Real Numbers

REAL NUMBERS CONTENTS  Number System  Euclid's Division Lemma or Euclid's Division Algorithm  The Fundamental Theorem of Arithmetic  The Fundamental Theorem of Arithmetic to Find H.C.F. and L.C.M. NUMBER SYSTEM (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 :  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....} A multiple is a number which is exactly divisible by another, 36 is a multiple of 2, 3, 4, 9 and 12.  Even Numbers : Integers which are multiples of 2 are even number (i.e.) 2,4, 6, 8............... are even numbers.  Odd numbers : Integers which are not multiples of 2 are odd numbers.  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 put in the form (ii) A rational number is either a terminating or non-terminating and recurring (repeating) decimal. p , where p and q are both q integers and q  0.  Prime and composite Numbers : All natural number 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 Type (2) : |–3| = 3 = – (–3) . Hence |a| = – a when a < 0 Eg : Hence, |a| = a, if a > 0 ; |a| = – a, if a < 0 1 = 0.333………. 3 = 0. 3  Irrational number : 3 10 0.33…… 9 10 9 1…… (i) A number is irrational if and only if its 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. (iii) If a and b are two real numbers, then either (i) a > b or (ii) a = b or (iii) a < b or 1 = 0.142857142857….. 7 = 0.142857 (iv) Negative of an irrational number is an irrational number. 7 10 0.14285…. 7 30 28 20 14 60 56 40 35 50 49 1…. (v) The sum of a rational number with an irrational number is always irrational. (vi) The product of a non-zero rational number with an irrational number is always an irrational number. (vii) The sum of two irrational numbers is not always an irrational number. (viii) The product of two irrational numbers is not always an irrational number. In division for all rationals of the form p (q  0), p & q are integers, two things can q happen either the remainder becomes zero or never becomes zero. Type (1) : Eg : 7 = 0.875 8 8 70 0.875 64 60 56 40 40 × This decimal terminating. expansion In both examples remainder is never becomes zero so the decimal expansion is never ends after some or infinite steps of division. These type of decimal expansions are called non terminating. In above examples, after Ist step & 6 steps of division (respectively) we get remainder equal to dividend so decimal expansion is repeating (recurring). So these are called non terminating recurring decimal expansions. Both the above types (1 & 2) are rational numbers. Types (3) : 0.875 is called If remainder is zero then decimal expansion ends (terminates) after finite number of steps. These decimal expansion of such numbers terminating. Eg :The decimal expansion 0.327172398……is not ends any where, also there is no arrangement of digits (not repeating) so these are called non terminating not recurring. These numbers are called irrational numbers. Eg. : 0.1279312793 rational terminating 0.1279312793…. rational non terminating or 0.12793 23 = 6 Ex.2 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. & recurring 0.32777 rational terminating 0.327 or rational non terminating 0.32777……. & recurring 0.5361279 rational  terminating 0.3712854043…. irrational non terminating 2 2.5 = non recurring 0.10100100010000 rational terminating 0.10100100010000…. irrational non terminating Real Numbers Irrational no. If remainder = 0 If remainder  0 & rem. = devidend Terminating & Non repeating Non terminating & repeating (recurring) Eg : 18 5  3.6 Eg : 1 3  0.33.... .. 2 5 5 is So, required numbers are 5 and 2 5 . Ex.3 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. If remainder  0 & rem.  any devidend  Irrational number between 2 3 = Non terminating non repeating 6 =6 Irrational number between 1/ 4 2 6 Eg : 0.671234….. =2 1/4 2 and 3 is 1/4 1/4 2 and 6 is 1/8 ×6 . Hence required irrational number are 61/4 and = 0. 3 Eg : 1.343634003908…... EXAMPLES  Ex.1 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 rational numbers, then number between them.  5 Similarly, irrational number between 2 and non recurring. Rational no.  Irrational number between 2 and 2.5 is ab is a rational 2  A rational number between 2 and 3 is 23 = 2.5 2 An irrational number between 2 and 3 is 21/4 × 61/8. Ex.4 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. Theorem : Let p be a prime number. If p divides a2, then p divides a, where a is a positive integer. Proof : Let the prime factorisation of a be as follows : This means that 2 divides b2, and so 2 divides b (again using Theorem with p = 2). a = p1p2…..pn, where p1,p2,…..pn are primes, not necessarily distinct. Therefore, a and b have at least 2 as a common factor. Therefore, But this contradicts the fact that a and b have no common factors other than 1. a2 = (p1p2…..pn) (p1p2 ….. pn) = p12 p 22 ….. p 2n . This contradiction has arisen because of our incorrect assumption that 2 is rational. Now, we are given that p divides a2. Therefore, from the Fundamental Theorem of Arithmetic, it follows that p is one of the prime factors of a2. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of a2 are p1, p2,…, pn. So p is one of p1, p2,……, pn. So, we conclude that (ii) Let us assume, to contrary, that 3 is rational. That is, we can find integers a and b a ( 0) such that 3  . b Now, since a = p1 p2 …… pn, p divides a. We are now ready to give a proof that irrational. Suppose a and b not having a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime. 2 is The proof is based on a technique called ‘proof by contradiction’. Ex.5 So, b 3  a . Prove that (i) 2 is irrational number (ii) 3 is irrational number Similarly 5 , 7 , 11 …... numbers. 2 is irrational. Squaring on both sides, and rearranging, we get 3b2 = a2. Therefore, a2 is divisible by 3, and by Theorem, it follows that a is also divisible by 3. are So, we can write a = 3c for some integer c. irrational Sol. (i) Let us assume, to the contrary, that rational. Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2. 2 is This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem with p = 3). So, we can find integers r and s ( 0) such r that 2  . s Therefore, a and b have at least 3 as a common factor. Suppose r and s not having a common factor other than 1. Then, we divide by the common a factor to get 2  , where a and b are b coprime. This contradicts the fact that a and b are coprime. But this contradicts the fact that a and b are coprime. This contradiction has arisen because of our incorrect assumption that 3 is rational. So, b 2 = a. So, we conclude that 3 is irrational. Squaring on both sides and rearranging, we get 2b2 = a2. Therefore, 2 divides a2. Now, by Theorem it following that 2 divides a. Ex.6 Prove that 7  3 is irrational So, we can write a = 2c for some integer c. Sol. Method I : 2 2 Substituting for a, we get 2b = 4c , that is, b2 = 2c2. Let 7  3 is rational number  7  3  p (p, q are integers, q  0) q  (2 7 )  p  7   3 q  rational)  7 is rational but this is contradiction that 7q  p is also integer q LHS = 7 is irrational 3 is also integer but this is contradiction that  2 7 is irrational 3 is irrational so our proved assumption is wrong that (7  3 ) is rational Theorem 1 :  7  3 is irrational proved. we know sum or difference of two rationals is also rational Let x be a rational number whose decimal expansion terminates. Then x can be p expressed in the form , where p and q are q coprime and the prime factorization of q is of the form 2n5m, where n, m are non-negative integers.  7  (7  3 ) (A) Numbers are terminating (remainder = zero) = Eg : 32 2 5 28 256  3   3  0.256 3 125 5 (2  5) 10 Eg : 9 9  22 36 36  0.36  2   25 5  2 2 ( 2  5) 2 (10) 2 Method II : Let 7  3 is rational 3 = rational but this is contradiction that  (7  3 ) is irrational Ex.7 1  7 2 ( division of two rational no. is also 7q  p 3 q Here p, q are integers  (ii) Let 2 7 is rational 3 is irrational proved. Prove that : (i) Sol. (i) Let 5 3 (ii) 2 7 are irrationals 5 is rational 3  5  =  3  3    OR 5 is rational Theorem 2 : ( product of two rationals is also rational) but this is contradiction that  So we can convert a rational number of the p form , where q is of the form 2n5m to an q a equivalent rational number of the form b where b is a power of 10. These are terminates. 5 is irrational proved. 3 5 is irrational p be a rational number, such that the q prime factorization of q is of the form 2n 5m, where n, m are non-negative integers. Then x has a decimal expansion which terminates. Let x = (B) Non terminating & recurring Eg : 1  0.142857 = 0.142857142857..... 7 = non terminating ( 7 cannot remove from denominator) Since denominator 7 is not of the form 2n 5m so we zero (0) will not show up as a remainder. (8) = terminating Theorem 3 : p be a rational number, such that the q prime factorization of q is not of the form 2n5m, where n, m are non-negative integers. Then, x has a decimal expansion which is non-terminating repeating (recurring). Let x = (9) 35 35  2 = terminating  50 100 (10) 77 7  11 7  11   210 7  30 7  2  5  3 = non terminating From the discussion above, we can conclude that the decimal expansion of every rational number is either terminating or nonterminating repeating.  EUCLID’S DIVISION LEMMA OR  EUCLID’S DIVISION ALGORITHM For any two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where 0  r < b. Eg : From given rational numbers check terminating or non terminating For Example 13 13 13  25 (13  32) (1)    3125 (5) 5 25  55 (10) 5 (i) Consider number 23 and 5, then: 23 = 5 × 4 + 3 = terminating Comparing with a = bq + r; we get: 17 17 17  53 17  125 (2)    8 23 ( 2  5) 3 (10) 3 a = 23, b = 5, q = 4, r = 3 and 0  r < b (as 0  3 < 5). = terminating 64 26 (3) ( we can not remove 7  455 5  7  13 & 13 from dinominator) non-terminating repeating ( no. is rational it is always repeating or recurring) 5 (4) 6 23 2 2 2    15 5  3 5 10 15 3 5 3 5 3 5  4   2 4 2 1600 2  10 ( 2  5)  10 10 6 = terminating 5 (ii) Consider positive integers 18 and 4. 18 = 4 × 4 + 2    For 18 (= a) and 4(= b) we have q = 4, r = 2 and 0  r < b. In the relation a = bq + r, where 0  r < b is nothing but a statement of the long division of number a by number b in which q is the quotient obtained and r is the remainder. Thus, dividend = divisor × quotient + remainder  a = bq + r (5) 29 29 = non terminating  343 (7) 3 (6) 23 23  5 23  5   235 2 ( 2  5) 3 (10) 3 The H.C.F. of two or more positive integers is the largest positive integer that divides each given positive number completely. = terminating i.e., if positive integer d divides two positive integers a and b then the H.C.F. of a and b is d. (7) 129 3  43  2 2 = 25  57  7 5 (2  5) 7  7 5  H.C.F. (Highest Common Factor) For Example  H.C.F. = Product of all common prime factors = 11. For any two positive integers a and b which can be expressed as a = bq + r, where 0  r < b, the, H.C.F. of (a, b) = H.C.F. of (q, r) and so on. For number 418 and 33 (i) 14 is the largest positive integer that divides 28 and 70 completely; therefore H.C.F. of 28 and 70 is 14. (ii) H.C.F. of 75, 125 and 200 is 25 as 25 divides each of 75, 125 and 200 completely and so on. 418 = 33 × 12 + 22  Using Euclid’s Division Lemma For Finding H.C.F. 33 = 22 × 1 + 11 and Consider positive integers 418 and 33. 22 = 11 × 2 + 0  H.C.F. of (418, 33) = H.C.F. of (33, 22) Step-1 = H.C.F. of (22, 11) = 11. Taking bigger number (418) as a and smaller number (33) as b express the numbers as a = bq + r   EXAMPLES  Ex.8  418 = 33 × 12 + 22 (i) 135 and 225 Step-2 Now taking the divisor 33 and remainder 22; apply the Euclid’s division algorithm to get: Using Euclid’s division algorithm, find the H.C.F. of [NCERT] (iii) 867 and 255 Sol.(i) Starting with the larger number i.e., 225, we get: 225 = 135 × 1 + 90 33 = 22 × 1 + 11 [Expressing as a = bq + r] Now taking divisor 135 and remainder 90, we get 135 = 90 × 1 + 45 Step-3 Again with new divisor 22 and new remainder 11; apply the Euclid’s division algorithm to get: Further taking divisor 90 and remainder 45, we get 90 = 45 × 2 + 0  22 = 11 × 2 + 0  Required H.C.F. = 45 38220 = 196 × 195 + 0 Since, the remainder = 0 so we cannot proceed further. Since, the remainder is 0 Step-5  The last divisor is 11 and we say H.C.F. of 418 and 33 = 11      (Step-3) H.C.F. = 51 (Ans.) Ex.9 Show that every positive integer is of the form 2q and that every positive odd integer is of the from 2q + 1, where q is some integer. Sol. According to Euclid’s division lemma, if a and b are two positive integers such that a is greater than b; then these two integers can be expressed as Common factors = 1 and 11 Prime factors of 33 = 3 and 11. 867 = 255 × 3 + 102 (Step-1) 102 = 51 × 2 + 0  Factors of 418 = 1, 2, 11, 19, 22, 38, 209 and 418 and, Prime factors of 418 = 2, 11 and 19. (Ans.) 255 = 102 × 2 + 51 (Step-2) (i) Using factor method: (ii) Using prime factor method: H.C.F. = 196 (iii) Given number are 867 and 255 Verification :  Highest common factor = 11 i.e., H.C.F. = 11 (Ans.) (ii) Starting with larger number 38220, we get: Step-4 Factor of 33 = 1, 3, 11 and 33. (ii) 196 and 38220 a = bq + r; where 0  r < b Now consider b = 2; then a = bq + r will reduce to a = 2q + r; where 0  r < 2, Ex.11 Show that one and only one out of n; n + 2 or n + 4 is divisible by 3, where n is any positive integer. Sol. i.e., r = 0 or r = 1 If r = 0, a = 2q + r  a = 2q a = bq + r; where q and r are positive integers and 0  r < b i.e., a is even and, if r = 1, a = 2q + r  a = 2q + 1 Let a = n and b = 3, then i.e., a is add; a = bq + r  n = 3q + r; where 0  r < 3. as if the integer is not even; it will be odd. r = 0  n = 3q + 0 = 3q Since, a is taken to be any positive integer so it is applicable to the every positive integer that when it can be expressed as r = 1  n = 3q + 1 and r = 2  n = 3q + 2 If n = 3q; n is divisible by 3 a = 2q  Consider any two positive integers a and b such that a is greater than b, then according to Euclid’s division algorithm: If n = 3q + 1; then n + 2 = 3q + 1 + 2  a is even and when it can expressed as = 3q + 3; which is divisible by 3 a = 2q + 1; a is odd.  n + 2 is divisible by 3 Hence the required result. If n = 3q + 2; then n + 4 = 3q + 2 + 4 Ex.10 Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer. Sol.  n + 4 is divisible by 3 Hence, if n is any positive integer, then one and only one out of n, n + 2 or n + 4 is divisible by 3. Let a is b be two positive integers in which a is greater than b. According to Euclid’s division algorithm; a and b can be expressed as a = bq + r, where q is quotient and r is remainder and 0  r < b. Taking b = 4, we get: a = 4q + r, where 0  r < 4 i.e., r = 0, 1, 2 or 3 r = 0  a = 4q, which is divisible by 2 and so is even. Hence the required result. Ex.12 Show that any positive integer which is of the form 6q + 1 or 6q + 3 or 6q + 5 is odd, where q is some integer. Sol. If a and b are two positive integers such that a is greater than b; then according to Euclid’s division algorithm; we have r = 1  a = 4q + 1, which is not divisible by 2 and so is odd. a = bq + r; where q and r are positive integers and 0  r < b. r = 2  q = 4q + 2, which is divisible by 2 and so is even. Let b = 6, then and r = 3  q = 4q + 3, which is not divisible by 2 and so is odd.   = 3q + 6; which is divisible by 3  Any positive odd integer is of the form 4q + 1 or 4q + 3; where q is an integer. Hence the required result. a = bq + r  a = 6q + r; where 0  r < 6. When r = 0  a = 6q + 0 = 6q; which is even integer When r = 1  a = 6q + 1 which is odd integer When r = 2  a = 6q + 2 which is even. When r = 3  a = 6q + 3 which is odd. When r = 4  a = 6q + 4 which is even. When r = 5  a = 6q + 5 which is odd. This verifies that when r = 1 or 3 or 5; the integer obtained is 6q + 1 or 6q + 3 or 6q + 5 and each of these integers is a positive odd number. Hence the required result. Ex.13 Use Euclid’s Division Algorithm to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Sol. Let a and b are two positive integers such that a is greater than b; then: a = bq + r; where q and r are also positive integers and 0  r < b Let a and b be two positive integers such that a is greater than b; then: a = bq + r; where q and r are positive integers and 0  r < b. Taking b = 3, we get: a = 3q + r; where 0  r < 3  Different values of integer a are 3q, 3q + 1 or 3q + 2. Cube of 3q = (3q)3 = 27q3 = 9(3q3) = 9m; where m is some integer. Cube of 3q + 1 = (3q + 1)3 = (3q)3 + 3(3q)2 ×1 + 3(3q) × 12 + 13 [Q (q + b)3 = a3 + 3a2b + 3ab2 + 1] Taking b = 3, we get: a = 3q + r; where 0  r < 3  The value of positive integer a will be 3q + 0, 3q + 1 or 3q + 2 i.e., 3q, 3q + 1 or 3q + 2. 2 = 9q2 = 3(3q2) = 3m; 3 where m is some integer. Square of 3q + 1 = (3q + 1) = 27q3 + 27q2 + 9q + 1 = 9(3q3 + 3q2 + q) + 1 = 9m + 1; where m is some integer. Cube of 3q + 2 = (3q + 2)3 Now we have to show that the squares of positive integers 3q, 3q + 1 and 3q + 2 can be expressed as 3m, or 3m + 1 for some integer m. Square of 3q = (3q) Sol. 2 = (3q)3 + 3(3q)2 × 2 + 3 × 3q × 22 + 23 = 27q3 + 54q2 + 36q + 8 = 9(3q3 + 6q2 + 4q) + 8 = 9m + 8; where m is some integer.  Cube of any positive integer is of the form 9m or 9m + 1 or 9m + 8. Hence the required result. = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1 for some integer m. Square of 3q + 2 = (3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1 = 3m + 1 for some integer m.  The square of any positive integer is either of the form 3m or 3m + 1 for some integer m. Hence the required result. Ex.14 Use Euclid’s Division Algorithm to show that the cube of any positive integer is either of the 9m, 9m + 1 or 9m + 8 for some integer m.  THE FUNDAMENTAL THEOREM OF  ARITHMETIC Statement : Every composite number can be decomposed as a product prime numbers in a unique way, except for the order in which the prime numbers occur. For example : (i) 30 = 2 × 3 × 5, 30 = 3 × 2 × 5, 30 = 2 × 5 × 3 and so on. (ii) 432 = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 24 × 33 or 432 = 33 × 24. (iii) 12600 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 7 = 2 3 × 3 2 × 52 × 7 In general, a composite number is expressed as the product of its prime factors written in ascending order of their values. (i) e.g., (i) 6615 = 3 × 3 × 3 × 5 × 7 × 7 = 33 × 5 × 72 (ii) 532400 = 2 × 2 × 2 × 2 × 5 × 5 × 11 × 11 × 11 2 540 2 270 540 divided by 2 gives 270 3 135 270 divided by 2 gives 135 3 45 135 divided by 3 gives 45 3 15 45 divided by 3 gives 15 5 15 divided by 3 gives 5  EXAMPLES  Ex.15 Consider the number 6n, where n is a natural number. Check whether there is any value of n  N for which 6n is divisible by 7. Sol. Since, 6 = 2 × 3; 6n = 2n × 3n 5 is a prime number and so cannot be further divided by any prime number   540 = 2 × 2 × 3 × 3 × 3 × 5 = 22 × 33 ×5 n  The prime factorisation of given number 6  6n is not divisible by 7. (ii) (Ans) We know, if any number ends with the digit zero it is always divisible by 5. 3 5313 7 1771 11 253  If 12n ends with the digit zero, it must be divisible by 5. This is possible only if prime factorisation of 12n contains the prime number 5. 23  (iii) i.e., prime factorisation of 12n does not contain the prime number 5.  There is no value of n  N for which 12n ends with the digit zero. (Ans)  USING THE FACTOR TREE    EXAMPLES  (ii) 21252 2 8232 2 4116 2 2058 3 1029 7 343 7 49 7 Ex.17 Find the prime factors of : (i) 540  21252 = 2 × 2 × 3 × 7 × 11 × 23 = 22 × 3 × 11 × 7 × 23. Now, 12 = 2 × 2 × 3 = 22 × 3  12n = (22 × 3)n = 22n × 3n 21252 2 10626 Ex.16 Consider the number 12n, where n is a natural number. Check whether there is any value of n  N for which 12n ends with the digit zero. Sol. 2 (iii) 8232   8232 = 2 × 2 × 2 × 3 × 7 × 7 × 7 = 23 × 3 × 73. Ex.18 Find the missing numbers a, b and c in the following factorisation: 2 a 2 1296 2 2520 2 b 2 648 2 1260 2 b 2 324 2 630 17 2 162 3 315 3 81 3 105 3 27 5 35 3 9 Can you find the number on top without finding the other ? Sol. c = 17 × 2 = 34 b = c × 2 = 34 × 2 = 68 and a = b × 2 = 68 × 2 = 136 3 i.e., a = 136, b = 68 and c = 34. (Ans) Yes, we can find the number on top without finding the others. Reason: The given numbers 2, 2, 2 and 17 are the only prime factors of the number on top and so the number on top = 2 × 2 × 2 × 17 = 136   EXAMPLES  i.e., L.C.M. (1296, 2520) = 45,360 H.C.F. = Product of common prime factors with lowest powers. = 23 × 32 = 8 × 9 = 72 i.e., H.C.F. (1296, 2520) = 72. and,  (ii) 1296 and 2520 and 7 13   L.C.M. = 17 × 52 = 17 × 25 = 425 In example 19 (i) : Product of given two numbers = 26 × 91 = 2366 91 13  L.C.M. = Product of each prime factor with highest powers. = 2 × 13 × 7 = 182. (Ans) i.e., L.C.M. (26, 91) = 182. (Ans) H.C.F. = Product of common prime factors with lowest powers. = 13. i.e., H.C.F (26, 91) = 13. (ii) Since, 1296 = 2 × 2 × 2 × 2 × 3 × 3 × 3 × 3 = 24 × 34 and, 25 = 5 × 5 = 52 = 1, as given numbers do not have any common prime factor. Sol. (i) Since, 26 = 2 × 13 and, 91 = 7 × 13 26 (Ans) and, H.C.F. = Product of common prime factors with lowest powers (iii) 17 and 25 2 (Ans) (iii) Since, 17 = 17 Ex.19 Find the L.C.M. and H.C.F. of the following pairs of integers by applying the Fundamental theorem of Arithmetic method i.e., using the prime factorisation method. (i) 26 and 91 L.C.M. = Product of each prime factor with highest powers = 24 × 34 × 5 × 7 = 45,360  USING THE FUNDAMENTAL THEOREM OF  ARITHMETIC TO FIND H.C.F. AND L.C.M.  7 2520 = 2 × 2 × 2 × 3 × 3 × 5 × 7 = 23 × 3 2 × 5 × 7 and, product of their L.C.M. and H.C.F. = 182 × 13 = 2366   Product of L.C.M and H.C.F of two given numbers = Product of the given numbers In example 19 (ii) : Product of given two numbers = 1296 × 2520 = 3265920 and, product of their L.C.M. and H.C.F. = 45360 × 72 = 3265920  L.C.M. (1296, 2520) × H.C.F. (1296, 2520) Sol. = 1296 × 2520  But in case of two co-prime numbers also, the product of the numbers is always equal to the product of their L.C.M. and their H.C.F. And, we know : H.C.F. (150, 100) = = Pr oduct of 150 and 100 L.C.M. (150,100) 150 100 = 50. 300 H.C.F. = 1; L.C.M. = 17 × 25 = 425; Ex.22 The H.C.F. and L.C.M. of two numbers are 12 and 240 respectively. If one of these numbers is 48; find the other numbers. product of numbers = 17 × 25 = 425 Sol. As, in case of co-prime numbers 17 and 25; and product of their H.C.F. and L.C.M. For any two positive integers : Their L.C.M. × their H.C.F. = Product of the number Pr oduct of the numbers  (i) L.C.M. = H.C.F. (ii) H.C.F. = Pr oduct of the numbers L.C.M. (iii) One number = L.C.M. H.C.F. Other number Ex.20 Given that H.C.F. (306, 657) = 9, find L.C.M. (306, 657) H.C.F. (306, 657) = 9 means H.C.F. of 306 and 657 = 9 Required L.C.M. (306, 657) means required L.C.M. of 306 and 657. For any two positive integers; their L.C.M. = Since, the product of two numbers = Their H.C.F. × Their L.C.M.  One no. × other no. = H.C.F. × L.C.M. = 1 × 425 = 425. Sol.  L.C.M. of 150 and 100 = 300 = 150 × 100 The given numbers 17 and 25 do not have any common prime factor. Such numbers are called co-prime numbers and their H.C.F. is always equal to 1 (one), whereas their L.C.M. is equal to the product of the numbers.   Since, the product of number 150 and 100 In example 19 (iii) :  L.C.M. (150, 100) = 300  Pr oduct of the numbers Their H.C.F. i.e., L.C.M. (306, 657) = 306 657 = 22,338. 9 Ex.21 Given that L.C.M. (150, 100) = 300, find H.C.F. (150, 100)  Other no. = 12  240 = 60. 48 Ex.23 Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 + 5 are composite numbers. Sol. Since, 7 × 11 × 13 + 13 = 13 × (7 × 11 + 1) = 13 × 78 = 13 × 13 × 3 × 2; that is, the given number has more than two factors and it is a composite number. Similarly, 7 × 6 × 5 × 4 × 3 + 5 = 5 × (7 × 6 × 4 × 3 + 1) = 5 × 505 = 5 × 5 × 101 The given no. is a composite number.