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Divisibility
REAL NUMBERS
We have been studying division of numbers for the last
many years. Let us recall the same division in a formal
manner.
Definition
A non-zero integer ‘a’ is said to divide an integer ‘b’, if
there exists an integer c such that
b = ac
Where
a
dividend
b
divisor
c
quotient
Example
If 6 and 24 are two integers. Then,
24 = 6 × 4
So, 6 divides 24 because there is an integer 4 which
when multiplied with 6, results into 24. However, 9
does not divide 24 because there do not exist an
integer such that
24 = 9 × c where ‘c’ is any integer
If ‘a’ is a non-zero integer, which divides an integer ‘b’,
then it is written as a | b. In other words, a|b means
that ‘a’ is factor of ‘b’ or ‘b’ is multiple of ‘a’. If ‘a’ does
not divide ‘b’, then it is written as a b .
There may be following possibilities: (i) −6 | 24 , because there exists an integer –4 such
that 20 =−6 × (−4) .
(ii) 6 | −24 , because there exists an integer –4 such
that −24 = 6 × (−4) .
(iii) −6 | − 24 , because there exists an integer 4 such
that −24 =−6 × 4 .
Note
1.
±1 divides every non-zero integer, i.e. ±1 | a for
every non-zero integer a.
2. 0 is divisible by every non-zero integer a, i.e. a | 0
for every non-zero integer a.
3. 0 does not divide any integer.
4. If a is non-zero integer and b is any integer, then
⇒ a | −b ⇒ −a | b
⇒ −a | −b
a|b
5. If a and b are non-zero integers, then
⇒
a = ±b
a | b and b | a
6. If a is a non-zero integer and b, c are any two
integers, then
 a|b ± c

for any integer x
⇒  a | bc
a | b and a | c
 a | bx

7. If a and c are non-zero integers and b, d are any
two integers, then
⇒
a | b and c | d
ac | bd
Or
⇒
ac | bc
a|b
Division Of Integers
Consider the division of one positive integer by another
positive integer.
Divide 83 by 6.
For this, we multiply 6 by the largest possible number
such that the result must not exceed 83. Largest such
number is 13. Then,
83 =6 × 13 + 5
Where
83
Dividend
6
Divisor
13
Quotient
5
Remainder
0 ≤ 5 < 6 i.e., remainder is greater than or equal to 0
and less than divisor
Example
(i) 25, 7
25 = 7 × 3 + 4,
0≤4<7
Q 7 goes in to 25 thrice and leaves remainder 4 
(ii) 7, 15
7 = 15 × 0 + 7,
0 ≤ 7 < 15
Q 15 is larger than 7. So, this relation is always possible 
(iii) 35, 5
35 = 5 × 7 + 0,
0≤0<5
Q 5 goes into 35 seven times and leaves no remainder 
So, we can say that for each pair of positive integers a
and b, we can find unique integers q and r satisfying
the relation
a = bq + r
Where 0 ≤ r ≤ b
Euclid’s Division Lemma
Let a and b be any two positive integers. Then, there
exist unique integers q and r such that
a = bq + r
Where 0 ≤ r < b
If b|a, then r = 0. Otherwise, r satisfies the stronger
inequality 0 < r < b.
Proof
Consider the arithmetic progression
...., a – 3b, a – 2b, a – b, a, a + b, a + 2b, a + 3b, .....
Where common difference is ‘b’ and the Arithmetic
Progression extends indefinitely in both directions.
Let smallest non-negative term of AP = r
Then, there exist a non-negative integer q such that
a – qb = r
⇒ a = bq + r
Here r is the smallest non-negative integer satisfying
the above result. Therefore,
0 ≤ r < b.
So,
a = bq + r
Where 0 ≤ r < b.
Now, we will prove that there is only one such pair of
numbers q and r
Let us assume that there is another such pair of nonnegative integers satisfying the above condition.
Let that pair of numbers be q 1 and r 1
Then,
a = bq 1 + r 1 , Where 0 ≤ r 1 < b
To prove that q and r is only pair of such numbers, we
will prove that q 1 and r 1 are no other numbers than q
and r, i.e. r 1 = r and q 1 = q
Here,
a = bq + r
[Given]
And
=
a bq1 + r1
[By assumption]
If b is not a divisor of a, then by Euclid’s division
lemma there exist positive integers q and r such that
a = bq + r
Where 0 < r < b
Here a and b may have common divisor, also b and r
may have common divisor. In such case, every common
divisor of b and r is a common divisor of a and b and
vice-versa
If a and b are the two positive integers such that
a = bq + r, then every common divisor of a and b
is a common divisor of b and r, and vice-versa.
⇒
r1 − r = bq − bq1
⇒
r1 − r= b (q − q1 )
Proof
Let c be a common divisor of a and b
Then,
c | a
⇒ a = cq 1 for some integer q 1
c | b
⇒ b = cq 2 for some integer q 2
Now,
=
a bq + r
⇒ r= a − bq
⇒
b | r1 − r
r cq1 − cq2q
⇒ =
⇒
r1 − r =
0
So,
bq + r= bq1 + r1
⇒
r1 = r
Q 0 ≤ r < b and0 ≤ r1 < b ⇒ 0 ≤ r1 − r < b 
……………(i)
Now,
−r1 =−r
⇒
a − r1 = a − r
⇒
bq1 = bq
[Multiplying both sides by (–1)]
[Adding a on both sides]
[∴a = bq + r, a = bq 1 + r 1 ⇒ bq = a – r, bq 1 = a –
r1]
⇒ q1 = q
……………(ii)
From (i) and (ii),
r1 = r
q1 = q
and
That means, there is no other pair of numbers
is
satisfying the condition. So, a = bq + r, 0 ≤ r < b
unique.
The above Lemma has been stated for positive integers
only. But, it can be extended to all integers whether
positive or negative. For this the statement is as
following: Let a and b be any two integers with a ≠ 0. Then, there
exist unique integers q and r such that
a = bq + r, where 0 ≤ r < | b |
Euclid’s Division Algorithm
If integer c divides another integer x, then c is called
divisor of x. Similarly, if an integer c divides integers
x 1 , x 2 , x n ..... then it is called a common divisor of x 1 ,
x 2 , x n ...... Infact, two or more integers may have lot of
common divisors. But, the largest common factor is
called Highest Common Factor (HCF) or Greatest
Common Divisor.
Example
1, 2, 3, 6 are common divisor of 12 and 18. Here 6 is
the largest common divisor of 12 and 18, so 6 is called
Highest Common Factor (HCF) or Greatest Common
Divisor (GCD) of 12 and 18.
Let a and b be two positive integers such that a > b.
r c (q1 − q2q)
⇒=
⇒
⇒
⇒
c | r
c | r and c | b
c is a common divisor of b and r
[Given c | b]
Hence, a common divisor of a and b is also a common
divisor of b and r. Conversely, Let d be a common
divisor of b and r. Then,
⇒ b = r1 d
For some integer r 1
d|b
d|r
⇒
r = r2 d
For some integer r 2
Now, we will prove that d is a common divisor of both a
and b.
Here,
=
a bq + r
⇒ =
a r1dq + r2d
⇒=
a d(r1q + r2 )
⇒
⇒
⇒
d| a
d | a and d | b
d is a common divisor of a and b
[Given c | b]
Note
1. The HCF of two or more positive integers always
exists and it is unique.
2. One (1) is the common factor of all the integers.
Application Of Euclid’s Division Algorithm
Euclid’s Division Lemma and Euclid’s Division Algorithm
can be used in many mathematical applications.
Let an integer be a = 117
Let another integer be b = 45
Now, by Euclid’s division lemma,
117 = 45 × 2 + 27
……………(i)
Q 45 )117(2 


90 


27 
Or
=
a bq1 + r1
Where q1 = 2 and q2 = 27
Here, common divisors of a = 117 and b = 45 are also
the common divisors of b = 45 and r1 = 27 and viceversa.
Now, again by Euclid’s division lemma,
45 = 27 × 1 + 18
……………(ii)
Q 27 )45(1


27 


18 
Or
=
b q2r1 + r2
5.
If r2 ≠ 0 , then repeat the above process till the
remainder r n is zero.
The divisor at this stage
rn −1
or the non zero
remainder at the previous stage will be the HCF of
a and b.
Example
Find the HCF of 210 and 55.
Given integers are 210 and 55.
Clearly, 210 > 55
Here, by Euclid’s division lemma to 210 and 55,
210 = 55 × 3 + 45
……………(i)
Where q2 = 1 and r2 = 18
Here, common divisor of r1 = 27 and r2 = 18 are also the
common divisors of b = 45 and r1 = 27 and vice-versa.
Also, common divisor of b = 45 and r1 = 27 are the
Here, remainder 45 ≠ 0, so again by Euclid’s division
lemma,
55 = 45 × 1 + 10
……………(ii)
common divisors of 117 and b = 45 and vice-versa.
Therefore, common divisors of r1 = 27 and r2 = 18 are
Here, remainder 10 ≠ 0, so again by Euclid’s division
lemma,
the common divisors of a = 117 and b = 45 and viceversa.
45 = 10 × 4 + 5
Now, again by Euclid’s division lemma,
27 = 18 × 1 + 9
Here, remainder 5 ≠ 0, so again by Euclid’s division
lemma,
10 = 2 × 5 + 0
…………(iv)
…………(iii)
Q



Or
=
r1 q3r2 + r3
18 )27(1

18 

9
Where q3 = 1 and r3 = 9 .
r1 = 27
and
r2 = 18 .
Also, the
common divisors of r1 = 27 and r2 = 18 are the common
divisors of b = 45 and r1 = 27 . Also, common divisor of
b = 45 and r1 = 27 are the common divisors of 117 and
b = 45.
Now, again by Euclid’s division lemma,
18 = 9 × 2 + 0
Or
=
r2 q4r3 + r4
Here, remainder 5 = 0.
∴ Divisor at this stage or the remainder at previous
state 5 is the HCF of 210 and 55.
The Fundamental Theorem Arithmetic
Here, common divisors of r2 = 18 and r3 = 9 are the
common divisors of
…………(iii)
…………(iv)
Where q4 = 1 and r4 = 0
Therefore r3 = 9 is the divisor of r2 = 18 and r3 = 9 . Also,
it is the greatest common divisor (HCF) of r 2 and r 3 .
So, r 3 = 9 is the greatest common divisor (HCF) of a =
117 and b = 45. Here r 3 = 9 is the last non-zero
remainder and it is the HCF or GCD of 117 and 45.
Here are the steps to find the HCF by using Euclid’s
division algorithm
Steps
1. If the given numbers are a and b, then obtain
whole numbers q 1 and r 1 such that
=
a bq1 + r1
0 ≤ r1 < b
2.
3.
If r 1 = 0, then b is the HCF of a and b.
If r 1 ≠ 0, then for b and r 1 , obtain two whole
numbers q 1 and r 2 such that =
b q1r1 + r2 .
4.
If r2 = 0 , then r 1 is the HCF of a and b.
If p is an integer such that p has only 1 and p itself as
its factors and p ≠ 1, then p is called a prime number,
like 2, 3, 5, 7, 11, 13, 19, 23, 29, 31, 37,.... are the
first few prime numbers.
Every composite number can be expressed
(factorised) as a product of primes and this
factorization is unique except for the order in
which the prime factors occur.
If a given positive integer is factorized, then its factors
will be prime or composite numbers. The composite
factors can again be further factorized into prime and
composite numbers. If this process is repeated, then
ultimately we will arrive at a stage where all the factors
are prime numbers. In other words, if we factorize any
number to its irreducible factors, then all the factors
are prime numbers. The Fundamental Theorem Of
Arithmetic is base on this conjecture, i.e. every positive
integer is either prime or it can be expressed as the
product of primes.
If we write the number as the product of powers of
primes and put them in a particular order whether
ascending or descending, then representation of the
number is unique. That means, every composite
number can be expressed as the products of powers of
primes in ascending or descending order in a unique
way.
Note
1. Every positive integer other than 1 is either prime
or composite.
2. 2 is the only even prime number
Example
Factorize the number 1176.
Proof
Let a be an integer.
Then, by Fundamental Theorem Of Arithmetic, integer a
can be factorized into its prime factors.
Let a = p 1 p 2 p 3 .....p n
Where
Prime factors of a
p 1 , p 2 , ...., p n
Now,
a = p1p2p3 ....pn
So,
1176 = 2 × 2 × 2 × 3 × 7 × 7
Or
⇒
a2 = (p1p2p3 ...pn ) (p1p2p3 ...pn )
⇒
a2 = p12p22p32 ....pn2
Here, p is prime number which divides a 2 . Therefore, p
is a prime factor of a 2 . But, the prime factorization of
any number is unique, so only prime factors of a 2 are
p 1 , p 2 , p 3 ,.....p n . Therefore, p is one of p 1 , p 2 ,
p 3 ,.....p n . That means,
p | p1p2p3 ...pn
⇒ p|a
Some Applications Of
Theorem Of Arithmetic
There are many applications
Theorem Of Arithmetic.
So,
1176 = 3 × 7 × 2 × 2 × 2 × 7
Or
So,
1176 = 7 × 2 × 3 × 7 × 2 × 2
The
Fundamental
Steps
1. Factorize each of the given positive integers and
express them as a product of powers of primes in
ascending order of magnitudes of primes.
2. For HCF, consider those prime factors which are
available in both the factorizations and identify
smallest (least) exponent of these common factors.
Multiply these factors to obtain HCF.
Or
For LCM, consider those prime factors which are
available in either of the factorizations and identify
largest (greatest) exponent of these common
factors. Multiply these factors to obtain LCM.
Relation Between HCF And LCM
The product of two integers is equal to the product if
their HCF and LCM. That means, if ‘a’ and ‘b’ are two
integers, then
HCF × LCM
a =
b
Theorems Related To Fundamental Theorem
Of Arithmetic
HCF × LCM
b =
a
Let p be a prime number and a be a positive
integer. If p divides a 2 , then p divides a.
of
Fundamental
1. Finding HCF And LCM Of Positive Integers
We use prime factorization method to find the HCF and
LCM of positive integers. For this, we use the
Fundamental Theorem Of Arithmetic in expressing the
given integers as the product of primes. In order to find
the HCF and LCM of two or more positive integers, we
may use the following algorithm.
In each factorization, the prime factors of 1176 are
same, although the order in which they appear are
different. Thus, the prime factorization of 1176 is
unique except for the order in which the primes occur.
In other words, if we factorize a number into its prime
factors, then order of factors may be different, but the
factors will be the same.
Fundamental Theorem of Arithmetic can be used to
derive other theorem as well which used in many
mathematical applications.
The
a × b = HCF × LCM
a×b
HCF =
LCM
a×b
LCM =
HCF
Or
Or
Or
Or
Proving Irrational Numbers
Example
Find the HCF and LCM of 90 and 144.
Using the prime factorization of 90 and 114
and
90 =2 × 32 × 5
We can prove irrational number of the form
‘n’ is a positive integer.
Example
114
= 24 × 32
For LCM, consider prime factors available in either with
their greatest exponents
Prime factors
Greatest exponents
2
4
3
2
5
1
∴ LCM is 24 × 32 × 51 = 16 × 9 × 5 = 720
2. Determining Type Of Decimal Form Of
Rational Number
We can use prime factorization to determine the nature
of decimal expansion of a rational number using The
Fundamental Theorem Of Arithmetic.
1.
Let
p
be a rational number, such that the prime
q
factorisation of q is of the form 2m × 5n , where m, n
p
are non-negative integers. Then
has a decimal
q
expansion which terminates.
Example
7 875
(i) = =
0.875
8 103
189 1512
(ii) = =
1.512
125
103
(iii)
2139 17112
×
=
1.7112
1250
10 4
2.
Let
⇒
( )
⇒
2=
17
17
(ii) = = 2.83333..........
6
2×3
 17 
 21 × 31 


1
= 0.142857142857..........
7
1
 71 
 
2
p
= 
 q
2
[Sq. both sides]
p2
q2
…………(i)
p2 = 2q2
So, 2 is factor of p
…………(ii)
Let p = 2a where ‘a’ is any integer
⇒
(p )
⇒
p2 = 4a2
⇒
2q2 = 4a2
⇒
2
q2 =
= ( 2a )
2
[Sq. both sides]
[From (i)]
2
4a
2
q2 = 2a2
So, 2 is also factor of q …………(iii)
From (i) and (iii)
2 is factor of both p and q. So, our assumption is
wrong.
⇒
∴
2 is an irrational number.
Note
Other type of cases to be solved differently. For
example,
(ii)
5
 31 
 
2
⇒
189 189 
125 = 53 


factorization of q is not of the form 2m × 5n , where
p
m, n are non-negative integers. Then,
has a
q
decimal expansion which is non-terminating but
repeating.
(ii)
Let
(i)
2139 
 2139
 1250 = 21 × 54 


2 is rational. So,
p
where p and q are integers with no other
2 =
q
common factor than 1
We assume
7 7 
 8 = 23 


p
be a rational number, such that the prime
q
Example
5
(i)
= 1.66666..........
3
2 is irrational
Prove
For HCF, consider prime factors available in both with
their smallest exponents.
Common prime factors
Least exponents
2
1
3
2
1
2
∴ HCF is 2 × 3 = 2 × 9 = 18
n where
3+2 5
1
2
Prove 5 has as rational number with
integer numerator and denominator
First rationalize the denominator and
convert it to case (i), then prove.
(ii)
2+ 5
First square both sides and convert it to
case (i), then prove.