Chapter 2 Division and Euclidean Algorithm
Chapter 2 Division and Euclidean Algorithm
SECTION A Division Algorithm
By the end of this section you will be able to

understand what is meant by the division algorithm

apply the division algorithm
A1 Introduction to Division Algorithm
We all know how to divide two numbers. For example 25 divided by 4 can be written as:
6 Rem 1
4
25
24
1
We can write this as
25  6  4  1
The number 6 is called the quotient and 1 the remainder. Let us place a geometric
interpretation of this example:
Figure 1
Of course the above can also be written as:
25  7  4  3
25  5  4  45
25  1 4  21
There are an infinite number of ways we can write 25 as a multiple of 4 plus a remainder.
This is of no use to us because we want a unique way of writing this. How can we achieve
this?
We place a restriction on the remainder – the remainder is greater than or equal to zero but
less than 4. With this restriction we have a unique way of dividing 25 by 4 which is our first
answer, 25  6  4  1 .
Example 1
Express the following numbers in terms of quotient and remainder:
(a) 27 by 5
(b) 365 by 7
(c) 159 by 3
Let us generalize the above results. Let a and b  0 be our given integers. Then there exists
a quotient q and a remainder r such that
a  bq  r
If we restrict our reminder r such that 0  r  b then our expression a  bq  r is unique.
We can place a geometric view on this as follows:
1
Chapter 2 Division and Euclidean Algorithm
2
Figure 2
We are now in a position to write down the general Division Algorithm.
Theorem (2.1). Division Algorithm.
Given integers a and b  0 there exists unique integers q called the quotient and r called the
remainder such that
a  bq  r
0r b
We need to prove this result. How?
By showing existence of q and r and also uniqueness of these integers.
Proof.
Existence
Let S be the set S  a  xb x  and a  xb  0 . We first show that the set S is nonempty. How?
Select an integer x so that a  xb  0 . Let x   a then
a  xb  a  a b

a 0
Because b 1
Hence the set S is non – empty. We can now apply the Well Ordering Principle:
Well Ordering Principle
Let S be a non-empty set of positive integers and zero. Then the set S has a least element.
This means that x  S , l  S such that x  l .
There exists a least element of the set S, say r which of course is an integer. We have r  S
which means there exists an integer x  q such that
r  a  qb  0
We also need to show that r  b . How?
By contradiction. Suppose r  b . Then r  b  0 which means it is a member of the set S
because S  a  xb x  and a  xb  0 . We are given that b  0 [positive] so
r b  r
This is a contradiction because r is the smallest element of S. Hence r  b .
Uniqueness
How do we prove that the integers q and r are unique?
Again by contradiction. Suppose there also exists integers q ' and r ' such that
a  bq ' r '
0 r'b
We already have
a  bq  r
0r b
Then a  bq  r  bq ' r ' where 0  r  b and 0  r '  b . Subtracting these two integers
gives
0  bq  bq '  r  r '
Re-arranging this gives and taking the modulus gives
r r'  b q q'
We have the inequalities 0  r  b and  b  r '  0 . Hence b  r  r '  b or equivalently
Chapter 2 Division and Euclidean Algorithm
3
r r' b
However b q ' q is a multiple of b so we must have q ' q  0  q '  q . Substituting this
into the above r  r '  b  q ' q  gives r  r '  0  r  r ' .
Hence the integers q and r are unique.
■
Example 2
Show that
a  a2  2
3
is an integer for every integer a .
Solution
How do we prove this result?
By applying the Division Algorithm with b  3 and any integer a:
a  3q  r
0r 3
This means the remainder r  0, 1 or 2 . We have
a  3q, 3q  1 and 3q  2
Substituting each of these into the given number
a  a2  2
3
Substituting a  3q :
3 q  9q 2  2 
3
Substituting a  3q  1 :
 3q  1   3q  1
2
2
3
 q  9q 2  2 
  3q  1 9q
2
 6q  1  2 
3
 3q  1  9q  6q  3
2


Substituting a  3q  2 :
 3q  2    3q  2 
2
3
2
3
3  3q  1  3q 2  2q  1
3
  3q  1 9q
2

a  a  2
 12q  4  2 
3
3  3q  1  3q  4q  2 
2
3
  3q  1  3q 2  2q  1
  3q  1  3q 2  4q  2 
2
Hence in each case
3
is an integer. For any every integer a we have
a  a2  2
integer.
Corollary (2.2).
Given integers a and b with b  0 there exists unique integers q and r such that
a  bq  r
0r  b
3
is an
Chapter 2 Division and Euclidean Algorithm
4
Proof.
If b is positive then we are done because we have the result by the Division Algorithm. If b is
negative, b  0 , then we consider b  0 . Applying the Division Algorithm to a and b  0
we have
a  b qr
0r  b
 b if b  0
Remember b  
.
b if b  0
In the case where b  b we can write the above expression
a  b q  r  bq  r

bq ' r
0r  b
Let q ' q
This completes our proof.
■
Example 3
Show that the cube of any integer is of one of the following forms:
9k , 9k  1, 9k  2
Solution
How do we prove this result?
By applying the Division Algorithm with b  9 :
n  9q  r
0r 9
Taking the cube of this gives
3
3
2
n 3   9q  r 
  9q   3  9q   3  9 q   r 3  9 k  r 3
By binomial
What values can r take?
r  0, 1, 2, 3,
r 0 , 1 , 2 , 3 ,
3
9 k
3
3
3
3
, 8
3
, 8
 0, 1, 8, 27  9  3 , 64  3  21  1, 125  3l  8,
3
r can only take the form 0, 1 and 8. Hence
n3  9k  r 3  9k , 9k  1 or 9k  8
Example 4
Show that 3a 2  1 is never a perfect square.
[Hint: The square of any integer is either of the form 3k or 3k  1 .]
Solution
By hint every square is of the form 3k or 3k  1 . Suppose 3a 2  1 is a perfect square. Then
3a 2  1  3k
3 a2  k   1
This is impossible because 3  integer   1 . Similarly we have
3a 2  1  3k  1
3 a2  k   2
We cannot have 3  integer   2 . Hence 3a 2  1 is never a perfect square.
Chapter 2 Division and Euclidean Algorithm
If
5