08 - Question bank - Real Numbers - Class 10 - Maths Class 10 Notes | EduRev

Crash Course for Class 10 Maths by Let's tute

Class 10 : 08 - Question bank - Real Numbers - Class 10 - Maths Class 10 Notes | EduRev

 Page 1


         REAL
 
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’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:   
    Let’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 Euclid’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 Euclid’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
 
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’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:   
    Let’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 Euclid’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 Euclid’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
     
 
 
 
EUCLID’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 Euclid’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. 
Euclid’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  
Euclid’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
 
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’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:   
    Let’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 Euclid’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 Euclid’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
     
 
 
 
EUCLID’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 Euclid’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. 
Euclid’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  
Euclid’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
 
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’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:   
    Let’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 Euclid’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 Euclid’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
     
 
 
 
EUCLID’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 Euclid’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. 
Euclid’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  
Euclid’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
 
EUCLID’S DIVISION ALGORITHM
1) Use Euclid’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:   
    Let’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 Euclid’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 Euclid’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
     
 
 
 
EUCLID’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 Euclid’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. 
Euclid’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  
Euclid’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-prime  
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