their rational parts, , 3 etc., , (viii) Surds of same order can be multiplied or, divided, , (ii) If is a surd then ‘n’ is known as order of surd, and ‘a’ is known as radicand., , (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., , (iii) Every surd is an irrational number but every, irrational number is not a surd., , Quadratic Surd:, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 11
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(x) If the product of two surds is a rational, number, then each one of them is called the, rationalising factor of the other., , =, , (xi) A surd consisting of one term only is called a, monomial surd., , , , (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., , Which is an irrational number., , 8 8 10, 4 4 15, 8, , 8, is not the, 3, square of any rational number, hence the, given expression is a surd., , Because the rational number, , 2 + 5 , 3 + 2, 2 − 3 etc. are binomial, surds., , (v) 3 12 ÷ 6 27 =, , (xiii)The binomial surds which differ only in sign, (+ or –) between the terms connecting them,, are called conjugate surds e.g. 3 + 2, , 108, 972, , ❖ EXAMPLES ❖, , (ii), , 64, , (v) 3 12 6 27, Sol. (i), , (vi), , (vi), , 5 3 25, , 45 = 9 5 = 3 5, Because the rational number 45 is not the, square of any rational number, hence 45 is, a surd., , =, , 20 45 =, , 30 30 =, , ( 30 ), , 2, , 5 ×, , =, , 3, , ( 8 ) ( 10 ) , ( 4 ) ( 15 ), , 3, , 25 =, , 3, , 5 25, , 5 5 5, 3, , 53 = 5, , (i) ( 3 5 )3, , (ii), , (i) ( 3 5 )3 = 5, , Sol., , (ii), , (i), , = 30, Sol. (i), , 3, , 3, , 64 =, , 3, , 3, , Hence,, , 64, , (Using Ist Law), 43 = 4, , (Using Ist Law), , 3, , 4x − 7 – 5 = 0 (ii), , 4, , 3x + 1 = 2, , 4x − 7 – 5 = 0, , , , 3, , 4x − 7 = 5, , ( 3 4x − 7 )3 = 53, , (iv) we have, , =, , 1, 1, =, 3, 9, , , , Ex.36 Find the value of x in each of the following:, , 900, , Which is a rational number and therefore, 20 45 is not a surd., , 8 10 ÷ 4 15 =, , 6 6 27, , Ex.35 Simplify the following :, , 45 is an irrational number., , (iii) We have, , 3 3 12, , , , Which is a rational number., 3, 5 × 3 25 is not a surd., , 8 is a rational number, hence 64 is not a surd., , Thus, , 3, , , , 64 = 8, , (ii), , 2, , 3 12 ÷ 6 27 is not a surd., , 45, 3, , ( 3 ) ( 12 ), ( 6 ) 27, 2, , =, , 1, is a rational number, therefore, 3, , Since, , (iv) 8 10 4 15, , 20 45, , (iii), , 6 27, , 6 6 27, , , , Ex.34. State with reasons which of the following are, surds and which are not :, , 3 12, , 3 3 12, , =, , and 3 − 2 or 2 + 5 and 2 – 5 are, conjugate surds., , (i), , 240, , 8, 3, , =, , 3, , 640, , , , 8 10, , 4x – 7 = 125, , 4 15, , 4x = 132, x = 33, , 2, , 8 8 10, , 2, , 4 4 15, , (ii), , 4, , [( n a )n = a], , 3x + 1 = 2, , ( 4 3x + 1 )4 = 24, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 12
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3x + 1 = 16, , (ii) Mixed Surd :, A surd which has a rational factor other than, unity, the other factor being irrational, is, called a mixed surd., , [( n a )n = a], , 3x = 15, x=5, Ex.37 Simplify each of the following:, , Sol., , (i), , 3, , (i), , 3, , 3·, , 3, , 3·, , 3, , (ii), , 4, 3, , 4 =, , 3, , Ex.43, , 128, , 3 4 = 12, [Using IInd Law], 3, , 2, , [Using IInd Law], =, , 3, , 43 ·, , 3, , 2, , = 4 3 2 [Using I,, , 3, , 43 = 4], , (i), Sol., , (i), , 4, , 8, 27, , (ii), 3, , 8, =, 27, , 3, , 3, , 3, , =, 4, , (ii), , 3, , 3888, 4, , 4, , =, , = (4×3)1/2 = 121/2 = 12, , 3888, 4, , (ii) 2., , 48, , 3, , =, , 2, 3, , 4, , Sol., , 4 3, , (i), , 4 3, , [Using IIIrd Law], (iv), 34 = 3 [Using Ist Law], , (ii), , (ii), , 3, 3 =, , 2 3, , Ex.40 Simplify :, Sol., , 2 3, , 5, , 3, , [Using IVth law], , 6, , 5, , [Using IVth law], , 5 4, , 3 4, , (2 ), , =, , 5, , Sol. (i), , ( 23 ) 4, , 3, , 2 =, , 5, , =, , 9, 32 = 18, 16, 2, , 3, 4, , 8 =, , 3, 8 =, 4, , 9, 8 =, 16, , 3, , 32, 2, , 3, 32, 4, , 2, , 3, 8, 4, , 9, 2, , Ex.45 Expressed each of the following as pure, surds2, 3 4 32, (i) . 3 108, (ii), 243, 2, 3, 2, 3, , 3, , 108 =, , 2, × (108)1/ 3, 3, 1/ 3, , 2 3 , = , 3 , , Using the above property, we have, 5 4, , 3, 32 =, 4, , =, , 12, , 5 =, , 32 =, , [Using Ist Law], , 3888, 48, , 81 =, , 2, , 3, 4, , Ex.39 Siplify each of the following, (i), , 4 = 2×41/3 = (23)1/3 ×41/3 = 81/3 ×41/3, , = (8 4)1/ 3 = (32)1/3 =, (iii), , 33, 4, , Ex.44 Express each of the following as a pure surd., (i) 2 3, (ii) 2. 3 4, 3, 3, (iii), (iv), 32, 8, 4, 4, , [Using IIIrd Law], , 27, , 23, , =, , 48, , 8, , v EXAMPLES v, , Sol. (i) 2 3 = 2 ×31/2 =(22)1/2 × 31/2 =41/2 ×31/2, , Ex.38 Simplify each of the following:, 3, , TypeI : On expressing of mixed surds into pure, surds, , 3, , (ii) 3 128 = 3 64 2 = 3 64, , 2 3 , 5 3 12, 2 4 5 are mixed surds., , 8., , × (108)1/3, , 1/ 3, , 8 , = , 27 , , Pure And Mixed Surds :, (i) Pure Surd :, A surd which has unity only as rational, factor, the other factor being irrational, is, called a pure surd., 5, , 1/ 3, , 8, , = 108 , 27, , , = (8×4)1/3, , 4, , Ex.41, , 3 , 2 , 3 are pure surds., , Ex.42, , 6 , 3 12 are pure surds., , × (108)1/3, , = (32)1/3 =, 1/ 4, , 3, , 32, , 3 4 32, 3, 32 , ., =, ×, , 2, 2, 243, 243 , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, , (ii), , Number system, , 13
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1/ 4, , 3 4 , = , 2 , , , , 1/ 4, , 81 32 , = , , 16 243 , , 32 , ×, , 243 , , 1/ 4, , 2, = , 3, , =, , 1/ 4, , 1/ 4, , 81 , = , 16 , , =, , ➢, , (i) a a + b, , 3, (ii) a b 2, , (iii) 2ab 3 ab, , 2ab., , a 3b 2, , (, , ab = (2ab), , ), , 3 1/ 3, , = (8, , Ex.50, , (ab), , a3b3.ab)1/3, , (iii), , Ex.51, , Ex.47 Express each of the following as mixed surd, in its simplest form:, (iii), , 5, , (v), Sol. (i), (ii), (iii), , 5, , (ii), , 3, , Ex.52, , 72 = 3 8 9 =, , 3, , =, , 2, , 2 3 9 = 3 23 ×, , 5, 288 = 5 32 9 = 25 9 =, , 5, , 3, , 9 = 23 9, , =, , 25 × 5 9 = 2 5 9, , 5, , 320 =, , 5, , 3210 =, , 5, , 25 10 =, , 5, , 25 × 5 10, , Ex.53, , = 2 5 10, (vi) 5. 135 = 5 27 5 = 5 3 5 = 5 3 × 5, 3, , 3, , =5×3×, Ex.48 Express, form :, , 4, , 3, , 3, , 3, , =, , [(13) 2 ]2, 132, , =, , 134, 132, , 53 2 4 7 2, 54 25 7, , 3, , 1, 5, , 4 −3, , , , 1, 2, , 5− 4, , 7 2−1, , 7, 7, =, Ans., 5 2 10, , (iv) (ab)p = apbp, , (iv) 1350 = 225 6 = 152 6 = 15 2 × 6 =15 6, (v), , if p q, if q p, , 53 2 4 49, 625 32 7, , =, , 80 = 16 5 = 4 5 = 4 × 5 = 4 5, , 5, , 132, , (vi) 5. 3 135, , 320, 2, , 3, , (169) 2, , 72, , (iv) 1350, , 288, , a p − q, =, , a q 1 / a q − p, , ap, , = 134–2 = 132 = 169 Ans., , v EXAMPLES v, , 80, , (121)3 33, = (112)3 × 11 = 116 × 111, 3, , = 117 Ans., 1/ 3, , 3, = (8 a4b4)1/3 = 8a 4 b 4, TypeII : On expressing given surds as mixed surds, in the simplest form., , (i), , (7)3 21, =?, 3, , (ii) (ap)q = apq, , 3, (ii) a b 2 = (a3) 1/3 × (b2)1/3 = (a3×b2)1/3, , (iii), , 44 4 5 = 4 4 5, , = 73 × 7 = 73+1 = 74 Ans., , a 3 + a 2b, , = (a3 + a2b)1/2 =, , 3, , 256 4 5, , SOME RULES FOR EXPONENTS, , Ex.49, , Sol. (i) a a + b = a × (a + b)1/2, = (a2)1/2 × (a + b) 1/2, = [a2× (a + b)]1/2, , 3, , 4, , (i) ap. ap = ap+q i.e. if base is same with different, or same powers are multiply each other then, powers are add., , Ex.46 Express each of the following as pure surd :, , =, , 4, , 256 5 =, , 4, , 1280 =, , Let a, b > 0 be real numbers & p, q are rational, numbes then, , 2, 3, , 4, , 4, , Sol., , 1/ 4, , 32 , ×, , 243 , , 3, , (63) 4 144, =?, 132 9, , =, , (9 7) 4 (12) 2, (11 12) 32, , =, , (32 7) 4 (3 2 2 ) 2, 11 2 2 3 32, , 5 = 15 3 5 ., , 1280 as mixed surd in its simplest, , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 14
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=, , (32 ) 4 7 4 32 (2 2 ) 2, 11 2 2 31+ 2, , =, , 38 7 4 32 2 4, 11 2 2 33, , =, , 2 4 38+ 2 7 4, 2 2 33 11, , =, , 2 4− 2 310−3 7 4, 11, , =, , 2 2 37 7 4, Ans., 11, , (v) a0 = 1, Ex.54, , (20 + 30 ) 52, 25, , =, =, , (1 + 1) .52, 52, , 2, 5, , (vi) a+p =, , 2 2, = =2, 50 1, , 1, , or a–p =, , −p, , a, , Ans., , 1, ap, , 2, =?, 4, , Ex.55, =, =, , Ex.56, , =, , 2− 2, , (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 , , CraveIITy-crave for best for JEE & NEET,NTSE&Olympiad , Kadipur,&Sultanpur (U.P.),ph: 8004894394, Number system, , 15
