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Real Numbers, , n, , class IX we, , studied about real, numbers. In this, chapter, we shall, We, , the, , begin with two important, , numbers,, , especially, , about irrational, continue our discussion on real numbers., results, namely Euclid's division lemma and, , fundamental theorem of arithmetic., , LEMMA, , A, , lemma is a, proven statement used for, , EUCLID'S DIVISION LEMMA, For, , any two given, and r such that, a, , Here, we call a, , positive integers a, , and b, there exist, , bq +r, where 0<r<b., as, dividend, b as divisor, q as, =, , CBSE 2009C], , quotient and r as remainder., , (divisor quotient) + remainder., x, , Suppose we divide 117 by 14. Then, we, get, 8 as, quotient and 5 as remainder., , Here dividend, , =117, divisor 14,, quotient, =, , and, , remainder 5., Clearly, 117 = (14 x 8) +5., , EXAMPLE1, SOLUTION, , unique whole numbers, , =, , Dividend, Example, , proving another statement., , =, , 14) 117(8, -112, , 8, , 5, , =, , A, , number when divided, 73 gives 34 as, remainder. Find the number.by, quotient and 23, Here divisor 73,, quotient 34 and remainder, By Euclid's division, lemma, we have, , as, , =, , =, , =, , dividend (divisor x, quotient) + remainder, x, (73 34) +23, (2482 +23) 2505., Hence, the required number, is 2505., ALGORITHM An, is, a, algorithm series of, for solving a certain type, well-defined steps which gives, of problem., =, , =, , =, , =, , EUCLID'S DIVISION, ALGORITHM It is, positive integers, say a and b with, , a, , technique to, , a>b, in the comnpute, following steps., 3, , the HCF, , a, , method, , of two given

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for, M, , a, , t, , h, , e, , m, , a, , t, , i, , c, , 10, , Class, , Real Numbers, , s, , School, , Secondary, renainder, , get, , we, , Step, , 1., , Step 2., , g, , the quotient, , r, , such that, , The, , and, , by b,, On, Sr<b., where0, a =bq +r,, a, , iivuding, , Ifr, , =, , 0, , b) =b., , then HCF(a,, , lfr#0, , apply, , then, , the, , EXAMPLE 2, sOLUTION, , Use, , be the required, , is O0., , EXAMPLE3, , to find, , the, , 1032), , HCF(272,, , 1032, , Since, , 1032, , divide, , get 3, 216 as, , So, by, , as, , using, , HCF, , of 272, , Step 2., , Since, , quotient, , Euclid's, , 272, , by, , get 1, , as, , 56, , as, , 255 =, , 8)48 (6, Step 3., , 48, , quotient and 48, , by Euclid's, Step 4, , Since the, 1, , as, , 2 +51., , 102, , as, , divide, , we, , 216, , by 56 to get, , lemma,, , we, , sOLUTION, , remainder 48 #0,, , remainder, and 0, , as, , Hence, HCF(867, 255) = 51., , EXAMPLE4, , get:, , 56 x3 +48., , as, , 51x 2+0., , remainder., , division, , quotient and 8, , Since the, , 102 x, , Since the remainder 51 +0, we divide 102 by 51 to get, , get:, , we, , divide 56, , by, , 48 to, , get, , 8 +0,, , we, , we, , get:, , divide 48, , We find HCF(196, 38220), using the following steps., Since 38220> 196,, , by, , 8 to, , get, , 6, , 195, , as, , we, , quotient and, , we, , get:, , divide 38220, 0, , as, , by, , 196 to, , remainder., , 196)38220 (195, 196, , by Euclid's division lemma, we get:, , 1862, , 38220, , 1764, , 196 x 195 +0., , Since the remainder is 0, so our procedure stops., , HCF(196, 38220), , remainder., , quotient, by Euclid's division lemma,, 48 8 x 6+0., , Use Euclid's algorithm to find HCF(196, 38220)., , get, , remainder., , 56 48x1 +8., , as, , 102, , The remainder is now 0, so our procedure stops., , by Euclid's division lemma,, Step 5, , 51)102(2, , by Euclid's division lemma, we get:, we, , lemma,, , 56 #0,, Since the remainder, , 216, , 204, , 2 as quotient and 0 as remainder., , 272 216x1 +56., , as, , 102) 255 (2, , by Euclid's division lemma, we get:, , 48, , b y Euclid's division, , 3, , 765, , 867 255 x3+ 102., Since the remainder 102 # 0, we, , remainder., , Step 3., , 255)867(3, , divide 255 by 102 to get 2 as, quotient and 51 as remainder., , 168, 48) 56 (1, , divide, , and, , Step 2., , 272)., , find HCF(867, 255), HCF(867, 255) using the following steps., , we get:, , 56)216 (3, , division, , we, , We find, , 816, , remainder, , 216 to, , quotient, , =, , by Euclid's division lemma,, , +216., 272 x3, , 216+0,, , HCF(56, 48) HCF(216, 56), HCF(272, 216) HCF(1032,, method is also known as, successive division method., Use Euclid's, to, algorithm, , Step 1. Since 867> 255, we divide 867, by 255 to get 3 as quotient and, , 216)272(1, 216, , and, , remainder., , the, , procedure stopS, , =, , 102 as remainder., , lemma, we get:, 1032, , 8), , 272) 1032 (3, , 272 to, , by, , and 1032., , following steps., , the, , we, , 272,, , >, , sOLUTION, , HCE, , algorithmn, Euclid's, , Step 1., , REMARK This, , ill the, , divisor, , We find, , r., , process, , will, , =HCF(48,, , so our, , 8., , =, , =, , and, , b, , division, , the, , Step 3., , 8, , become 0,, , now, , =, , lemma, , to, , remainder, , Continue, , The last, , Note, , remainder has, , Hence, HCF(272, 1032), , =, , 196., , 980, 980, , 0

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10, for Class, , Real, , bers, , M a t h e m a t i c s, , Secondary, , School, , THEOREM4, , DIVISION, , OF, , A P P L I C A T I O N S, , sOME, Shiow, , THEOREM1, , that every, , PROOF, , On, , be, , n, , When, , Casel., , When, , I., , In this, Hence, every, , THEOREM12, , =, , PROOF, , n, , Then,, , be, , n, , +, , =, , n, , =, , some, , clearly odd., , 211 and every, of the form, integer is, ., , integer, , for, , THEOREM3, , integer, , some, , (3m, , +, , 1), , 3m, , of the form, , or, , (3m, , +, , 1), , or, , integer is of, , Clearly,, and 1, , or, , (4m, , (4m, (4m + 2)., , be, , 5), , Euclid's division lemma, show that the square of any positive, is either of, the form 3mm or (3m + 1) for sonne integer 1, [CBSE 2008, , an, , arbitrary positive integer., , Then, by Euclid's division lemma, we have, n = 3 q + r, where 0 S r < 3 ., , we, , (3tm, , +, , 2), for, , the form 3m, , n2, , some, , or, , integer, , (3m +1), , or, , CaseI., , 1., , (3m, , +, , ., , 9q2 + +6qr, , (i), where 0, , r < 3., , When r=0., Putting r =0 in (i), we get:, , 2), , 9q2 3 (3q), , n, , is, , oftheform (4m +1), , or, , (4m +3), , +, , +, , 2), , 1), , or, , (4m, , are even, , (4m +1), , for some integer m., , or, , Case l ., , =, , When r =, , 3m, where, , =, , m, , =, , 3q, , is, , an, , integer., , n*= (9q, r, , be the, , 1., , Putting r=1 in (i), we get:, +1, +1 +69) 3 (34+2q) +1 3(3q+2), , Case ll., , When, , r=2, , Puttingr, +, , 2), , or, , (4m, , and since, , n, , +, , is, , 3)., odd,, , so n, , # 4m, , =, , = 3m +1, where n = (3q* + 2q) is an integer., , remainder., , (4m +3), for some integer m., is of the form (4m + 1) or (41 +3), Hence, any positive odd integer, =, , n, , +, , ., , 4m and, , n, , Let, , or (6m, , remainder., , n =4m + r, where 0 Sr<4, 4m, , Using, integer, , PROOF, , be the, , r, , for some integer m., integer., Let n be a n arbitrary odd positive, the quotient and, On dividing n by 4, let m be, w e have:, So, by Euclid's division lemma,, =, , (6m +4) or (6m + 5)., , 6in, (6m +2), (6m, , On dividing n by 3, let q be the quotient and r be the remainder., , Show that any positive odd integer, , n, , =, , n, , or, , m., , division lemma,, , or, , THEOREM5, , integer, (2m +1) for some, , 3m +r, where 0Sr<3., 3m, , 611+r, where r =0,1, 2, 3, 4, 5, 6in or (6m 1) or, +, (6m +2) or (61m +3), , +, 4) give even values of n., Thus, when n is odd, it is of the form, (61 +1) or (6m +3), for some integer m., , an, , n, , 1, , +r, where 0Sr<6, or, , even., , integer is, any positive, , have:, , 6111, , 1), for some, , even, , we, , 1, , But,, , is, , positive, , by Euclid's, , a, , 1, , 2m +1., , Thus, any positive, , PROOF, , be the, , arbitrary poaitive integer., and, let m be the quotient, On dividing n by 3,, have:, Let, , 1, , some, , Euclid's division lemma,, , remainder., , r, , odd, , integer is of the form (61m+ 1) or, m., integer, Let be, given posilive odd integer., On, dividing n by 6, let m be the quotient and r be the remainder., Then, by, , 0Sr<2., , is clearly, , case, 1, , (3m +2) for, , PROOF, , positive, , (6m +3) or (6nm +, 5) for, , have:, , we, , form, is of the, positive odd integer, Show that, , mteger., , some, , Show that, every, , integerni., , (211, , case,, n, , that every, , 211., , =, , n, , In this, Case, , where, , 2m+r,, , =, , n, , 1 2 1 or, , is, , wlere m, , lemma,, , division, , and, form 2m, , (21+1),, , by 2,, , n, , Euclid's, , by, , even, , integer., arbitrary positive, and, quotient, the, be, let m, , an, , dividing, , Then,, , positive, , integer is of tihe, , form, is, odd integer ofthe, , positive, , Let, , LEMMA, , EUCLID'S, , n=, =, , =, , 2 in, , (i),, , (9q +4+ 12), , we, , =, , 3n1+1, where, , m, , get:, , 3(3q+4q+1) +1, (3q* +4q+ 1) is, =, , an, , integer., , is of the form 3m, of any positive integer, Hence, the square, 1., (31+1) for some integer, , or

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Secondary School Mathematics for Class 10, THEOREM6, , 9, , Real Nunmbers, , Using Euclid's division lemma,, , show that the cube of any posttive, nteger is of the forn1 9m or (9m + 1) or (9m +8) for sonme iiteger ni., , ience,, , one, , only, , and, , out of, , one, , n, (11, , 1) and, , +, , +2) is divisible, , (n, , by 3., PROOF, , Let, , n, , ICBSE 2009C], , be, , arbitrary positive integer., On, dividing by 3, let g be the quotient and rbe the remainder. So,, an, , Show that, , THEOREM 8, , Twlhere, , n, , by Euclid's division lemma, we have, n, 3q+r, where 0 Sr<3., (3 +r)' 27q* +r+9qr (3q +r), =, , -, , Case I., , PROOF, , (27q+27qr +9qr) +r, , .(i), where 0 Sr, , <, , n= 27q, , =, , 9(3q*), , m, , =, , l., , 3q is an integer, , (27q+27q, , +9)+1, , 9m +1, where, r, , let g be the quotient and, , m, , =, , =9q(3q3 +34, , +, , q(3q +3q +1) is, , If, , So,, , int eger., , =2 in, , n, , =, , 3q +1, , PROOF, , On, , Show that, , =, , m, , q(3q 6 +4), +, , is, , an, , integer., , and, , only one out of n, (n +1) and (n +2) is divisible, by 3, where n is any positive integer., dividing n by 3, let q be the quotient and r be the remainder., , n, , =, , n, , Case ., , Case Il., , =, , one, , only, , and, , =, , (3q +1), , 4), , +, , 3q, , =, , out, , is divisible, , of n,, , =, , then (n +2), , clearly divisible by 3., , What do you, , which, , 1),, , is, , +, , which, , 3(q +2),, , 6, , n, , +2,, , n, , by, , 3., , +4, , is divisible, , by, , 3., , mean, , by Euclid's division lemma?, a s remainder., by 61l gives 27 as quotient and 32, , number when divided, , YA, Find the number., , J, , By what, , number should 1365 be divided to, , =, , (3q +3) 3(q +1),, , In this case, (n +2) is divisible, , =, , Using Euclid's division algorithm, find, , which is, , 7. Show that any, , quotient, , and 32, , as, , 504, , the HCF of, , and 1188, is either, , 9 6 0 and 1575., even or, , odd., +, , 3), , 1), , or, , (6m, , (4m +1), , or, , (4m +3),, , is of the form (6m, , +, , positive odd integer, , is of the form, , where m is some integer., , by3., , In this case, (1 +1) is divisible by 3., , as, , or (6nm + 5), where m is some integer., , ANSWERS (EXERCISE 1A), , Case Il1. If n = (3q+2) then (n+ 1) = (3q +3) =3(q +1), which is, , clearly divisible by 3., , get 31, , remainder?, , positive integer, odd integer, that, positive, any, 6.Show, , If n = 3q then n is clearly divisible by 3., n=, , (1, , one, , 5. Show that every, , 3+r, where r =0,1 or 2, 3q orn (3q + 1) or n (3q+2)., If, , then (n +4), , 3q +2, , in this case,, , A05 and 2520, , Then, n = 3q +r, where 0 S r < 3, 1, , t+, , EXERCISE 1A, , of any positive integer is of the form 9m or (9m+1), , one, , 3q+3- 3(q, , (n +2), , then, , (i), we get:, , or (9m1 +8) for some integer m., THEOREMM7, , be the remainder., , r, , is divisible by 3., , 1) +1, an, , Hence,, , 9m +8, where, , Hence, the cube, , 3,, , 3q or n = 3q + 1 or n = 3q + 2., , =(27+54+ 36q) +8 =9q(3q +6q +4)+8, =, , by, , ICBSE 2008C), , If n = 3q then n is divisible by 3., , Case IlI. When, , Case ll. When r =2., , Putting, , 4 is divisible, , t, , n, , So, in this case, (1 +2) is divisible by 3., , Putting r =1 in (i), we get:, F, , 2,, , 1 +, , divisible by 3., , When r = 1., , =, , by 3,, , 3q+ r, where 0 Sr <3, , =, , of n,, , positive integer., , 3q+r, wherer = 0,1,2, , Case, , 91m, where, , out, , 1, , Case I., , =, , n, , one, , Then, =, , 3., , Putting r=0 in (i), we get:, , is any, , dividing, , =, , When r = 0., , Case Il., , On, , n, , only, , and, , one, , 2. 1679, , 3. 43, , 4. (i) 45, , (i), , 36, , (ii), , 15

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10, , Secondary School Mathematics for Class 10, , HINTS TO sOME SELECTED QUESTIONS, 3. Let the, , required divisor be, Then, dividend =(divisor, quotient), 1365= (, , 5. Let, , On, , n, , be, , 1, 1, , +, , remainder, , 21, , by, , n, t, , r,, , or, , n, , 2, let, , m, , be the, , wvhere 0, =, , 2m+1, , Also,, , quotient and be the remainder., [by Euclid's division lemma), , ***********, , (ii) 69 = 3 x 23, , iv) 234= 2 x3x3x13, , theorem of arithmetic, know from the fundamental, , (ii) 105, , 4", , can never, , be verified, , 0., , uohere, any number of the form 6",, with the digit 0., , But, 6"= (2 x 3)", (2", prime factors of 6"., , x, , =, , n, , E, , N, , we, , 3"), , shows that 2 and 3, , fundamental theorem, , know from the, , can never, , are, , end, , the only, , of arithmetic, , that the prime factorization of each number is unique., , 3 x 5 x7, , So, 5 is not a factor of 6"., Hence, 6", , easily, , digit, , If 6" ends with 0 then it must have 5 as a factor, , (v) 462 = 2 x3x7x 11, , can, , end with the, , sOLUTION, , (vi) 651 =3 x 7x31, The above factorization, , factor, , only prime, , Show that, , Every composite number can be uniquely, product of primes, except for the order in which these prime factors, , (i) 12 2x2x3, , factor., , as a, , that 2 is the, , EXAMPLE 3, , Also,, , Examples, , we, , Hence,, , for some integer ni., , FUNDAMENTAL THEOREM OF ARITHMETIc, OCcurs., , 22 shows, , So, 5 is not a factor of 4"., , r, , S0, n is either even or odd., , expressed as a, , =, , that the prime factorization of each number is unique., , r<2, , ********, , (2)", , =, , of 4", , x 31) +32. Find x., , 2m, , =, , If 4" ends with 0 then it must have 5, , But, 4", , arbitrary positive integer, , an, , dividing, , Then,, , sOLUTION, , r., , x, , 11, , Real Numbers, , by actual, , division., , EXAMPLE4, , can n e v e r, , Find the HCF and, , end with the, , of, , LCM, , digit, , 126 and 156, , 0., , using prime factorization, , method., , sOLVED, EXAMPLE1, , Show that each, , EXA MPLES, , of the follorwing, , (i) 5 x 11 x 13 + 13, SOLUTION, , is, , a, , SOLUTION, , 5x 11, , 2, x, , 13+13, , 13, , 2, , 63, , 78, , (5 x 11 +1) (13 x 56)., Clearly, it shows that the given number has more than, two factors. Hence, it is a, composite number., , (ii) 6 x 5 x4, , =, , x, , x3x2x1+5, , =, , =, , =, , 5, , (5, , x, x, , (6, , x4x 3 x2x 1+1), , Clearly, it shows that the given number has, factors. Hence, it is a composite number., , Show that any number of, the form 4",, digit 0., , 39, , 3, , 145)., more, , than, , two, , EXAMPLE 2, , 156, , 126, , (ii) 6 x 5 x 4 x3x2x1+5, , We have, , (i), , We have:, , composite number:, , n, , EN, , can never, , 126, and 156, , end with the, , 13, , (2x3 x3x7) (2 x32x 7), =, , (2 x 2 x3x 13) = (22 x3x13)., , HCF(126, 156) = product of common terms with lowest, , power, =, , (2 x3) (2 x 3) =6, =

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12, , Secondary School Mathematics for Class 10, and LCM(126, 156) = product of prime factors with higest, , EXAMPLE6, SOLUTION, , If one of the, , 1449., is 23 and their LCM is, , For two numbersa and b, we know that:, , (a x b) = {HCF(a, b)} x (LCM(a, b)}., , 3276., HCF = 6 and LCM = 3276., , Herea, , =, , 161, HCF =23 and LCM, , (161 x b) = (23 x 1449)b=, , We have:, , 1449., , =, , And, we have to find b., , Find the HCF and LCM of 612 and 1314 using prime factorization, method., , SOLUTION, , of two numbers, , The HCF, , numbers is 161,find the other., , power, = (2- x 3* x 7x 13) = (4 x9 x7x 13), , EXAMPLE5, , 13, , Real Numbers, , (23, , 1449), , 161, , 207, , Hence, the other number is 207., , 612, , 2, , 306, , EXAMPLEE7, , Given that HCF(252, 594) = 18, find LCM(252, 594)., , sOLUTION, , We have:, , 1314, , LCM, , 2, , 153, , 3, , 18, , 3, , 219, , 73, , 17, , (2x3x3x73), , (2 x, , =, , EXAMPLE 8, SOLUTION, , 3 73), x, , 1314), , =, , (2 x 3), , =, , product, , (2 *9), of, , =, , (2, , x, , 3, , x, , 17, , prime factors, x, , 73), , 148) 185 (1, , We find HCF(148, 185), which is 37., , 148, 37) 148 (4, , =, , highest, , EXAMPLE 9, sOLUTION, , Hence, HCF = 18 and LCM = 44676., , AN IMPORTANT PROPERTY, , Product of two given numbers = product of their HCF and LCM., , Find the HCF and LCM, , of 108, 120 and 252 using prime factorization, , By prime factorization,, , (2x 3), , 120, , (2 x3 x5), , 252, , we, , get:, , 2, , 2 120, , 108, , 54, , 108, , 2, , 27, 3, , 9, , 60, , 2 30, 315, , =, , product, , of, , common, , The above result is true for two numbers, , only., , 3, , 63, , 321, , terms with lowest, , power, CAUTION, , 2 252, 2 126, , (22 x 3 x7), , HCF(108, 120, 252), , Thus, (a x b) = HCF(a, b) x LCM (a, b)., , of the given fraction is, , method., , (4 x9x 17x 73), , = 44676., , 0, , Hence, the simplest form, with, , 148, , 148 1487-3, , 18, , power, =, , 148, simplest form of i85, , of the given fraction by 37., , power, =, , Find the, , So, we divide the numerator and denominator, , HCF(612, 1314) = product of common terms with lowest, , and LCM(612,, , 8316., , -, , Hence, LCM(252, 594) = 8316., , 612 (2x 2 x3x3x 17)= (2 x3*x 17), and 1314, , their HCF, , (252 x 594), , 657, , 51, , 3, , product of two given numbers, , = (2 x 3) = (4 x 3) = 12.

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14, , =product, , LCM(108, 120, 252), , of prime factors with highest, , 527, , x 3, , x5 x 7) = 7560., , Hence, the, , EXAMPLE 13, , (245-5), , divides, , 2, , 1032, , 60, , 2, 2, , 516, 258, , 30, , 3, , 129, , So, the required number is HCF(240, 1032). 2240, , 2 120, , Now, 240= (2* x3 x 5), , 2, , 2, , and 1032 = (2 x3 x 43)., , 3 15, , HCF(240, 1032) = (23 x 3) = 24., , and, , 240, , =, , 5, , Find the, , largest, , Clearly,, , the, , required, , (129-3), , number divides, , =, , capacity, , 3, , (2 3, , HCF(126, 540), , =, , of, , common terms, , number, , divides, , (398-7) 391,, , Clearly,, , maximum number, , army, , of columns, , 48, , 2 612, , the maximum number of, , 48, , 2 24, , 2 306, , (2 x 3)., =, , (22, , x, , number of, , A sweetseller has 420, , 3), , (4, , =, , 3), , x, , columns, , kaju burfis, , =, , 12., , 12., , =, , and 150 badam, , burfis., , He wants, , the same number,, to stack them in such a way that each stack has, How, the, area, many, of these can be, the, least, and they take up, tray., of, in each stack? How many stacks are formed?, , SOLUTION, , 2 420, , 2, , 210, , 75, , Now, 420, , 3 105, , 25, , 150, , (2 x 3 x5 x 7), , 5, , 425, , 5 85, , 17, , 535, , (2x 3 x5)., , HCF(420, 150), , 17 391, , 2 150, , Maximum number of burfis in each, stack = HCF(420, 150)., , and, , 23, , (5x 17),, , an, , to march in the, , placed, , required number = HCF(391, 425, 527)., , 425, , of columns. What is the, , required, , EXAMPLE15, , (436-11) =425 and (542-15) =527 exactly., Now, 391 = (17 x 23),, , nunber, , HCF612, 48), , with lowest, , EXAMPLE12 Find the largest number that will divide 398, 436 and 542, leaving, remainders 7, 11 and 15 respectively, required, , behind, , same, , 5, , Hence, the required number is 18., , the, , 21 litres., , =, , contingent of 612 members is to march, are, a parade. The two groups, , of 48 members in, , and, , = (2 x 3*) = (2 x 9) = 18., , Clearly,, , container, , required, , band, , 315, , power, , sOLUTION, , An army, , of the, , Now, 612 =(22 x 32 x 17), , 135, , 5)., , product, , 3 21, , columns =HCF(612, 48)., , 45, , 540 = (2 x2 x 3 x 3 x 3 x 5 ), x, , sOLUTION, , 2 540, 2 270, , 3 63, , Now, 126 = (2 x3 x3 x 7) =(2 x3 x 7), , x, , 3 63, , x3x7)., , (5, , 735, , 2 126, , (2 x 33 x 7), =, , 5, , 3 147, , HCF(504, 735) = (3 x 7) = 21., , leaving, , 126 and, , 2 126, , required number =HCF(126, 540)., , =, , 2 252, , in which they can nmarch?, , (545 5 ) = 540 exactly., , and, , 2504, , factors, we get:, , and 735, , remainders 3 and 5 respectively., SOLUTION, , Resolving 504 and 735 into prime, , 504, , EXAMPLE 14, , number which divides 129 and 545,, , of, , either tank an exact number of times., SOLUTION, , 43, , Hence, the required number is 24., , EXAMPLE 11, , Two tanks contain 504 and 735 litres of milk respectively., milk, which can m e a s u r e the, axmum, a container, , capacity of, , Find the largest number which divides 245 and 1037, eavng, remainder 5 in each case., , Clearly, the required number, (1037-5) = 1032 exactly., , sOLUTION, , required, , 17., , =, , number is 17., Find the, , HCF = 12 and LCM = 7560., , EXAMPLE10, , (17 x 31)., , HCF(391, 425, 527), , power, = (2, , 15, , Real Numbers, , Mathematics for Class 10, , Secondary School, , =, , (2, , x, , 3, , x, , 5), , =, , 30., , maximum number of burfis in each stack = 30., number o f stacks =, , =, , (14 + 5) =, , 19.

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16, , Secondary School Mathematics for Class T10, , EXAMPLE 16, , Real Numbers, , Ravi and Sikha drive around Circular, takes, sports ficld. Ravi, n, minutes to take one round, while Sikha, round, the, completes, l6, , 0114tes. If both start at the same point, at the same lmme and, , go m the same direction, after how much time will they meet at the, , starting point?, SOLUTION, , ., , The HCF of two numbers is 145 and their, numbers is 725, find the other., ., , LCM(16, 20) = (2, , The, , 6. Is it, , =(2 x 5)., x 5) = (16 x 5), , HCF of two numbers is 18 and their, , pOssible, , 80., , HINT, , of class, , can, , be distributed, , equally, , among students, , of section, , A, , Find, , So, the number of these books must be a multiple of 30 as well, as that of 28., , have, , LCM is 760?, two numbers whose HCF is 18 and, , always divides, , LCM, , completely., , the, , largest number, , 368, (iv) 496, , (ii) 1095, 1168, , 473, (ii) 645, , the, , 5 and 7, , which divides 438 and 606,, , same, , largest number which, , divides 320 and 457,, , leaving, , remainder, , leaving, , remainders, , respectively., , 91 leaves the, least number which when divided by 35, 56 and, , remainder 7 in each, , case, , 11. Find the smallest number which when divided, , by, , 28 and 32 leaves, , remainders 8 and 12 respectively., , Consequently, the required number is LCM(30, 28)., , 12, , Now, 30 =2 x3 x5, , Find the smallest number which when increased, , by, , 17 is, , exactly, , divisible by both 468 and 520., , 28 =2x 7., LCM(30, 28) = product of prime factors with highest power, , (2, , HCF, , 10. Find the, , equally among the students of section A or B., , x3x5x 7) (4 x3x5x 7), Hence, the required number of books =420., =, , is 12960. Find their, , product, , 6 in each case., , 9Find, , Clearly, he required number of books are to be distributed, , and, , to, , (6), , or, , section B., SOLUTION, , ot the, , 7. Find the simplest form of:, , Ina school there are two sections, nanmely A and B,, X. There, are 30 students in section A and 28 students in section B. Find the, minimum number of books required for their class library so that, , they, , one, , Give reason., , Hence, both will meet at the starting point after 80 minutes., , EXAMPLE17, , LCM is 2175. If, , LCM., , Required number of minutes = LCM(16, 20)., Now, 16 2' and 20, , The HCF of two numbers is 23 and their LCM is 1449. If one of the, , numbers is 161, find the other., , =, , =, , 420., , 13), , Find the, , greatest, , number of four, , digits, , which is, , exactly divisible by 15,, , 24 and 36., 14. In, , a, , seminar, the number of, , mathematics, , of, , rooms, , are, , 60,, , required,, , 84 and 108, , if in each, , in Hindi, English and, Find the minimum number, , participants, , respectively., , room,, , the, , same, , number of, , participants are, , to be seated and all of them being in the same subject., , enaEXERCISE 1B, 1., , Using prime factorization,, (i) 36, 84, , (iv) 144, 198, , find the HCF and LCM of:, , (ii) 23, 31, , ii) 96, 404, , (v) 396, 1080, , (vi) 1152, 1664, , same. How many stacks will be there?, , HCF x LCM = product of given numbers., , 2. Using prime factorization, find the HCF and LCM of:, , 6i) 8,9,25, iv) 24, 36, 40, , be divided into, pieces of timber 42 m, 49 m and 63 m long have to, of each, the, same length. What is the greatest possible length, of, planks, plank? How many planks are formed?, Find the greatest possible length which can be used to measure exactly, the lengths 7 m, 3 m 85 cm and 12 m 95 cm., , 16. Three, , In each case, verify that:, , (i) 12, 15, 21, , (ii) 17, 23, 29, , (v) 30, 72, 432, , (vi) 21, 28, 36, 45, , 336,, , English, mathematics and science books containing, a, way that, 240 and 96 books respectively have to be stacked in such, is the, stack, of, each, the, and, height, all the books are stored subjectwise, , 15. Three sets of, , (17)

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19, , 18, , Real Numbers, , Secondary School Mathematics for Class 10, , 18., , Find the maximum number of students among whom 1001 pens and, , pencils can be distributed in such a way that each student gets the, same number of, pens and the same number of pencils., the ceilin8 ot a, Find the least number of, square tiles required to pave, , ., , =1,, , (iii) HCF, , 910, , HCF =6, LCM, , (v), 3. 207, , Three, least, , measuring, , length, , rods, , are, , of cloth that, , 64 cm, 80, , cm, , and 96, , be measured, , can, , an, , cm, , in, , length., , Find the, , exact number of times,, , using any of the rods., 21. An, , electronic device makes, , a, , beep after, , every 60 seconds., , Another, , device makes a beep after every 62 seconds. They beeped together, 10 a.m. At iwhat time will they beep together at the earliest?, , at, , HCF, , (vi), , HCF, , not, , 9. 45, , 14. 21, , 15. 14, , 19. 814, , 20., , =, , =, , 4,, , LCM, , 1,, , LCM, , 360, , =, , 1260, , =, , divide LCM exactly, , 23, , (v) 31, , (ii), , 21. 10:, , 18. 91, , 17. 35 cm, , 16. 7 m, 22 planks, , 9.6 m, , 13. 9720, , 12. 4663, , 11. 204, , 10. 3647, , 8. 24, , 12 hrs, , 22. 8:7:, , 31 hrs, , 23., , 16 times, , 330, 660), 24. From last blank to first (55, 165,, , HINTS TO sOME SELECTED QUESTIONS, Required, , number, , 9., , Required, , number, , 10., , Required, , number, , 8., , 11., , Here,, , together?, 12., , Required, , HCF(432, 600), , =, , 24., , =, , HCF(315, 450), , =, , 45., , =, , {LCM(35, 56, 91), , =, , 20 and (32-12), , (28 8), required, , 24. Find the missing numbers in the following factorization., , (iv), , 5. 720, , )5, , 7.(, , 48 seconds, 72 seconds and 108 seconds respectively. If they all change, , 10, 12 minutes respectively. In 30 hours, how many times do they toll, , 2160, , =, , 4. 435, , 22. The traffic lights at three different road crossings change after every, , simultaneously at 8 a.m., then at what time will they again change, simultaneously?, 43 Six bels commence tolling together and toll at intervals of 2, 4, 6, 8,, , 11339, , =, , No, since HCF does, , 6., , room 15 m 17 cm long and 9 m 2 cm broad., 0., , LCM, , number, , 7), , +, , (LCM (28, 32), , =, , digits, , =, , 20), , -, , 17), , ={LCM(468, 520), , number, , 13. Greatest number of four, , (3640 +7), , =, , 3647., , =, , 20., , =, , =, , =, , (224 20), , (4680- 17), , 204., , =, , =, , 4663., , 9999., , LCM(15, 24, 36) = 360., , 2, , On, , 33, , dividing 9999 by 360,, required, , number, , 14. Maximum number of, , remainder, , participants, , 279., , =, , (9999 279), , =, , =, , 9720., , in each, , room, , Minimum number of rooms required ={, , 55, , 15. HCF(336, 240,, , 96), , Number of stacks, , 5, , 11, , 16., , Required length, , =, , ANSWERs (EXERCISE 1B), =12,, , (i) HCF, , =252, , LCM, , (ii) HCF = 4, LCM = 9696, , (ii) HCF = 1, LCM = 713, , iv) HCF =18, LCM = 1584, , (v) HCF 36, LCM 11880 (vi) HCF 128, LCM 14976, (i) HCF 1, LCM 1800, (ii) HCF =3, LCM = 420, =, , 2., , 17., , =, , =, , =, , =, , Required length, , So,, , =, , we, , =, plank in, , =( +, , HCF(700, , HCF(60, 84, 108), , 2)-21, (7+5+2) =14., , metres, , =, , HCF(42, , m,, , 49, , m,, , 63 m), , 0 ) = (6+7 +9) = 22., , cm,, , 385, , cm,, , 1295, , cm), , =35 cm., , 18. Maximum number of students = HCF(1001, 910) =91, 19. Side of each square tile = HCF(1517 cm, 902 cm) = 41 cm., , Required number of tiles, , 12., , =, , make stacks of 48 books each., , (, , of each, , Number of planks, , 1., , 48., , =, , =, , 1517, x 902, 41x, 4 1 ) = 814., , =, , 20. Required length = LCM(64 cm, 80 cm, 96 cm) = 960 cm = 9.6 m., , =, , 7, , m.

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20, ., , Secondary School Mathematics for Class, , nterval ot beeping together = LCM (60 seconds, 62 seconds) = I, , 6Umin, bU, , 21, , Real Numbers, , 10, 860 secons, , AN IMPORTANT TEST, , 31 min., , Let, , be the simplest form of a given rational number, , So, they will beep together again at 10:3I a.m., .22, , interval of change = LCM (48 seconds, 72 seconds, 108 seconds) = 432 seconus, , (2" X5") for some non-negative integers, is a terminating decimal., , then, , and n, , m, , P, , (1) It q, = 7 min 12 seconds., , Required, , time of simultaneous change, , =, , 8:7: 12hrs., , (i) Ifq +(2", , 23. LCM of 2, 4,6, 8, 10, 12 - 120, After every 2 hours, , they, , Required umber of, , toll, , tinmes, , =( 1 ) times, , SOLVED EXA MPLES, , 16 times., , =, , EXAMPLE 1, RATIONAL NUMBERS, , Counting, , numbers 1,, , 2, 3, 4,..., etc.,, , are, , knoWn, , All, , counting, , numbers, , together, , with 0, , form, , the collection of, , whole numbers., , All, , ...,, , etc.,, , whole numbers., , are, , counting numbers, negatives, , of counting numbers and 0 form, , the, , collection of all integers., Thus,.,4,-3,-2, -1, 0, 1, 2, 3,, RATIONAL NUMBERS, , q+0, , are, , The numbers, , ...,, , etc.,, , are, , where p and q are, , integers, and, , rational number when, , in, , of the form, , (i) 0, , none, , of 2 and 5 is, , So, the given rational, , Clearly, (22x 5), , Every, , is in its, , simplest form., x, , 5")., , terminating, , a, , decimal., , 62, Now 25)2x5)(2x5(0, 1000, , expressed, , (ii) The given, Now, 50, So, the, , AN IMPORTANT OBSERVATION, a, , factor of 31., , = 0.062., , decimal form., , To Test Whether a Given Rational Number is, , a, , is of the form (2", , number is, , called rational numbers., , RATIONAL NUMBERS IN DECIMAL FORM, , (iv) 17, , (ii) 1000, , 35, 1, (i) The given number is x, , So, the given, , integers., , decimal form is expressible either in terminating or in nonterminating repeating, , Terminating or Repeating Decimal, , Let x be a rational number whose simplest form is, , where p and q are integers, , and q # 0. Then,, (i), , SOLUTION, , Clearly,, , Thus, 0, 1, 2,3, 4, 5,, INTEGERS, , each of the following rational, Without actual division, show that, decimal., Express each in decimal form., numbers is a terminating, , 31, ( 2x55, , as natural, , numbers., WHOLE NUMBERS, , decimal., , 5") thenis a nonterminating repeating, , together, ******, , NATURAL NUMBERS, , x, , x is a, terminating decimal only when q is of the form (2" x 5") for some, non-negative integers m and n., , number is, , =, , (2, , given, , Clearly,, , 50, , =, , x, , 5), , and, , none, , of 2 and 5 is, , rational number is in its, , (2", , (2 x 5, , x, , 5"),, , simplest form., , where, , So, the given number is a terminating, , m, , 66=, , 100, , i ) x isa nonterminating repeating decimal, if q# (2" x 5")., , (iii) The given number, , is, , = 0.66., , 41, , 1000, , =, , 1 and, , decimal., , Now, 50 2 5 5 2 33, xx2, 5 ) 2 66, 5), , (10), , factor of 33., , a, , n, , =, , 2.

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22, , Also,, , Now, 1000 = (8 x 125) = (28 x 5)., , Clearly,, , of 2 and 5 is, , none, , factor of 41., , a, , So, the given number is in its, 1000, , Now,, , (2", , =, , x, , simplest, , form., , number is, , given, , is, , Now, 343, a, , (2, , =, , x3x 5) (2" x 5")., decimal., nonterminating repeating, , a, , 53, (ii) Given numberis43, , (2" X5),, , where m = 3 and n = 3., , S0, the, , 90, , 90, , which is of the form, , 5), , 23, , Real Numbers, , Secondary School Mathematics for Class 10, , 53, , is in its, , 343, , terminating decimal., , and 7 is not, , 7, , Also, 343 = 7, , simplest, , a, , factor of 53., , form., , + (2" x 5")., , 4, And, 100n, 0.041., , 343, (iv) The given number is, , And, 625, , =, , 5* and 5 is, , not, , 625, , =, , of the form (2mx5"), where, , 5' is, , So, the given, , number is, , a, , terminating, , s-, , Now,, , (5 x, , 2), , decimal., nonterminating repeating, =, , 6., , 10190-, , form., So, the given number is in its simplest, , Now,, , a, , 66, (iv) Given number is 80 and HCF(66, 180), , factor of 17., , a, , is, , m, , =, , 0,, , n, , 4., , Now, 30, , decimal., , =, , (2, , x3x, , 5) and, , none, , of 2, 3, 5 is, , a, , factor of 11., , 30 is in its simplest form, Also, 30 = (2x3x 5) # (2"x 5")., , (10), , 2, , 27, 10000=, 0.0272., , 30, , and hence, , 66, 180, , is, , a, , nonterminating repeating, , decimal., , EXAMPLE 2, , Without actual division, show that each, , of the following, , rational, , nontermninating repeating decimal., 66, 121, (iv) 180, (ii) 90 (ii) 53, 343, , numbers is, , a, , (23 x3x 7), sOLUTION, , 15, , The decimal expansion, , 43, 25, , 2, , factor of 121., is in its, , So,, , simplest form., , x 7*) # (2" x 5")., 1s a, , decimal., nonterminating repeating, , ICBSE 2009, , EXAMPLE4, , sOLUTION, , 0.0215., 2 x 2x510 10000, , it will terminate after 4, , places of decimals., , Express each of the following as a, (a) 0.6, , 12175), , (a) Let, , (6) 1.8, , rational number in, , x=0.6. Then,, , 10x 6.666., On subtracting (i) from (ii), we get, , And, 90 = (2 x3*x 5)., , Clearly, n o n e of 2, 3 and 5 is a, , 90, , is in its, , factor of 17., , simplest form., , simplest form., , (c) 0.16, , x = 0.666., , (ii) Given number is 90, , 5, will terminate, , We have:, , 121, (23 x3x 7), , Clearly, none of 2, 3 and 7 is a, So, the given rational number, , of the rational number, , after how many places of decimals?, SOLUTION, , (i) Given number, , And, (23 x 3, , EXAMPLE3, , 9x=6x=8, Hence, 0.6 =, , . i), , (i)

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M, , a, , t, , h, , e, , m, , a, , t, , i, , c, , s, , 25, , for Class 10, , School, , Real Numbers, , Secondary, , (b), , 1.8. Then,, , x=, , Let, , T, , ., , 1.888..., , 10x, , ., , 18.888., , (i), On subtracting, , ), , EXAMPLE 7, , Show that, , primefactors of, , the, say about, , (i), , of the following, , each, , (11),, , from, , we, , get, , (c), , x, , Let, , 1, , =, , nonterminating, , . (i), , 10x= 1.6666..., , (ii), , 16.6666.., , =, , (i), On subtracting, , from, , (ii),, , we, , EXAMPLE 8, , Decide wlhether the, Give, , get, SOLUTION, , Clearly,, , or, , 5, , given, , or, , and, , is, , prime, , the, , terminating, , of its, , factors, , is, , 32.123456789, , decimal., , So,, , denominator are, , a, , and the, it is rational, , other than 2, , number 0.12012001200012..., , and, , a, , both., , number, , repeating, , given, , the, , nonterminating, , 90x15*=, , 2, , support your, , to, , reason, , 23.123456789, , rational, , factors of its, , prime, , (i), , r = 0.1666..., , And, 100x, , the, , (ii) Clearly,, , 0.16. Then,, , =, , (ii) 32.123456789, , number, , (1) Clearly, the given, decimal. So, it is, , sOLUTIONN, , denominator are, , 1.8, , their, , you, , can, , denoninators?, , i ) 23.123456789, , 9=17x=-1, , Hence,, , What, is rational., , numbers, , or, , 5 also., , is rational or, , not., , answer., , number, , is, , 0.12012001200012..., , nonrepeating, , decimal., , So, it is, , a, , noot, , rational., , 0.16, EXERCISE 1C, , EXAMPLE 5, sOLUTION, , Express 0.32, Let, , x, , as a, , fraction, , in simplest form., , rational, , the following, show that each of, 1. Without actual division,, each in decimal form., , 0.32. Then,, , 100r, , On, , numbers is, , (i), , x = 0.3232..., , subtracting (i) from (ii),, , we, , Express, , (iv) 15, 16000, , (iii) 171, S00, , )2x5, , get, , 19, , 17, (v) 320, , 99x 32x=99, , terminating, , 23, , (ii), , 32.3232..., , a, , decimal., , (vi) 3125, , show that each of the, 2. Without actual division,, decimal., numbers is, , Hence, 0.32=, , a, , following, , nonterminating repeating, , 73, 129, 2x3x5)) (23x5x7, (vi)7, (i), , EXAMPLE 6, , Express 0.254, , SOLUTION, , Let, , as a, , fraction, , in, , simplest form, , x = 0.254. Then,, , x =0.2545454.., 10x 2.545454., and, , 1000x = 254.545454., , On subtracting (i) from (ii), we get, , ., , ., , (i), (ii), , (iii), , 3., , Express, , each of the, , 0.254 55, , as a, , fraction in, , simplest, , (ii) 0.24, , (iv) 35, (vii) 455, form., , iv) 0.12, , 0.365, , ANSWERS (EXERCISE1C, , 990x 252x=, 4, , following, (ii) 2.4, , (i) 0.8, , 2.24, 1. (i)0.115, , i) 0.192, , (vi) 0.00608, , rational, , ii) 0.21375, , (iv) 0.009375, , (v) 0.053125

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Secondary School, , Mathematics, , for Class 10, , 27, , Real Numbers, , 181, , (vi) 495, , If, , Type3, , m, , is a positive integer, , which is not, , a, , perfect, , cube then Vm, , is irrational., , HINTS TO sOME SELECTED, , 1., , THEOREM 1, , xTO2=0.21375., PROOF, , 15, , a, , prime, , number and, , Let p be, , a, , prime, , number and, , We know that every, , be, , a, , positive integer. Ifp, , be, , a, , positive integer, , a, , divides a, , a, , such that p, , positive integer, , can, , be, , expressed, , as, , the, , product of primes., , (vi) 147, , 2 ()357, , 32, (3x7), , Let a, , where, , Pn, , =PP2, , P, p 2 , P,, , are, , not, , primes,, , necessarily all, , distinct. Then,, , (vi) 55x7K13, , a=(PP2.. P,)(PiP2, , Pr), , a =(pipp), , ****, , Now, p divides a, , IRRATIONAL NUMBERS, IRRATIONAL NUMBERS, , The numbers which when, , expressed, , in decimal form, as, , p is, , are, , 0.1010010001..., , is, , a, , nonterminating, , prime factor of a, p r i m e factors of a are pi, Pzr, , and, , Thus, (p divides a*), , nonrepeating decimal. So, it is irrational., 0.3030030003., etc., are all irrational., 0.2020020002,, (Gi), (i) 0.12112111211112.. is irrational,, 0.13113111311113... is irrational, and so on., , (iv) 0.232232223... is irrational,, 0.343343334... is irrational, and so on., , If m is a positive integer which is not a perfect square then Vm, is irrational., , Thus, 2, 3, V5, V6, V7, /8, 10, 11, etc., are all irrational., , Using the, , above result,, , THEOREM2, PROOF, , Prove that, , /2, , b, , (p divides a)., , =, , we can, , prove the following., CBSE 2008, 09], , is irrational., , be rational and let its, , If possible, let 2, Then, a and, , Pr, , a=PhP.. Pl, , p dividesa, , is, , irrational., , (i) Clearly,, , a, , p is one ot Pi, P2..P, , irrational, , decimaBs are known, expressible as nonterminating and nonrepeating, numbers., decimal, Note that every nonterminating and nonrepeating, Examples, , Tpe 2., , Let p be, , divides, , (), , Type 1, , all irrational., , then show that p divides a., , =009375, T0, i T 1 0 0053125, 0, xs I, T000 1000.00608., (vi 3125 312 25002, (iv), , are, , sOME RESULTS ON IRRATIONALS, , 0192, , (ii), , etc.,, , n is irrational, while i s rational, , Type 4, , T00.115, , (i, , V2, V3, V5, V6,, , Thus,, , QUESTIONS, , are, , integers having, , simplest form, , no common, , be, , factor other than 1,, , and b # 0., , Now,2-2-, , on squaring both sides], , 2b=a, 2 divides a [, , (i), , 2 divides 2b1

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29, Real Numbers, , 28, , Secondary, , School Mathematics for Class, , 10, , PROOF, , Ifpossible, let, , /5, , and b, , are, , Let a = 2c for some integer c., , Then,, , Putting a =2c in (i),, , and b # 0., , 2b, , get:, , we, , 4c b=2c2, 2 divides, , b, , 2 divides, , 2 is prime and, , Thus, 2 is, , a, , common, , b., factor of a and, , the fact that, , contradicts, , But, this, , and b have, , a, , THEOREM3, , /3, , is, , common, , by assuming, , that, , V2, , factor, , 5b, , and b, , are, , no, integers having, , form be, , simplest, , common, , =, , divides a, , 3, , divides a, , >, , 5 divides, , a]., , =, , factor other than, , : 5 is prime, , 1,, , Thus,, , is, , 5 is, , a, , common, , factor of, , The contradiction arises, , i), , Hence, 5, , a, , b>5 divides b]., , and 5 divides, , and b., a, , and b have, , no common, , factor, , and 3 divides a * 3 divides, , a]., PROOF, , integer c., (i), we get:, , by assuming, , that, , v5, , is rational., , is irrational., , Prove that 11 is irrational., , THEOREM5, , prime, , 5 divides 5], , other than 1., , 3 divides 3b], , 3c in, , If, , possible,, , Then,, , let, , 11, , be rational and let its, , simplest, , form be, , =3c2, 3, , [ 3 is prime and, a common, , 3c2, , 3 divides, , 3 divides, , b 3, , divides, , b]., , a, , and b have, , no common, , integers having, , no c o m m o n, , factor other than 1,, , by assuming, , 11-, , factor, , lon squaring both sides], , 11b2 =a2, , ... (i), , :11 divides 111, , 1 1 divides a, , 1 1 is prime and 11 divides a, , other than 1., , is irrational., , are, , 11 divides a, , factor ofa and b., , The contradiction arises, , and b, , Now, 11, , divides b, , this contradicts the fact that, , a, , and b # 0., , 3 divides b [, , Hence, 3, , a, , some, , 3b2 9c2, , But,, , and 5 divides, , 5 dividesb, , .., , a, , [:3, , Thus, 3 is, , prime, , 5c2, 5 divides b [, , b, , a, , lon squaring both sides], , 3, , 3 c for, a, , 5 divides 5b], , But, this contradicts the fact that, , 3b, , Putting, , 25c2, , [CBSE 2008, '09C, , rational and let its, , Now, v3-3-, , a, , . (i), , Putting a = 5c in (i), we get:, , and b # 0., , Let, , 5 is, , :, , is rational., , irrational., , If possible, let V3 be, a, , no, , is irrational., , Prove that, , Then,, , on squaring both sides], , 5b2 =a2, 5 divides a, , b]., , factor other than 1,, , Let a = 5c for some integer c., arises, , contradiction, , Hence, 2, , no c o m m o n, , integers having, , 5 divides a, , other than 1., , The, , divides, , 2, , b, , a, , form be, be rational and let its simplest, , Now, V5 5 -, , 2c1, , 2divides, , :, , 2 dividesb, , PROOF, , 2divides b]., , primeand divides b, , T 2 is, , CBSE 2008, '09], , Prove that V5 is irrational., , THEOREM4, , 2 divides a, , that V3 is rational., , Let a 1lc for some positive integer c., , 11 divides a].

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10, Secondary School Mathematics for Class, , P'utting a = 1lc in (i), we get:, , 11b= 121c, , THEOREM7 f a is, , = 11c?, , 11 divides b, 1l divides b, is prime and, b., factor of a and, , 11 divides, , But, this, , a, , common, , the fact that, , contradicts, , other than1., , Thecontradiction, , arises, , no, , and b have, , a, , by assuming, , common, , b], , THEOREM6, PROOF, , fp is prime, , Let p be, , a, , prime, , I(a+/b)-al is rational, , number and it, , having, , no c o m m o n, , vp, , is, , let, , possible,, , where, , vp =, , factor other than, , 1, and, , irrational., , to, , prove that, , by assuming, , that (a, , +, , /b) is, , rational., , m, , and, , n, , # 0., , 1, , are, , integers, , THEOREM 8, PROOF, , both, , fa, , is, , a nonzero, , rational and, , Vb, , is irrational then show that, , a, , Vb is, , irrational., Let a be a nonzero rational and let /b be irrational., , Then,, , sides], , we, , have, , to, , show that, , a/b, , is irrational., , If possible, let avb be rational., .. i), p divides, , [, , p divides m, , Then,, , p], , aVb=where, , p and q, , are n o n z e r o, , integers, having, , no, , common factor other than 1., , divides m, , [pis primeand p, , divides m*, , p, , divides, , Now,a/b-V-, , m]., , Let m = pg for some integer q., , But, p and ag, , Putting m = pg in (i), we get:, , p divides *, p divides n, p is, a common, , m, , and, , But, this contradicts the fact that, , other than1., The contradiction arises, , Hence, p is irrational., , p divides, , [, , m, , Thus, from (i),, , pa], , prime and p divides * p, , factor of, , are, , both rational and aq #0., , aq 1S rational., , pn=pq n*=pq*, , Thus, p is, , have, , difference of rationals is rational], , [, , be rational., , pn*= n, , p, , we, , Hence,(a +Vb) is irrational., Vp, , lon squaring, , Then, p=uP=, , is, , Vb is rational., This contradicts the fact that /b is irrational., , that 11 is rational., , that, , Vb), , +, , (at b) is rational and a is rational, , The contradiction arises, , nunber then prove, , be, Let its simplest form, , a, , factor, , irrational., Hence, 11 is, a, , is irrational then prove that (a, , b, , be rational and b be irrational. Then,, (a+ /b) is irrational., If possible, let (a + Vb) be rational. Then,, Let, , PROOF, , I1l, Thus, 11 is, , rational and, , rrational., , I Idivides 11c], , 11 divides b, , 31, , Real Numbers, , follows that /b is rational., , This contradicts the fact that Vb is irrational., divides, , The contradiction arises by assuming that a/b is rational., , n]., , Hence, ab is irrational., , n., , and, , it, , n, , have, , no common, , factor, , sOLVED, , by assuming that p, , is rational., , EXAMPLES, , EXAMPLE 1, , Show that (2+/3) is an irrational number., , sOLUTION, , If possible, let (2 + /3) be rational. Then,, , [CBSE 2008C. '09]

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Secondary School Mathematics for Class, , (2+ 3), , is rational, 2 is rational, , I(2+V3) -2) is rational, [, 3, , x3/5) is rational, product of two, , rationals is, difference of, , sOLUTION, , 5, , is rational., , by assuming, , contradiction, , Hence,, , EXAMPLE2, , rational], , (2, , Show that, , +, , 3), , Ifpossible, let, , rational., , CBSE 2008], , is irrational., , (4-V3), , 4 is rational,, , is rational., , The contradiction arises, , /3) is, , be rational. Then,, , (4-/3), , EXAMPLE5, , Showthat, , SOLUTION, , We have, , 2, , If, is, difference ofrationals rational], , The, , Hence,, , EXAMPLE3, SOLUTION, , the fact that, , contradiction arises, , (4-V3), , v3 is, , possible, let, , by assuming, , 23, , is, , rational, so, , Then,, , EXAMPLE6, , ( x2/3) is rational, , sOLUTIONN, , product of, , 3, , two, , rationals is, , rational], , is rational., , This contradicts the fact that, , v2, , v3, , is rational., , is rational, , [, , is rational, , product of rationals is rational], , by assuming, Hence, 2/3 is irrational., , EXAMPLE 4, , Show that 3/5 is irrational., , SOLUTION, , If possible, let 3/5 be rational. Then,, , that 2/3 is rational., , arises by assuming, , that, , 2, , is, , is irrational., , Prove that (3+5/2) is irrational., , CBSE 2009], , If possible, let (3 +52) be rational. Then,, , (3+5 2) is rational, 3 is rational, 3+5v2) -3] is rational, , is irrational., , The contradiction arises, , 3/5 is rational,is rational, , 7 V2, , This contradicts the fact that v2 is irrational., , is rational, , rational,, , be rational. Then, from (i),, , Since, the contradiction, , rational., , (1), , v 2 is rational., , that (4-v3) is rational., , is irrational., , possible, let 2/3 be, , V2, , (2x/2), , irrational., , Show that 2/3 is irrational., If, , a, , Now, 2 is rational,, , 3 is rational., contradicts, , that 3/5 is rational., , is irrational., , is rational, , 14-(4-3)) is rational, , This, , by assuming, , Hence, 3/5 is irrational., , is irrational., , (4-3), , rational], , rationals is, , This contradicts the fact that V5 is irrational., , v3 is irrational., This contradicts the fact that, that (2 +, arises, The, , 33, , Real Numbers, , 10, , difference of two rationals is rational], , 5 / 2 is rational, , x52) isrational, :, , product of two rationals is rational], , 2 is rational., This contradicts the fact that v2 is irrational.

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10, , Secondary School Mathematics for Class, , by assuming, rational, hence (3 +5/2) is irrational., , Since the contradiction arises, , EXAMPLE7, SOLUTION, , (5-2 3) is rational and 5, I(5-2V3)-5} is rational, , -2, , 1. Define, , 2., , oftwo, , rationals is, , rational], , sOLUTION, , Let, , us, , product of two, , rationals is, , rational], , Let, , (2, , t, , V3), , =, , v2, , (ii) 3.1416, , or, , numbers., , irrational, (iv) 3.142857, , (ii) t, , (vi) 2.040040004. (vii) 1.535335333 ., (x)V3, , 3. Prove that each of the following numbers is irrational., , is irrational., , V3, that (5-2 3 ) is rational., assuming, by, , (v) (5+3/2), , is irrational., , suppose that, , (ii) irrational numbers (iii) real, , 2-3), , CBSE 2008], , CBSE 2009), , (iv) (2+5), , cBSE 2008C], , (CBSE 2008, , (vi) 3/7, , (i) 6, , The contradiction arises, , (V2+ 3), , numbers, , (viii) 3.121221222.. (ix) /21, , This contradicts the fact that, , Prove that, , 22, , (v) 5.636363 ., , 3 is rational., , EXAMPLE8, , (i) rational, , Classify the following numbers as rational, (), , 3 is rational, , (5-2/3), , is a rational number lying between, , 10, , EXERCISE 1D, , is rational, , -2-2/3) is rational, , Hence,, , =, , (CBSE 2008, , difference, , :, , 1.5, , and 3., , be rational. Then,, , possible, let (5-2/3), , I, , Clearly,, , that (3*5v2) is, , Show that (5-2/3) is irrational., If, , 35, , Real Numbers, , via2-3W5), , (vi), , CBSE 2010], , is irrational., , (V2, , /3), , +, , a, where, , a, , (ix) (3+ V5), , is rational., , 4., , is rational., , Prove thatisirrational., , .(1), , Then, v2 = (a -/3)., , HINT, On squaring both sides of (i), we get, , 5. (i) Give an example of two irrationals whose sum is rational, , 2 a+3-2ay32a3 =a*+1., (ii), , 3-, , (ii) Give, HINT, , an, , example of, , two, , irrationals whose, , product, , is rational., , () Take (2 + V3) and (2-V3)., , This is impossible, as the right-hand side is rational, while v3, , (i) Take (3+ 2) and (3-V2)., , is irrational., , 6. State whether the given statement is true or false., , This is a contradiction., , Since the contradiction arises by assuming that (/2+/3) is, rational, hence (2 +3) is irrational., EXAMPLE9, , Write a rational number betveen 2, , sOLUTION, , We have, 2, , = 1.414.. and 3, , and V3., , i), , The sum of two rationals is, , always rational., , (ii) The product of two rationals is always rational., (i) The sum of two irrationals is always an irrational., , CBSE 2008], , iv) The product of two irrationals is always an irrational., , (v) The sum of a rational and an irrational is irrational., = 1.732., , .., , (vi) The product of a rational and an irrational is irrational.

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Secondary School Mathematics for Class, , 10, , 37, , Real Numbers, , 7. Prove that (2/3-1) is an irrational number., , ICBSE 2010], , 8. Prove that, , (4-5 2) is an irrational number, , ICBSE 2010], , 9. Prove that (5-2/3) is an irrational number., , ICBSE 2010], , 12. Show that there is no value of n for which (2" x 5") ends in 5., 13. Is it possible to have two numbers whose HCF is 25 and LCM is 520?, 14. Give an example of two irrationals whose sum is rational., , 15. Give an example of two irrationals whose product is rational., , 16. If a and b are relatively prime, what is their LCM?, , 10. Prove that 5/2, is irrational., , 17. The LCM of two numbers is 1200. Show that the HCF of these numbers, 11., , Provethatis irrational., , cannot be 500., , Why?, , Short-Answer Questions, 18. Express 0.4 as a rational number in simplest form., , HINT, , 19. Express 0.23 as a rational number in simplest form., , ANSWERS (EXERCISE1D), 2., , 6., , (i) rational, , (ii) rational, , (ii) irrational, , (iv) rational, , (v) rational, , (vi) irrational, , (vii) irrational, , (vii) irrational, , (ix) irrational, , (x) irrational, , (i) True, , (ii) True, , (i) False, , 20) Explain why 0.15015001500015.. is, 2Show that, , 2, , an, , is irrational., , 22. Write a rational number between 3, (iv) False, , (v), , True, , (vi) True, , irrational number., , and 2., , Explain why 3.1416 is a rational number., , 2, , ANSWERS (EXERCISE 1E), , EXERCISE 1E, , 3. (23 x3x5) 4. 1, , Very-Short-Answer Questions, , 9. b, , 1. State Euclid's division lemma., , 10. 6, , 5. ab, , 8. 1, , 6. 42, , (2"x 5"), where m and n are some nonnegative integers, 13. No, , 11. 0.0365, , 17. since 500 is not, , 16. ab, , a, , factor of 1200, , 2. State fundamental theorem of arithmetic., , 18.19.9 22.1.8, , 3. Express 360 as product of its prime factors., , 4. Ifa and b are two prime numbers then find HCF{a, b)., , HINTS TO SOME SELECTED QUESTIONS, , 5. Ifa and b are two prime numbers then find LCM(a, b)., 6. If the product of two numbers is 1050 and their HCF is 25, find their, , 7. A number having at least 3 factors is called a composite number., , 3X5, , 365, , 365, , 11 (555x5 1365 100, , LCM., , 7. What is a composite number?, , 00365., , 12. (2x 5'") = (2 x 5)" = 10", which always ends in a zero., , 8. Ifa and bare relatively prime then what is their HCF?, 13. HCF always divides the LCM, 9., , If the rational number, , has, , a, , terminating decimal expansion, what is, , the condition to be satisfied by b?, 10. Simplify:, , 11. Write the, , CBSE 2010, , 2/5, , 15. (3+ 2) and (3-2)., 22., , Clearly, /3, 3, , expansion of, , completely., , (2+3)and (2-/3), , 20. Given number is nonterminating and nonrepeating decimal., , (2/45+320), decimal, , ICBSE 2008), , 14., , 2x 5, , ICBSE 2009], , 23., , =, , 1.732..., , . So, we may, , take 1.8, , as, , the, , required, , rational, , number between, , and 2., , Clearly, it is a nonterminating repeating decimal., *****************.

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Real Number, , Secondary School Mathematics for Class 10, , 12. A number when divided by 143 leaves 31 as remainder. What will be, the remainder when the same number is divided, 13?, , MULTIPLE-CHOICE QUESTIONS (MCa), Choose the correct, , .Which of the, , a, , =(2x 3, , 5') and b (2'x, =, , HCF of, , 33, , x, , (2'x3*, , x, , (b), , (a) 30, 4. LCM of (2', , 5),(22, , x, , 3, , x, , x, , (2', , 5, , (a) 40, , x, , 7) is, one, , ot the, , numbers is 54, what is the other number?, , (b) 45, , 6., , The, , product, , (c), , following, , is, , an, , (d), , 9, , 81, , is 5. The LCM of the, of two numbers is 1600 and their HCF, , (a)22, , (b) 3.1416, , (c) 3.14116, , (d) 3.141141114 ., , (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , 16. 2.13113111311113.. is, (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , numbersis, (b) 1600, , (a) 8000, , 7. What is the, , largest, , (d) 1605, , (c) 320, , number that divides each, , one, , of 1152 and 1664, , exactly?, (b) 64, , (a) 32, 8. What is the, , (c) 128, , (d) 256, , remainders, largest number that divides 70 and 125, leaving, , 5 and 8 respectively?, ) 9, , (a) 13, 9. What is the, , (c) 3, , largest number that divides 245 and 1029,, , (d) 585, , leaving remainder, , 5 in each case?, , (a), , 15, , (b), , 16, , (d))5, , irrational number?, , 15. 2.35 is, , (d) 1680, , (c) 1120, , their LCM is 162. If, 5. The HCF of two numbers is 27 and, , (a) 36, , of the, , (c) 3, , is, , (d) 105, , (c) 60, x, , Which, , (b) 1, , 14. t is, , 3*x 5) and (2x3x5'x 7), , (b) 560, , ?, , =, , (d) 540, , (c) 360, , 48, , 5) and, , (d) (32,62), , 5) then 1ICF(a, b), , (b) 180, , (a) 90, , 3., , 13., , (c) (31,93), , (b) (18, 25), x, , by, , a)0, , ofthe following questions:, , tollowing is a pair of co-primes?, , (a) (14, 35), 2. If, , in each, , answer, , 39, , ) 9, , (d) 5, , 1095 15, 10. The simplest form ot 168, , 17. The number 3.24636363 ... is, (a) an integer, , (b) a rational number, , (c) an irrational number, , (d) none of these, , 18. Which of the following rational numbers is expressible as a terminating, decimal?, , a)165, , b), , (d) 1625, 462, , (c), , 19. The decimal expansion of the rational number, , will terminate, , after, (a) one decimal place, , (b) two decimal places, , (c) three decimal places, , (d) four decimal places, , 14753, 20. The decimal expansion of the number5will, terminate after, 1250, , (a)26, , (b), , (c)6, , d)1, , 11. Euclid's division lemma states that for any positive integers a and b,, there exist unique integers q and r such that a = bq + r, where r must, , satisfy, (a) 1 <r<b, , (b) 0<rsb, , (c) 0Sr<<b, , (d) 0<r<b, , (a) one decimal place, , (b) two decimal places, , (c) three decimal places, , (d) four decimal places, , 21. The number 1.732 is, , (a) an irrational number, , (b) a rational number, , (c) an integer, , (d) a whole number

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Real Numbers, , Secondary School Mathematics for Class 10, , A number when divided by 143 leaves 31 as remainder., What will be, the remainder when the same number is divided by 13?, , MULTIPLE-CHOICE QUESTIONS (MCQ), , (a), Choose tie correct answer in each of the following questions:, , 13., , (b) 1, , 0, , )3, , is, , Which of the following, , an, , 1. Which of the following is a pair of co-primes?, , 2. If, , a, , =, , (2, , x, , 3', , x, , 5i) and b (2' x3, , (2 3 5),(22, x, , x, , 3*, , 5), , x, , and, , (2'x, , 3, , x, , 5), , and, , (2', , x, , 5, , x, , (2x3 x5'x 7), , (d) 105, , (d) 1680, , numbers, , 6., , The, , product, , (c), , one, , (c), , largest, , number that divides each, , one, , exactly?, 8. What is the, , largest number that divides 70, (c), , (b) 9, , and 125,, , leaving, , leaving remainder, (d), , (c) 9, , 5, , unique integers, , (c)16, q and, , r, , such that, , a, , =, , bqtr,, , where, , r, , must, , (b) 0<rsb, , 19. The decimal, , rational numbers is, , expressible as a terminating, , 2027, c)25, , (b), , expansion, , of the rational number, , (d) 462, will, , terminate, , after, , (a) one decimal place, , (b) two decimal places, , (c) three decimal places, , (d) four decimal places, , (a) one decimal place, , (b) two decimal places, , (c) three decimal places, , (d) four decimal places, , 21. The number 1.732 is, , (a), , satisfy, (a) 1<r<b, , (d) none of these, , 4753 will terminate after, 20. The decimal expansion of the number950, , b5, , 11. Euclid's division lenmma states that for any positive integers a and b,, there exist, , (c) an, Which of the following, , (d) 585, , 3, , 1095 is, , (b), , (d) none of these, (b) a rational number, , irrational number, , (a)165, , 10. The simplest form of i168 is, , (a)26, , (b) a rational number, , irrational number, , remainders, , 5 in each case?, (b) 16, , (d) none of these, , decimal?, , 9. What is the largestnumber that divides 245 and 1029,, , (a) 15, , an, , (a) an integer, , 18., , 5 and 8 respectively?, , (a) 13, , (c), , of 1152 and 1664, , (d) 256, , (c) 128, , (b) 64, , (a))32, , (b) a rational number, number, , is, 17. The number 3.24636363..., , (d) 1605, , (c) 320, , an irrational, , (a) an integer, , numbers is, , 7. What is the, , (d) none of these, , 16. 2.13113111311113... is, , (d) 81, , (c) 9, , (b) 1600, , irrational number, , (a) an integer, ot the, , LCM of the, their HCF is 5. The, of two numbers is 1600 and, , (a) S000, , an, , 15. 2.35 is, (c) 1120, , (b) 45, , (b) a rational number, , (a) an integer, , is, , LCM is 162. If, numbers is 27 and their, 5. The HCF of two, is 54, what is the other number?, , (a) 36, , 14. T is, , 7) is, , (b) 560, , (a) 40, , ?, , =, , 5, , (d) 3.141141114, , (c) 3.1416, , (d) 540, , (c) 60, , (b) 48, , (a) 30, 4. LCM of, , x, , b), , (c) 360, , (b) 180, , (a) 90, 3. HCF of, , 5) then HCF(a,, , x, , =, , (d), , irrational number?, , (b) 3.1416, , (d) (32, 62), , (c) (31, 93), , (b) (18, 25), , (a) (14, 35), , 39, , (c) 0sr<b, , (d) 0<r<b, , an, , irrational number, , (c) an integer, , (b) a rational number, (d) a whole number

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Real Numbers, , Secondary School Mathematics for Class 10, 22a and b are two positive integers such that the least prime factor of a, Is 3 and the least prime factor of b is 5. Then, the least prime factor of, , (a+b) is, , product of two numbers, theirHCF, , =, , (b) 3, , (c), , (d) 8, , 5, , 9., , is, , -, , 320., , Required, , number, , =, , HCFI(245-5), (1029 -5)1, , =, , HCF(240, 1024), , 1168, , 11. On dividing a by b, let q be the quotient and r be the remainder., , (c) a terminating decimal, , Then, we have:, , (d) a nonterminating repeating decimal, , a= bq + r, where 0 S r < b., , 12. Let the given number when divided by 143 give q as quotient and 31 as remainder., , is, , Then, number = 143q+31 = (13 x 11g+ 13*2+51 = 13 z (114+2)+5., , rational number, , (a) a fraction, , (b), , a, , (c) an irrational number, , (d), , none, , So, the same number when divided by 13 gives 5 as remainder, , of these, , 13. 3.141141114 ... isa nonterminating, nonrepeating decimal. So, it is irrational., , 25. (2+ 2) is, (b), , (a) an integer, , (d), , (c) an irrational number, , a, , 14. n is an irrational number., , rational number, , none, , 26. What is the least number that is divisible, , by, , 15. 2.35 =2.353535, , of these, , 16. 2.1311311131113, , . ., , is, , a, , nonterminating, nonrepeating decimal. So,, , 17. The number 3.24636363.. is a nonterminating repeating decimal., , (d) 5040, , (c) 2520, , (b) 1260, , (a) 100, , which is a repeating decimal., , 2.35 is rational., , all the natural numbers, , from 1 to 10 (both inclusive)?, , So, it is a rational number., , ANSWERs (MCQ), 1. (b), , 2. (b), , 10. (d), , 11. (c), 20. (d), , . (c), , 12. (d), 21. (b), , 18., , 4. (d), , 5. (d), , 6. (c), , 7. (c), , 13. (d), , 14. (c), , 15. (b), , 16. (c), , 22. (a), , 23. (b), , 24. (c), , 25. (c), , 8. (a), 17. (b), 26. (c), , 9. (b), , 2027, 625, , 2027, , (5 x 20, , So, it is expressible as a terminating decimal., , 18. (c), , 185, , 19., , 185, -, , 1.85., , (10)2100 = 1.85., , So, it will terminate after 2 decimal places., , HINTS TO sOME SELECTED QUESTIONS(MCQ), 2. HCF(a, b) = product of common terms with lowest power, , = (2 x3 x 5) = (4 x9x5) = 180., 3. HCF, , 20 1, , 14753x8 14753 x8, (10), , So, it will terminate after 4 decimal places., , 21. 1.732=000which is a rational number., , product of common terms with lowest power, = (2 x 3 x 5) = (4 x3x5) = 60., , 4. LCM, , 16., , 1095, , (a) a rational number, , 19. (b), , =, , 10. 1ICFH1095, 1168) = 73., , (b) an irrational number, , 24, , 1600, , 7. Required number = tHCF(1152, 1664) = 128., , 8. Required number =11CFI(70-5), (125-8)) = HCF(65, 117) = 13., , (a) 2, 23. 2, , 6 . LCM, , 41, , product of prime factors with highest power, (2x3x 5x 7) = (16 * 3x5 x 7) = 1680., , 22. Clearly, 2 is neither a factor of a nor that of b., , a and b are both odd., , Hence, (a +b) is even., least prinme factor of (a + b) is 2., , 5. Other number, , HCF XLCM= 27l6= 81., given number, , it is, , irrational.

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42, , Real Numbers, , Secondary School Mathematics for Class 10, , 3, , 7.Let x=be a rational number such that q # (2", , Here,, , is, , And, the, , rational and, , product, , of, , a, , 2, , is, , irrational., , rational and, , an, , irrational, , is, , 5") then x hasa, , Nonterminating repeating decimal expansion., , irrational., , 8.A nunnber wlhich can be expressed, decimal is an irrational number., , 2 and henceis irrational., , 2, 3, V5, V6, V7, V8, V10,..,, , 25. The sum of a rational and an irrational is irrational., , as a, , nonterminating, , T, e, etc., are, , all, , and, , nonrepeating, , irrational numbers., , Here, 2 is rational and v2 is irrational., (2+2) is irrational., 26., , FORMATIVE ASSESSMENT (UNIT TEST), , Required number LCM|1, 2,3, 4, 5, 6,7, 8,9, 10, =, , = LCM|1, 2, 3,2,5, 2 x 3, 7, 2,3,2 x 51, (1, , x2x3x5 x7) (8 x9 x5x 7), =, , =, , 50 Marks, MCQ (1 mark), , 2520., * * * * * ' * * *, , ., , ., , * * * * * * * * * * ' *, , 71 is, 1. The decimal representation of 50, , *****, , (a) a terminating decimal, , (b) a nonterminating, repeating decimal, , sUMMARY OF RESULTS, , (c) a nonterminating and nonrepeating decimal, 1. Euclid's Division Lemma, , (d) none of these, , Given positive integers a and b, there exist whole numbers q and r satisfying, a, , =, , 2. Which of the, , bq +r,wvhere 0r<b., , 2. Euclid's Division Algorithm, Step 1., , (a) 1, =, , b. If, , r *0, apply, , Euclid's division lemma to, , b and r., , 4., , Every composite number can be expressed as a product of primes, and this, factorization is unigue, apart from the order in which the prime factors occur, 4.Ifp is prime and p divides a then p divides a, where a is a positive integer., , lfwe put, , it in the, , number which, , can, , (d) 4, , (b) 1.42, , (c) 0.141, , (d) None of these, , digit 0., , 6. The HCF of two numbers is 27 and their LCM is 162. If one of the number, is 81, find the other., , are, , be, , simplest formthen, , expressed, q, , is a terminating decimal., , irrationals., 8. Find the, , 6. Let x be a rational, , (c) 3, , Short-Answer Questions (2 marks), , 7. Examine whether, etc.,, , (b) 2, , 5. Show that any number of the form 4",n EN can never end with the, , 3. The Fundamental Theorem of Arithmetic, , 2, V3, /5, 6, V7,, , (d), , 0.68+0.73 = ?, , (a) 1.41, , Step 3. Continue the process till the remainder is zero. The divisor at this, stage is HCF(a, b)., , 5. To prove that, , (c)5, , be the remainder if (31-1) is divided by 9?, , where 0 r<b., , Step 2., , 80, , 3. On dividing a positive integer n by 9, we get 7 as remainder. What will, , Apply the division lemma to find q and r such that a = bq + r,, , If r=0 then HCF, , (b)9, , (a), , According to this, we find the HCF of two, , positive integer a and b with a >b in following steps., , following has a terminating decimal expansion?, , =(2", , x, , as a, , decimal., , terminating, some non-negative, , 5") for, , simplest, , form of, , 148, , 185, , 9. Which of the following numbers are irrational1?, , (a) 2, , (b) V6, , (e), , (2, , integers m and n., , (c) 3.142857, , (d) 2.3, , (g)0.232332333... (h) 5.2741

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44, , Secondary School Mathematics for Class 10, , 10. Prove that, , (2+ V3) is irrational., Short-Answer Questions (3 marks), 11. Find the HCF and LCM of 12, 15, 18, 27., 12. Give an example of two irrationals whose sum is rationa, , 13. Give prime factorization of 4620., 14. Find the HCF of 1008 and 1080, , 15. Find the HCF and LCM of, 16. Find, , by prime factorization, , method., , 10 and g1, g 77, , the largest number which divides 546 and 764, leaving, , remainders, , 6 and 8 respectively., , Long-Answer Questionms (4 marks), 17. Prove that 3, , is an irrational number., , 18. Show that every, , form (4q +1), positive odd integer is of the, , or, , (44, , +, , 3), , for some integer 9., 19. Show that, , one, , and, , is divisible by 3,, +, only one out of n, (11+ 2) and (n 4), , where n is any positive integer., 20. Show that (4 +3/2) is irrational., , ANSWERS (UNIT TEST), 1., , (b), , 2., , (b), , 3., , (b), , 4., , (b), , 9. 2,3/6, r, 0.232332333..., 12. (2+/3) and (2- /3), 14. 72, , 6. 54, , 7. No, , 8., , 11. HCF = 3, LCM = 540, , 13. (2 x 3 x 5 x7X 11), , 15.HCF= RLCM, , 16. 108