Page 1 REAL EUCLIDâ€™S DIVISION ALGORITHM 1) Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 Solution: STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + STEP 2: Since the remainder 144 0 1 3 6 2 1 4 4 x 9 + 6 6 STEP 3: Since the remainder 1 4 4 6 6 x 2 + 1 2 = STEP 4: Since the remainder 6 6 1 2 x 5 + 6 = STEP 5: Since the remainder 1 2 6 x 2 + 0 = The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. 2) Show that any positive odd integer is of the form Solution: Letâ€™s start with taking a, where a is a positive odd integer. We apply the division algorithm with and b =8 Since 0 8 r = < the possible remainders are That is, a can be 8q or 8q+1 or 8q+2 or 8 However, since a is odd a cannot be 8 Therefore any odd integer can only be of the form 3) A fruit seller has 56 red apples and has the same number, and they take up the least area of the tray. that can be placed in each stack for this purpose? Solution: To find the solution we have to first find the HCF of 56 and 32 maximum number of apples in each stack take up least amount of space. We use Euclidâ€™s division algorithm to find the HCF of 56 and 32 56 = 32 x 1 + 24 Since, 24 0 ? ? 32 = 24 x 1 + 8 8 0 ? 24 = 8 x 3 + 0 So the HCF of 56 and 32 is 8. Therefore the fruit seller has to make s REAL NUMBERS S DIVISION ALGORITHM Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + 144 0 ? we apply the division lemma to 1362 and 144 1 3 6 2 1 4 4 x 9 + 6 6 = Since the remainder 66 0 ? we apply the division lemma to 144 and 66 1 4 4 6 6 x 2 + 1 2 = Since the remainder 12 0 ? we apply the division lemma to 66 and 12 6 6 1 2 x 5 + 6 Since the remainder 6 0 ? we apply the division lemma to 12 and 6 1 2 6 x 2 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7 is a positive odd integer. We apply the division algorithm with the possible remainders are 0,1,2,3,4,5,6 and 7 +2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or 32 green apples. She wants to stack them in such a way has the same number, and they take up the least area of the tray. What is the maximum number of that can be placed in each stack for this purpose? irst find the HCF of 56 and 32. Then this number will give us the maximum number of apples in each stack and as a result the number of stacks will be least and they will hm to find the HCF of 56 and 32 32 x 1 + 24 24 x 1 + 8 + 0 Therefore the fruit seller has to make stacks of 8 for both kinds of apples. : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, we apply the division lemma to 1362 and 144 e apply the division lemma to 144 and 66 we apply the division lemma to 66 and 12 we apply the division lemma to 12 and 6 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the +7, where q is some integer is a positive odd integer. We apply the division algorithm with a +7 where q is the quotient. (since they are divisible by 2) , or 8 7 q+ in such a way that each stack What is the maximum number of apples Then this number will give us the and as a result the number of stacks will be least and they will Page 2 REAL EUCLIDâ€™S DIVISION ALGORITHM 1) Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 Solution: STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + STEP 2: Since the remainder 144 0 1 3 6 2 1 4 4 x 9 + 6 6 STEP 3: Since the remainder 1 4 4 6 6 x 2 + 1 2 = STEP 4: Since the remainder 6 6 1 2 x 5 + 6 = STEP 5: Since the remainder 1 2 6 x 2 + 0 = The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. 2) Show that any positive odd integer is of the form Solution: Letâ€™s start with taking a, where a is a positive odd integer. We apply the division algorithm with and b =8 Since 0 8 r = < the possible remainders are That is, a can be 8q or 8q+1 or 8q+2 or 8 However, since a is odd a cannot be 8 Therefore any odd integer can only be of the form 3) A fruit seller has 56 red apples and has the same number, and they take up the least area of the tray. that can be placed in each stack for this purpose? Solution: To find the solution we have to first find the HCF of 56 and 32 maximum number of apples in each stack take up least amount of space. We use Euclidâ€™s division algorithm to find the HCF of 56 and 32 56 = 32 x 1 + 24 Since, 24 0 ? ? 32 = 24 x 1 + 8 8 0 ? 24 = 8 x 3 + 0 So the HCF of 56 and 32 is 8. Therefore the fruit seller has to make s REAL NUMBERS S DIVISION ALGORITHM Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + 144 0 ? we apply the division lemma to 1362 and 144 1 3 6 2 1 4 4 x 9 + 6 6 = Since the remainder 66 0 ? we apply the division lemma to 144 and 66 1 4 4 6 6 x 2 + 1 2 = Since the remainder 12 0 ? we apply the division lemma to 66 and 12 6 6 1 2 x 5 + 6 Since the remainder 6 0 ? we apply the division lemma to 12 and 6 1 2 6 x 2 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7 is a positive odd integer. We apply the division algorithm with the possible remainders are 0,1,2,3,4,5,6 and 7 +2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or 32 green apples. She wants to stack them in such a way has the same number, and they take up the least area of the tray. What is the maximum number of that can be placed in each stack for this purpose? irst find the HCF of 56 and 32. Then this number will give us the maximum number of apples in each stack and as a result the number of stacks will be least and they will hm to find the HCF of 56 and 32 32 x 1 + 24 24 x 1 + 8 + 0 Therefore the fruit seller has to make stacks of 8 for both kinds of apples. : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, we apply the division lemma to 1362 and 144 e apply the division lemma to 144 and 66 we apply the division lemma to 66 and 12 we apply the division lemma to 12 and 6 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the +7, where q is some integer is a positive odd integer. We apply the division algorithm with a +7 where q is the quotient. (since they are divisible by 2) , or 8 7 q+ in such a way that each stack What is the maximum number of apples Then this number will give us the and as a result the number of stacks will be least and they will REAL NUMBERS EUCLIDâ€™S DIVISION ALGORITHM 4) Prove that one of every four consecutive integers is divisible by 4 Solution: Let the four consecutive integers be n,n Consider a = bq + r, where 0= < Let a=n and b=4. Hence r can have values 0 CASE 1: Let r=0 ? 4 n q = CASE 2: Let r=1 ? 4 n q = CASE 3: Let r=2 ? 4 n q = CASE 4: Let r=3 ? 4 n q = +3 then Thus in every case one of the four numbe : 5) Use Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 for some integer m. Solution: Let n be any positive integer. Consider Let a=n and b=3. Hence r can have values 0,1 or 2. CASE 1: Let r=0 then n =3q ? = CASE 2: Let r=1 then 3 1 n q = + CASE 2: Let r=2 then 3 2 n q = + Therefore 2 n is always of the form REAL NUMBERS S DIVISION ALGORITHM Prove that one of every four consecutive integers is divisible by 4. Let the four consecutive integers be n,n+1,n+2 and n+3. 0 r b = < can have values 0, 1, 2 or 3. n q which is divisible by 4 : n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4 n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = + n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = + Thus in every case one of the four numbers is divisible by 4. Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 Let n be any positive integer. Consider a=bq+r where 0 r b = < can have values 0,1 or 2. 2 2 (3 ) n q ? = 2 9q = 2 3(3 ) q = 3m = where m = 3q 3 1 n q = + 2 2 (3 1) n q ? = + 2 9 6 1 q q = + + 2 3(3 2 ) 1 q q = + + 3 1 m = + where 2 3 2 m q q = + 3 2 = + 2 2 (3 2) n q ? = + 2 9 12 4 q q = + + 2 2 9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + + 3 1 m = + where is always of the form 3m or 3m+1 REAL NUMBERS 4 1 3 4 4 4( 1) which is divisible by 4 : 4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 : 4 3 1 4 4 4( 1) which is divisible by 4 Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 r b = < m = 3q 2 9 12 3 1 3(3 4 1) 1 = + + + = + + + where 2 3 4 1 m q q = + + Page 3 REAL EUCLIDâ€™S DIVISION ALGORITHM 1) Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 Solution: STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + STEP 2: Since the remainder 144 0 1 3 6 2 1 4 4 x 9 + 6 6 STEP 3: Since the remainder 1 4 4 6 6 x 2 + 1 2 = STEP 4: Since the remainder 6 6 1 2 x 5 + 6 = STEP 5: Since the remainder 1 2 6 x 2 + 0 = The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. 2) Show that any positive odd integer is of the form Solution: Letâ€™s start with taking a, where a is a positive odd integer. We apply the division algorithm with and b =8 Since 0 8 r = < the possible remainders are That is, a can be 8q or 8q+1 or 8q+2 or 8 However, since a is odd a cannot be 8 Therefore any odd integer can only be of the form 3) A fruit seller has 56 red apples and has the same number, and they take up the least area of the tray. that can be placed in each stack for this purpose? Solution: To find the solution we have to first find the HCF of 56 and 32 maximum number of apples in each stack take up least amount of space. We use Euclidâ€™s division algorithm to find the HCF of 56 and 32 56 = 32 x 1 + 24 Since, 24 0 ? ? 32 = 24 x 1 + 8 8 0 ? 24 = 8 x 3 + 0 So the HCF of 56 and 32 is 8. Therefore the fruit seller has to make s REAL NUMBERS S DIVISION ALGORITHM Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + 144 0 ? we apply the division lemma to 1362 and 144 1 3 6 2 1 4 4 x 9 + 6 6 = Since the remainder 66 0 ? we apply the division lemma to 144 and 66 1 4 4 6 6 x 2 + 1 2 = Since the remainder 12 0 ? we apply the division lemma to 66 and 12 6 6 1 2 x 5 + 6 Since the remainder 6 0 ? we apply the division lemma to 12 and 6 1 2 6 x 2 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7 is a positive odd integer. We apply the division algorithm with the possible remainders are 0,1,2,3,4,5,6 and 7 +2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or 32 green apples. She wants to stack them in such a way has the same number, and they take up the least area of the tray. What is the maximum number of that can be placed in each stack for this purpose? irst find the HCF of 56 and 32. Then this number will give us the maximum number of apples in each stack and as a result the number of stacks will be least and they will hm to find the HCF of 56 and 32 32 x 1 + 24 24 x 1 + 8 + 0 Therefore the fruit seller has to make stacks of 8 for both kinds of apples. : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, we apply the division lemma to 1362 and 144 e apply the division lemma to 144 and 66 we apply the division lemma to 66 and 12 we apply the division lemma to 12 and 6 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the +7, where q is some integer is a positive odd integer. We apply the division algorithm with a +7 where q is the quotient. (since they are divisible by 2) , or 8 7 q+ in such a way that each stack What is the maximum number of apples Then this number will give us the and as a result the number of stacks will be least and they will REAL NUMBERS EUCLIDâ€™S DIVISION ALGORITHM 4) Prove that one of every four consecutive integers is divisible by 4 Solution: Let the four consecutive integers be n,n Consider a = bq + r, where 0= < Let a=n and b=4. Hence r can have values 0 CASE 1: Let r=0 ? 4 n q = CASE 2: Let r=1 ? 4 n q = CASE 3: Let r=2 ? 4 n q = CASE 4: Let r=3 ? 4 n q = +3 then Thus in every case one of the four numbe : 5) Use Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 for some integer m. Solution: Let n be any positive integer. Consider Let a=n and b=3. Hence r can have values 0,1 or 2. CASE 1: Let r=0 then n =3q ? = CASE 2: Let r=1 then 3 1 n q = + CASE 2: Let r=2 then 3 2 n q = + Therefore 2 n is always of the form REAL NUMBERS S DIVISION ALGORITHM Prove that one of every four consecutive integers is divisible by 4. Let the four consecutive integers be n,n+1,n+2 and n+3. 0 r b = < can have values 0, 1, 2 or 3. n q which is divisible by 4 : n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4 n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = + n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = + Thus in every case one of the four numbers is divisible by 4. Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 Let n be any positive integer. Consider a=bq+r where 0 r b = < can have values 0,1 or 2. 2 2 (3 ) n q ? = 2 9q = 2 3(3 ) q = 3m = where m = 3q 3 1 n q = + 2 2 (3 1) n q ? = + 2 9 6 1 q q = + + 2 3(3 2 ) 1 q q = + + 3 1 m = + where 2 3 2 m q q = + 3 2 = + 2 2 (3 2) n q ? = + 2 9 12 4 q q = + + 2 2 9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + + 3 1 m = + where is always of the form 3m or 3m+1 REAL NUMBERS 4 1 3 4 4 4( 1) which is divisible by 4 : 4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 : 4 3 1 4 4 4( 1) which is divisible by 4 Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 r b = < m = 3q 2 9 12 3 1 3(3 4 1) 1 = + + + = + + + where 2 3 4 1 m q q = + + REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS 6) Find the HCF and LCM of the following pairs of numbers and verify that HCF numbers. a)39 and 169 b) 436 and 594 Solution: a) 39 and 169 Factors of the two numbers are 39 169 HCF(39,169) = 13 LCM(39,169) = 3 x 13 x 13=507 HCF x LCM = 13 x 507= 6591 Product of the numbers =39 x 169=6591 HCF x LCM= Product of the numbers b)436 and 594 Factors of the two numbers are 436=2 x 2 x 594=2 x 3 x 3 x 3 x 11 HCF(436,594) = 2 LCM(436,594) = 2 x 2 x 3 x 3 x 3 x 11 x 109=129492 HCF x LCM = 2 x 129492=258984 Product of the two numbers = 436 x 594= 258984 ? HCF x LCM= Product of the numbers 7) If HCF(90,225)=15, find LCM(90,225) Solution: We know that HCF x LCM ? HCF(90,225) LCM(90,225) = ? 15 LCM(90,,225) = 90 ? LCM 8) Find whether 8 n can end with the digit 0 for any natural number. Solution: The prime factors of 8 are 2.Therefore the prime factors of 8 Any digit ending with 0 must have 5 as one of its prime factors. But the prime factors of 8 are 2. Therefore 8 n cannot end with 0. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS Find the HCF and LCM of the following pairs of numbers and verify that HCF x LCM=Product of the two b) 436 and 594 are 39 = 3 x 13 169 = 13 x 13 3 x 13 x 13=507 = 13 x 507= 6591 Product of the numbers =39 x 169=6591 HCF x LCM= Product of the numbers are 436=2 x 2 x 109 594=2 x 3 x 3 x 3 x 11 2 x 2 x 3 x 3 x 3 x 11 x 109=129492 = 2 x 129492=258984 Product of the two numbers = 436 x 594= 258984 HCF x LCM= Product of the numbers LCM(90,225) : We know that HCF x LCM = product of the two numbers LCM(90,225) = product of (90,225) LCM(90,,225) = 90 225 LCM(90,225) = 90 x 225 15 20250 15 = =1350 ? LCM(90,225) = 1350 can end with the digit 0 for any natural number. factors of 8 are 2.Therefore the prime factors of 8 n will also be 2. Any digit ending with 0 must have 5 as one of its prime factors. But the prime factors of 8 are 2. cannot end with 0. LCM=Product of the two will also be 2. Page 4 REAL EUCLIDâ€™S DIVISION ALGORITHM 1) Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 Solution: STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + STEP 2: Since the remainder 144 0 1 3 6 2 1 4 4 x 9 + 6 6 STEP 3: Since the remainder 1 4 4 6 6 x 2 + 1 2 = STEP 4: Since the remainder 6 6 1 2 x 5 + 6 = STEP 5: Since the remainder 1 2 6 x 2 + 0 = The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. 2) Show that any positive odd integer is of the form Solution: Letâ€™s start with taking a, where a is a positive odd integer. We apply the division algorithm with and b =8 Since 0 8 r = < the possible remainders are That is, a can be 8q or 8q+1 or 8q+2 or 8 However, since a is odd a cannot be 8 Therefore any odd integer can only be of the form 3) A fruit seller has 56 red apples and has the same number, and they take up the least area of the tray. that can be placed in each stack for this purpose? Solution: To find the solution we have to first find the HCF of 56 and 32 maximum number of apples in each stack take up least amount of space. We use Euclidâ€™s division algorithm to find the HCF of 56 and 32 56 = 32 x 1 + 24 Since, 24 0 ? ? 32 = 24 x 1 + 8 8 0 ? 24 = 8 x 3 + 0 So the HCF of 56 and 32 is 8. Therefore the fruit seller has to make s REAL NUMBERS S DIVISION ALGORITHM Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + 144 0 ? we apply the division lemma to 1362 and 144 1 3 6 2 1 4 4 x 9 + 6 6 = Since the remainder 66 0 ? we apply the division lemma to 144 and 66 1 4 4 6 6 x 2 + 1 2 = Since the remainder 12 0 ? we apply the division lemma to 66 and 12 6 6 1 2 x 5 + 6 Since the remainder 6 0 ? we apply the division lemma to 12 and 6 1 2 6 x 2 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7 is a positive odd integer. We apply the division algorithm with the possible remainders are 0,1,2,3,4,5,6 and 7 +2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or 32 green apples. She wants to stack them in such a way has the same number, and they take up the least area of the tray. What is the maximum number of that can be placed in each stack for this purpose? irst find the HCF of 56 and 32. Then this number will give us the maximum number of apples in each stack and as a result the number of stacks will be least and they will hm to find the HCF of 56 and 32 32 x 1 + 24 24 x 1 + 8 + 0 Therefore the fruit seller has to make stacks of 8 for both kinds of apples. : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, we apply the division lemma to 1362 and 144 e apply the division lemma to 144 and 66 we apply the division lemma to 66 and 12 we apply the division lemma to 12 and 6 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the +7, where q is some integer is a positive odd integer. We apply the division algorithm with a +7 where q is the quotient. (since they are divisible by 2) , or 8 7 q+ in such a way that each stack What is the maximum number of apples Then this number will give us the and as a result the number of stacks will be least and they will REAL NUMBERS EUCLIDâ€™S DIVISION ALGORITHM 4) Prove that one of every four consecutive integers is divisible by 4 Solution: Let the four consecutive integers be n,n Consider a = bq + r, where 0= < Let a=n and b=4. Hence r can have values 0 CASE 1: Let r=0 ? 4 n q = CASE 2: Let r=1 ? 4 n q = CASE 3: Let r=2 ? 4 n q = CASE 4: Let r=3 ? 4 n q = +3 then Thus in every case one of the four numbe : 5) Use Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 for some integer m. Solution: Let n be any positive integer. Consider Let a=n and b=3. Hence r can have values 0,1 or 2. CASE 1: Let r=0 then n =3q ? = CASE 2: Let r=1 then 3 1 n q = + CASE 2: Let r=2 then 3 2 n q = + Therefore 2 n is always of the form REAL NUMBERS S DIVISION ALGORITHM Prove that one of every four consecutive integers is divisible by 4. Let the four consecutive integers be n,n+1,n+2 and n+3. 0 r b = < can have values 0, 1, 2 or 3. n q which is divisible by 4 : n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4 n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = + n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = + Thus in every case one of the four numbers is divisible by 4. Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 Let n be any positive integer. Consider a=bq+r where 0 r b = < can have values 0,1 or 2. 2 2 (3 ) n q ? = 2 9q = 2 3(3 ) q = 3m = where m = 3q 3 1 n q = + 2 2 (3 1) n q ? = + 2 9 6 1 q q = + + 2 3(3 2 ) 1 q q = + + 3 1 m = + where 2 3 2 m q q = + 3 2 = + 2 2 (3 2) n q ? = + 2 9 12 4 q q = + + 2 2 9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + + 3 1 m = + where is always of the form 3m or 3m+1 REAL NUMBERS 4 1 3 4 4 4( 1) which is divisible by 4 : 4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 : 4 3 1 4 4 4( 1) which is divisible by 4 Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 r b = < m = 3q 2 9 12 3 1 3(3 4 1) 1 = + + + = + + + where 2 3 4 1 m q q = + + REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS 6) Find the HCF and LCM of the following pairs of numbers and verify that HCF numbers. a)39 and 169 b) 436 and 594 Solution: a) 39 and 169 Factors of the two numbers are 39 169 HCF(39,169) = 13 LCM(39,169) = 3 x 13 x 13=507 HCF x LCM = 13 x 507= 6591 Product of the numbers =39 x 169=6591 HCF x LCM= Product of the numbers b)436 and 594 Factors of the two numbers are 436=2 x 2 x 594=2 x 3 x 3 x 3 x 11 HCF(436,594) = 2 LCM(436,594) = 2 x 2 x 3 x 3 x 3 x 11 x 109=129492 HCF x LCM = 2 x 129492=258984 Product of the two numbers = 436 x 594= 258984 ? HCF x LCM= Product of the numbers 7) If HCF(90,225)=15, find LCM(90,225) Solution: We know that HCF x LCM ? HCF(90,225) LCM(90,225) = ? 15 LCM(90,,225) = 90 ? LCM 8) Find whether 8 n can end with the digit 0 for any natural number. Solution: The prime factors of 8 are 2.Therefore the prime factors of 8 Any digit ending with 0 must have 5 as one of its prime factors. But the prime factors of 8 are 2. Therefore 8 n cannot end with 0. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS Find the HCF and LCM of the following pairs of numbers and verify that HCF x LCM=Product of the two b) 436 and 594 are 39 = 3 x 13 169 = 13 x 13 3 x 13 x 13=507 = 13 x 507= 6591 Product of the numbers =39 x 169=6591 HCF x LCM= Product of the numbers are 436=2 x 2 x 109 594=2 x 3 x 3 x 3 x 11 2 x 2 x 3 x 3 x 3 x 11 x 109=129492 = 2 x 129492=258984 Product of the two numbers = 436 x 594= 258984 HCF x LCM= Product of the numbers LCM(90,225) : We know that HCF x LCM = product of the two numbers LCM(90,225) = product of (90,225) LCM(90,,225) = 90 225 LCM(90,225) = 90 x 225 15 20250 15 = =1350 ? LCM(90,225) = 1350 can end with the digit 0 for any natural number. factors of 8 are 2.Therefore the prime factors of 8 n will also be 2. Any digit ending with 0 must have 5 as one of its prime factors. But the prime factors of 8 are 2. cannot end with 0. LCM=Product of the two will also be 2. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS 9) Find the HCF and LCM of the following numbers using prime factorisation method a) 9, 27 and 108 b) 16, 56 and 144 Solution: a) 9, 27 and 108 Factors of the numbers are 9 = 3 x 3 27 = 3 x 3 x 3 108 = 2 x 2 x 3 x 3 x 3 HCF(9,27,108)=3 x 3=9 LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108 ? HCF=9 & LCM=108 b) 16, 56 and 144 Factors of the numbers are 16 = 2 x 2 x 2 x 2 56 = 2 x 2 x 2 x 7 144 = 2 x 2 x 2 x 2 x 3 ?HCF (16,56,144)=2 x 2 x 2=8 ?LCM (16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008 ? HCF = 8 & LCM = 1008 10) Prove that 16, 343 and 225 are relatively Solution: Factors of the numbers 16 =2 x 2 x 2 x 2 343 = 7 x 7 x 7 225 = 3 x 3 x 5 x 5 ?HCF(16,343,225)=1 That is they do not have any common factors ? They are relatively prime. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS Find the HCF and LCM of the following numbers using prime factorisation method 108 b) 16, 56 and 144 3 x 3 3 x 3 x 3 2 x 2 x 3 x 3 x 3 LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108 2 x 2 x 2 x 2 2 x 2 x 2 x 7 2 x 2 x 2 x 2 x 3 x 3 HCF (16,56,144)=2 x 2 x 2=8 (16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008 and 225 are relatively prime. =2 x 2 x 2 x 2 = 7 x 7 x 7 = 3 x 3 x 5 x 5 That is they do not have any common factors REAL NUMBERS Find the HCF and LCM of the following numbers using prime factorisation method Page 5 REAL EUCLIDâ€™S DIVISION ALGORITHM 1) Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 Solution: STEP 1: Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + STEP 2: Since the remainder 144 0 1 3 6 2 1 4 4 x 9 + 6 6 STEP 3: Since the remainder 1 4 4 6 6 x 2 + 1 2 = STEP 4: Since the remainder 6 6 1 2 x 5 + 6 = STEP 5: Since the remainder 1 2 6 x 2 + 0 = The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. 2) Show that any positive odd integer is of the form Solution: Letâ€™s start with taking a, where a is a positive odd integer. We apply the division algorithm with and b =8 Since 0 8 r = < the possible remainders are That is, a can be 8q or 8q+1 or 8q+2 or 8 However, since a is odd a cannot be 8 Therefore any odd integer can only be of the form 3) A fruit seller has 56 red apples and has the same number, and they take up the least area of the tray. that can be placed in each stack for this purpose? Solution: To find the solution we have to first find the HCF of 56 and 32 maximum number of apples in each stack take up least amount of space. We use Euclidâ€™s division algorithm to find the HCF of 56 and 32 56 = 32 x 1 + 24 Since, 24 0 ? ? 32 = 24 x 1 + 8 8 0 ? 24 = 8 x 3 + 0 So the HCF of 56 and 32 is 8. Therefore the fruit seller has to make s REAL NUMBERS S DIVISION ALGORITHM Use Euclidâ€™s division algorithm to find the HCF of 1362 and 6954 : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, 6 9 5 4 1 3 6 2 x 5 1 4 4 = + 144 0 ? we apply the division lemma to 1362 and 144 1 3 6 2 1 4 4 x 9 + 6 6 = Since the remainder 66 0 ? we apply the division lemma to 144 and 66 1 4 4 6 6 x 2 + 1 2 = Since the remainder 12 0 ? we apply the division lemma to 66 and 12 6 6 1 2 x 5 + 6 Since the remainder 6 0 ? we apply the division lemma to 12 and 6 1 2 6 x 2 + 0 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the HCF of 1362 and 6954 is 6. odd integer is of the form 8q+1, or 8q+3, or 8q+5, or 8q+7 is a positive odd integer. We apply the division algorithm with the possible remainders are 0,1,2,3,4,5,6 and 7 +2 or 8q+3 or 8q+4 or 8q+5 or 8q+6 or 8q+7 where cannot be 8q or 8q+2 or 8q+4 or 8q+6 (since they are integer can only be of the form 8 1 q+ , or 8 3 q+ , or 8 5 q+ , or 32 green apples. She wants to stack them in such a way has the same number, and they take up the least area of the tray. What is the maximum number of that can be placed in each stack for this purpose? irst find the HCF of 56 and 32. Then this number will give us the maximum number of apples in each stack and as a result the number of stacks will be least and they will hm to find the HCF of 56 and 32 32 x 1 + 24 24 x 1 + 8 + 0 Therefore the fruit seller has to make stacks of 8 for both kinds of apples. : Since 6954>1362 we apply division lemma to 6954 and 1362 to get, we apply the division lemma to 1362 and 144 e apply the division lemma to 144 and 66 we apply the division lemma to 66 and 12 we apply the division lemma to 12 and 6 The remainder has now become zero, so our procedure stops. Since the divisor at this stage is 6, the +7, where q is some integer is a positive odd integer. We apply the division algorithm with a +7 where q is the quotient. (since they are divisible by 2) , or 8 7 q+ in such a way that each stack What is the maximum number of apples Then this number will give us the and as a result the number of stacks will be least and they will REAL NUMBERS EUCLIDâ€™S DIVISION ALGORITHM 4) Prove that one of every four consecutive integers is divisible by 4 Solution: Let the four consecutive integers be n,n Consider a = bq + r, where 0= < Let a=n and b=4. Hence r can have values 0 CASE 1: Let r=0 ? 4 n q = CASE 2: Let r=1 ? 4 n q = CASE 3: Let r=2 ? 4 n q = CASE 4: Let r=3 ? 4 n q = +3 then Thus in every case one of the four numbe : 5) Use Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 for some integer m. Solution: Let n be any positive integer. Consider Let a=n and b=3. Hence r can have values 0,1 or 2. CASE 1: Let r=0 then n =3q ? = CASE 2: Let r=1 then 3 1 n q = + CASE 2: Let r=2 then 3 2 n q = + Therefore 2 n is always of the form REAL NUMBERS S DIVISION ALGORITHM Prove that one of every four consecutive integers is divisible by 4. Let the four consecutive integers be n,n+1,n+2 and n+3. 0 r b = < can have values 0, 1, 2 or 3. n q which is divisible by 4 : n q +1 then 3 n+ = 4 1 3 4 4 4( 1) q q q + + = + = + which is divisible by 4 n q +2 then 2 n+ = 4 2 2 4 4 4( 1) q q q + + = + = + n q +3 then 1 n+ = 4 3 1 4 4 4( 1) q q q + + = + = + Thus in every case one of the four numbers is divisible by 4. Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3 Let n be any positive integer. Consider a=bq+r where 0 r b = < can have values 0,1 or 2. 2 2 (3 ) n q ? = 2 9q = 2 3(3 ) q = 3m = where m = 3q 3 1 n q = + 2 2 (3 1) n q ? = + 2 9 6 1 q q = + + 2 3(3 2 ) 1 q q = + + 3 1 m = + where 2 3 2 m q q = + 3 2 = + 2 2 (3 2) n q ? = + 2 9 12 4 q q = + + 2 2 9 12 3 1 3(3 4 1) 1 q q q q = + + + = + + + 3 1 m = + where is always of the form 3m or 3m+1 REAL NUMBERS 4 1 3 4 4 4( 1) which is divisible by 4 : 4 2 2 4 4 4( 1) + + = + = + which is divisible by 4 : 4 3 1 4 4 4( 1) which is divisible by 4 Euclidâ€™s division lemma to show that the square of any positive integer is either of the form 3m or 3m+1 r b = < m = 3q 2 9 12 3 1 3(3 4 1) 1 = + + + = + + + where 2 3 4 1 m q q = + + REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS 6) Find the HCF and LCM of the following pairs of numbers and verify that HCF numbers. a)39 and 169 b) 436 and 594 Solution: a) 39 and 169 Factors of the two numbers are 39 169 HCF(39,169) = 13 LCM(39,169) = 3 x 13 x 13=507 HCF x LCM = 13 x 507= 6591 Product of the numbers =39 x 169=6591 HCF x LCM= Product of the numbers b)436 and 594 Factors of the two numbers are 436=2 x 2 x 594=2 x 3 x 3 x 3 x 11 HCF(436,594) = 2 LCM(436,594) = 2 x 2 x 3 x 3 x 3 x 11 x 109=129492 HCF x LCM = 2 x 129492=258984 Product of the two numbers = 436 x 594= 258984 ? HCF x LCM= Product of the numbers 7) If HCF(90,225)=15, find LCM(90,225) Solution: We know that HCF x LCM ? HCF(90,225) LCM(90,225) = ? 15 LCM(90,,225) = 90 ? LCM 8) Find whether 8 n can end with the digit 0 for any natural number. Solution: The prime factors of 8 are 2.Therefore the prime factors of 8 Any digit ending with 0 must have 5 as one of its prime factors. But the prime factors of 8 are 2. Therefore 8 n cannot end with 0. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS Find the HCF and LCM of the following pairs of numbers and verify that HCF x LCM=Product of the two b) 436 and 594 are 39 = 3 x 13 169 = 13 x 13 3 x 13 x 13=507 = 13 x 507= 6591 Product of the numbers =39 x 169=6591 HCF x LCM= Product of the numbers are 436=2 x 2 x 109 594=2 x 3 x 3 x 3 x 11 2 x 2 x 3 x 3 x 3 x 11 x 109=129492 = 2 x 129492=258984 Product of the two numbers = 436 x 594= 258984 HCF x LCM= Product of the numbers LCM(90,225) : We know that HCF x LCM = product of the two numbers LCM(90,225) = product of (90,225) LCM(90,,225) = 90 225 LCM(90,225) = 90 x 225 15 20250 15 = =1350 ? LCM(90,225) = 1350 can end with the digit 0 for any natural number. factors of 8 are 2.Therefore the prime factors of 8 n will also be 2. Any digit ending with 0 must have 5 as one of its prime factors. But the prime factors of 8 are 2. cannot end with 0. LCM=Product of the two will also be 2. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS 9) Find the HCF and LCM of the following numbers using prime factorisation method a) 9, 27 and 108 b) 16, 56 and 144 Solution: a) 9, 27 and 108 Factors of the numbers are 9 = 3 x 3 27 = 3 x 3 x 3 108 = 2 x 2 x 3 x 3 x 3 HCF(9,27,108)=3 x 3=9 LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108 ? HCF=9 & LCM=108 b) 16, 56 and 144 Factors of the numbers are 16 = 2 x 2 x 2 x 2 56 = 2 x 2 x 2 x 7 144 = 2 x 2 x 2 x 2 x 3 ?HCF (16,56,144)=2 x 2 x 2=8 ?LCM (16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008 ? HCF = 8 & LCM = 1008 10) Prove that 16, 343 and 225 are relatively Solution: Factors of the numbers 16 =2 x 2 x 2 x 2 343 = 7 x 7 x 7 225 = 3 x 3 x 5 x 5 ?HCF(16,343,225)=1 That is they do not have any common factors ? They are relatively prime. REAL NUMBERS THE FUNDAMENTAL THEOREM OF ARITHEMATICS Find the HCF and LCM of the following numbers using prime factorisation method 108 b) 16, 56 and 144 3 x 3 3 x 3 x 3 2 x 2 x 3 x 3 x 3 LCM(9,27,108)=2 x 2 x 3 x 3 x 3= 108 2 x 2 x 2 x 2 2 x 2 x 2 x 7 2 x 2 x 2 x 2 x 3 x 3 HCF (16,56,144)=2 x 2 x 2=8 (16,56,144)=2 x 2 x 2 x 2 x 3 x 3 x 7=1008 and 225 are relatively prime. =2 x 2 x 2 x 2 = 7 x 7 x 7 = 3 x 3 x 5 x 5 That is they do not have any common factors REAL NUMBERS Find the HCF and LCM of the following numbers using prime factorisation method REAL NUMBERS IRRATIONAL NUMBERS 11) Prove that 7 is irrational. Solution: Let 7 be rational. ? There exists integers a, b such that are co-prime. 7b a ? = Squaring both sides we get, 2 2 7b a = Therefore 2 a is divisible by 7. Hence a is also divisible by 7. ? There exists an integer c such that 2 2 7 49 b c = and hence 2 2 b c Therefore a and b have a common factor 7. This is a contradiction, since a and b are co Hence our assumption 7 Therefore 7 is irrational 12) Prove that 6 - 5 is irrational Solution: Let us assume that 6 5 - that 6 5 a b - = where a and b are co Rearranging the equation we get Since a and b are integers But this contradicts the fact that Therefore our assumption that Therefore 6 5 is irrational - 13) Prove that 3 4 is irrational. Solution: Let us assume that 3 4 is rational. Therefore there exists integers a and b( where a and b are co-prime i.e. 3 3 4 a b = . Therefore 3 a Therefore there exists an integer c such that 3 3 16 b c ? = Hence Therefore a and b have a common factor 4 3 Therefore 4 is irrational REAL NUMBERS IRRATIONAL NUMBERS There exists integers a, b such that 7 a b = where a and b Squaring both sides we get, is divisible by 7. Hence a is also divisible by 7. integer c such that 7 a c = 2 2 7 (7 ) b c ? = 2 2 7 b c = which means 2 b is divisible by 7 and b is also divisible by 7 a and b have a common factor 7. This is a contradiction, since a and b are co 7 is rational is wrong. Therefore 7 is irrational 6 5 is rational. Therefore there exists integers a and b where a and b are co-prime. Rearranging the equation we get 6 5 a b - = also b0, 6 a b - is rational. Therefore 5 is rational But this contradicts the fact that 5 is irrational. Therefore our assumption that 6 5 - is rational is wrong. Therefore 6 5 is irrational is rational. Therefore there exists integers a and b( prime Taking cubes on both sides we get,t 3 4 a b = 3 a is divisible by 4 Hence a is also divisible by 4 Therefore there exists an integer c such that a=4c. 3 3 3 4 (4 ) 64 b c c ? = = Hence 3 b is divisible by 4 and hence b is divisible by 4. e a and b have a common factor 4. This is a contradiction, since a and b are co Therefore 4 is irrational where a and b( 0 b? ) is divisible by 7. Hence a is also divisible by 7. is divisible by 7 and b is also divisible by 7 a and b have a common factor 7. This is a contradiction, since a and b are co-prime. is rational. Therefore there exists integers a and b( 0 b? ) such is rational is rational. Therefore there exists integers a and b( 0 b? ) such that 3 4 a b = 3 3 a b divisible by 4 3 3 3 b c c and hence b is divisible by 4. . This is a contradiction, since a and b are co-primeRead More

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