MTH4105 Introduction to
Mathematical Computing
Chapter 9 supplement
School of Mathematical
Sciences
© 2015 by Dr Francis J. Wright
Division and divisibility
For some non-examinable optional extra intellectual stimulation, this supplement
explores the division operation and how it leads to the important relation of divisibility.
Division and divisibility: numer, denom
In general, the result of dividing two integers or two polynomials is a fraction, which
2
2
x
consists of a numerator divided by a denominator, e.g.
or
. Interactively, you
x C1
3
can select the numerator or denominator of a fraction in the same way that you can
select any other text, but Maple also provides functions for this purpose, called
numer and denom, which work like this:
2
2
2
x
x
2
= 2, denom
= 3, numer
= x2, denom
= x C1.
numer
3
x C1
x C1
3
Division becomes much more interesting if we require that the result must be in the
original set, i.e. the result of dividing two integers must be an integer and the result of
dividing two polynomials must be a polynomial. In general, this division will not be
possible unless we allow a remainder to be left over.
In Maple, divisibility functions that apply only to integers have names that begin with
"i" to distinguish them from those that apply to polynomials. We have already seen
this with ifactor and factor and we will see some more examples below.
Integer division: iquo, irem
Integer division is a straightforward generalization of the division of natural numbers
considered in Mathematical Structures.
Let a, b be integers. We say that b divides a, or a is divisible by b, which we write
symbolically as b a, if there is an integer q such that a = b q. We call a, b, q
respectively the divisor, dividend and quotient. Note that this divisibility test is a
relation and the value of the expression b a is either true or false. It is clear from the
definition that b 0 for all b because 0 = b$0, but 0 § a unless a = 0 because
a s 0$q for any finite q.
In general, the relation a = b q does not hold, but provided b s 0 it is always true
that a = b q Cr for some integer r, which is what would be left over if we were to
divide a by b, so we call r the remainder. We can impose conditions on r that make q
and r unique.
Page 1 of 9
28-Nov-2015
Integer quotient and remainder are implemented in Maple as the functions iquo and
irem such that q = iquo a, b and r = irem a, b satisfy the equation a = b q Cr and
the conditions that r ! b and the sign of r is the same as the sign of a make the
values of q and r unique. For example:
> a, b d 7, 3
a, b := 7, 3
(1.2.1)
> q, r d iquo a, b , irem a, b
q, r := 2, 1
> evalb a = b q Cr , evalb
r ! b , evalb sign r = sign a
true, true, true
> a, b dK7, 3
> q, r d iquo a, b , irem a, b
> evalb a = b q Cr , evalb
(1.2.2)
(1.2.3)
a, b := K7, 3
(1.2.4)
q, r := K2, K1
(1.2.5)
r ! b , evalb sign r = sign a
true, true, true
(1.2.6)
There appears to be no way to use the vertical bar notation actively for divisibility, but
we could implement an infix integer divisibility predicate and test it like this:
> `&divides` d b, a /is irem a, b = 0
&divides := b, a /is irem a, b = 0
(1.2.7)
> 3 &divides 7, 3 &divides 6
false, true
(1.2.8)
Quiz
Given two integers a, b, which of the following evaluates to true if and only if a is an
integer multiple of b?
is iquo b, a = 0
No. iquo b, a is the integer quotient of b divided by a. This is the wrong division
function and the division is the wrong way around!
> iquo 3, 9 , iquo 3, 10
0, 0
(1.2.1.1.1)
is iquo a, b = 0
No. iquo a, b is the integer quotient of a divided by b. This is the wrong division
function.
> iquo 9, 3 , iquo 10, 3
3, 3
(1.2.1.2.1)
is irem b, a = 0
No. irem b, a is the integer remainder of b divided by a. This division is the
wrong way around!
> irem 3, 9 , irem 3, 10
3, 3
(1.2.1.3.1)
is irem a, b = 0
Yes. irem a, b is the integer remainder of a divided by b. If this is zero then b
divides a.
> irem 9, 3 , irem 10, 3
0, 1
(1.2.1.4.1)
Page 2 of 9
28-Nov-2015
Exercises
Assign the number 12345 to the variable n. It is obvious that 5 divides n. (Why?)
Use irem to confirm that 5 is a divisor of n and then use iquo to assign the quotient
of n divided by 5 back to the variable n. It is now obvious (even more so than
before) that 3 divides n. (Why?) Use irem to confirm that 3 is a divisor of n and then
use iquo to assign the quotient of n divided by 3 back to the variable n. Show that
n is now prime and hence write down the prime factorization of 12345.
>
Solution
> n d 12345
n := 12345
It is obvious that 5 divides n because the last digit of n is 5.
> irem n, 5
0
(1.2.2.1.1)
(1.2.2.1.2)
> n d iquo n, 5
n := 2469
(1.2.2.1.3)
It is obvious that 3 divides n because the sum of the digits, which is equal to
6 C6 C9, is divisible by 3.
> irem n, 3
0
(1.2.2.1.4)
> n d iquo n, 3
n := 823
(1.2.2.1.5)
true
(1.2.2.1.6)
> isprime n
Hence the prime factorization of 12345 is 3 # 5 # 823. Check:
> ifactor 12345
3 5 823
Suppose that the variable f evaluates to a vulgar fraction, e.g. f =
(1.2.2.1.7)
7
. Write an
3
1
, except that the nearest
3
1
one can get in Maple output is an expression sequence such as 2, . Generate
3
this expression sequence as the integer quotient of the numerator and
denominator followed by the integer remainder of the numerator and denominator
divided by the denominator to give a proper fraction. Define a function called
mixed that performs the conversion of a vulgar fraction to a mixed fraction
represented as an expression sequence, and test it for a few different vulgar
fractions.
expression that converts f to mixed fraction form, e.g. 2
>
Solution
> fd
7
3
f :=
7
3
Page 3 of 9
(1.2.2.2.1)
28-Nov-2015
The integer part is give by
> iquo numer f , denom f
2
(1.2.2.2.2)
1
3
(1.2.2.2.3)
The proper fraction part is give by
irem numer f , denom f
>
denom f
This is the obvious definition of the function mixed:
> mixed d f/ iquo numer f , denom f
,
irem numer f , denom f
denom f
mixed := f/ iquo numer f , denom f
,
(1.2.2.2.4)
irem numer f , denom f
denom f
This is a more succinct definition that applies an expression as a mapping:
irem
> mixed d f/ iquo,
numer f , denom f
denom f
irem
mixed := f/ iquo,
numer f , denom f
(1.2.2.2.5)
denom f
> mixed
> mixed
7
3
2,
1
3
(1.2.2.2.6)
3,
1
2
(1.2.2.2.7)
7
2
Integer greatest common divisor and least common multiple: igcd,
ilcm
The greatest common divisor (or highest common factor) of two integers a and b,
normally written in mathematics as gcd a, b , is the unique nonnegative integer d
such that
• d a and d b;
• if e is any integer such that e a and e b, then e d.
Although it is not immediately obvious from this definition, gcd a, b is the largest
integer that divides both a and b unless a = b = 0. Integer gcd is implemented in Maple
as the function igcd, which works like this:
> igcd K15, 9
3
(1.3.1)
Note that
> igcd 0, 0
0
(1.3.2)
Page 4 of 9
28-Nov-2015
It is clear from the definition that we can replace integers by their absolute values
when computing gcds. Euclid’s Algorithm computes gcd a, b for two nonnegative
integers a and b as follows:
a) if b = 0 then gcd a, b = a;
b) if b s 0, divide a by b to find q and r such that a = b q Cr and 0 % r ! b. Then
calculate gcd b, r ; the result is equal to gcd a, b .
This is a recursive algorithm or function definition because it refers back to itself: it
computes a gcd in terms of a simpler gcd. We will consider recursive functions in
more detail next week. But this particular algorithm just involves repeatedly computing
remainders and returning the last non-zero remainder. For example, to compute
gcd K15, 9 first replace both numbers by their absolute values to give gcd 15, 9 .
Then compute
> irem 15, 9
6
(1.3.3)
> irem 9, 6
3
(1.3.4)
0
(1.3.5)
> irem 6, 3
and return 3 as the result.
A concept related to the greatest common divisor is the least common multiple or lcm
of two integers, which is the smallest nonnegative integer that is a multiple of both of
them. It is used primarily to put a sum of fractions over a common denominator, which
should be the lcm of the two denominators; any other common denominator would be
bigger than necessary. Integer lcm is implemented in Maple as the function ilcm and
works like this:
> ilcm K15, 9
45
(1.3.6)
ab
Note that, generally, lcm a, b =
, e.g.
gcd a, b
ab
> ilcm a, b =
igcd a, b
a =K15, b = 9
21 = 21
(1.3.7)
The gcd and lcm of more than two integers are defined in the obvious way and the
Maple functions igcd and ilcm accept any number of arguments. You can also use the
Maple functions gcd and lcm with integer arguments, but they only accept two inputs.
They are intended primarily for use with polynomials (see the Maple help for details)
and any arguments after the second are not used for input.
Quiz
You are given the following chain of computations:
> irem 49, 15
4
(1.3.1.1)
> irem 15, 4
3
(1.3.1.2)
> irem 4, 3
Page 5 of 9
(1.3.1.3)
28-Nov-2015
1
(1.3.1.3)
> irem 3, 1
0
(1.3.1.4)
What is the greatest common divisor of 49 and K15, i.e. the value of gcd 49,K15 ?
Is it G4,G3,G1, 0?
It is C1 because it is the last non-zero value in the remainder sequence,
starting from the absolute values.
> igcd 49,K15
1
(1.3.1.1.1)
Exercise
Use successive remainder computation (Euclid's Algorithm) to show that 1234 and
115 are coprime, i.e. that their gcd is 1, and check this directly using the Maple
igcd function.
>
Solution
> irem 1234, 115
84
(1.3.2.1.1)
31
(1.3.2.1.2)
22
(1.3.2.1.3)
9
(1.3.2.1.4)
4
(1.3.2.1.5)
1
(1.3.2.1.6)
1
(1.3.2.1.7)
> irem 115, 84
> irem 84, 31
> irem 31, 22
> irem 22, 9
> irem 9, 4
> igcd 1234, 115
Synoptic exercises
Attempt the following in the boxes provided or in a new Maple worksheet (or
document).
Question 1
Compute q = iquo a, b , r = irem a, b and verify that a = b q Cr, r ! b and r has
the same sign as a for a few random pairs of integers a, b. The neatest way to do
this is to use rand and a loop to run through a few tests. Review Chapter 7 as
necessary. Take care to avoid the case b = 0 because division by 0 is an error.
Your answers
>
My answers
> R d rand K9 ..9 :
to 3 do
Page 6 of 9
28-Nov-2015
a, b d R , R ;
if b = 0 then next end if;
# this is one fairly elegant way to avoid division by 0
q, r d iquo a, b , irem a, b ;
is a = b q Cr , is r ! b , is r a R 0
end do
a, b := 5, 3
q, r := 1, 2
true, true, true
a, b := K4, K8
q, r := 0, K4
true, true, true
a, b := 2, K6
q, r := 0, 2
true, true, true
(1.4.1.1.1)
Question 2
Implement Euclid's Algorithm to compute the gcd of the values of the variables a
and b using a loop as follows. Review Chapter 7 as necessary.
Assign the values 1234 and 115 to the variables a and b. Then in a loop, while b is
nonzero, replace a by b and replace b by the remainder in the division of a by b.
Use parallel assignment or introduce temporary variables as necessary. When the
loop terminates, display the value of a, which should be the gcd required. Modify
your code so that it always returns a nonnegative gcd, and test it with some
negative input values.
Your answers
>
My answers
> a, b d 1234, 115;
while b s 0 do
a, b d b, irem a, b
end do:
gcd = abs a
a, b := 1234, 115
gcd = 1
(1.4.2.1.1)
Question 3
This question is about prime factorization of an integer using a loop. Review
Chapter 7 as necessary.
Assign the integer 123 to the variable n and use ifactor to find the complete prime
factorization of n. Display the result neatly. Assign 0 to the variable p. Write a loop
that executes while p % n. In the loop, increment p to the next larger prime and if p
divides n then print out an appropriate message. Do all of the above in the same
execution group. In this way, you should be able to compute the prime
factorization of 123.
Copy the code you have written to a new execution group and modify it so that it
Page 7 of 9
28-Nov-2015
constructs a sequence of the prime factors and displays this sequence at the end
of the loop, without displaying any intermediate values.
Copy the code you have written to a new execution group, replace 123 by 10!, and
execute the code. Note that the code runs for a long time and does not find the
powers of the prime factors, only the primes themselves.
Copy the code you have written to a new execution group and modify it so that,
instead of testing whether p divides n, while p divides n it collects each p that
divides n into a sequence of factors and then replaces n by the quotient of n divided
by p. Note that this code runs very much faster.
Optionally, instead of collecting multiple copies of the same factor, compute its
power and display the full prime factorization elegantly.
Your answers
>
My answers
> n d 123 : n = ifactor n ;
pd0:
while p % n do
p d nextprime p ;
if irem n, p = 0 then print p, `is a factor of`, n end if
end do:
123 = 3 41
3, is a factor of, 123
41, is a factor of, 123
(1.4.3.1.1)
> n d 123 : n = ifactor n ;
p d 0 : S d NULL :
while p % n do
p d nextprime p ;
if irem n, p = 0 then S d S, p end if
end do:
`The prime factors of`, n, `are`, S
123 = 3 41
The prime factors of, 123, are, 3, 41
> n d 10!: n = ifactor n ;
p d 0 : S d NULL :
while p % n do
p d nextprime p ;
if irem n, p = 0 then S d S, p end if
end do:
`The prime factors of`, n, `are`, S
3628800 = 2 8 3 4 5 2 7
The prime factors of, 3628800, are, 2, 3, 5, 7
(1.4.3.1.2)
(1.4.3.1.3)
> n d 10!: n = ifactor n ;
p d 0 : S d NULL :
while p % n do
Page 8 of 9
28-Nov-2015
p d nextprime p ;
while irem n, p = 0 do
S d S, p;
n d iquo n, p
end do
end do:
`The prime factors are`, S
8
4
2
3628800 = 2
3
5
7
The prime factors are, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 5, 5, 7
(1.4.3.1.4)
> n d 10!: n = ifactor n ;
p d 0 : S d NULL :
while p % n do
p d nextprime p ;
for pow from 0 while irem n, p = 0 do
n d iquo n, p
end do;
if pow O 0 then S d S, p, pow end if; # don't include zero powers
end do:
`The prime factors are`, S;
mul `` s1
s
2
, s in S
3628800 = 2 8 3 4 5 2 7
The prime factors are, 2, 8 , 3, 4 , 5, 2 , 7, 1
2
8
3
4
Page 9 of 9
5
2
7
(1.4.3.1.5)
28-Nov-2015