IB MATH HIGHER OPTION
DISCRETE MATHEMATICS
There is no doubt that the options in the Higher syllabus are a
stern test of both your knowledge and understanding of more
advanced mathematics than you will meet in the core syllabus.
This revision booklet has been designed to cover most aspects of
the option syllabus along with a variety of examples which show
how the concepts can be applied. But I cannot hope to cover every
aspect of every topic in this booklet, and it is to be regarded as a
supplement to such text books and notes as you have been given
during the course.
Many candidates fall down on the option questions because they
have not put enough preparation in. The practice questions after
each section test your knowledge of the syllabus: you should try to
answer all of them. If you have difficulty with any of them, talk to
your teacher or fellow students, read your notes and text books,
and then come back and try again. And do practice on as many
past paper questions as you have time for. Be aware that any
papers sat prior to May 2012 will not contain questions on
Recurrence Relations, the Pigeon-hole Principle and Chinese
Postman problems with four vertices of odd degree.
The following ideas should help you answer questions with more
accuracy and more confidence:
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The option syllabus does not progress from the core
syllabus – it is very much a stand-alone topic. However,
questions will be designed to start at a reasonably low level
before beginning to stretch you quite considerably
There is a great deal to remember in this option: all sorts of
theorems and techniques in the number theory, methods
and algorithms in the graph theory. Learn, learn and learn
again; it is a waste of marks if you fail to answer a question
simply because you haven't learnt the method properly.
You will be required to do proofs. Make sure you can
construct proofs so that you cover every possibility, even the
obvious ones.
Do read the questions in the exam very carefully. Often
there are clues and hints as to how to proceed. Additionally,
you can easily go wrong if you didn't pick up on an important
point. And when you have completed a question, check that
you have answered exactly what was asked.
IBDP Matematics HL – Discrete Mathematics Option
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Elementary Number Theory
Pigeon-hole Principle
This can be simply stated. If I have n pigeon-holes and N pigeons
in them, and if N>n then there must be at least one pigeon-hole
with more than one pigeon in it.
Use the pigeon-hole principle to prove that in a school of 380 there
will be at least two people with the same birthday.
Let the days of the year be the pigeon holes. We have 366 pigeon holes
and 380 students so there must be at least 2 who share the same
birthday.
Six distinct numbers are taken from the digits 1 to 10. Prove that
two of them must be add up to 11.
There are 5 pairs of numbers that will sum to 11, {1,10},....{5,6} . Let
these be the 5 pigeon-holes, If I select 6 numbers and place them in their
matching pigeon-hole then there must be at least one with 2 numbers in.
Factors and division
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Much of the emphasis in this section is on proof. The basic
principles are not hard to understand, but it is important that you
do understand them thoroughly (and their associated proofs) so
that you can then build more complex proofs on this base.
The notation a|b stands for “a divides b”; that is, a is a factor of b.
We can rewrite this as the equation b=ax for some x ∈ ℤ.
Theorem
a|b and a|c ⇒ a|(bx + cy) where x, y ∈ ℤ.
For example, 3 is a factor of both 12 and 21, and it is also a factor
of 5 × 12 + 7 × 21. This seems fairly obvious because we must be
able to take 3 out of the expression as a common factor:
5 × 12 + 7 × 21 = 3(5 × 4 + 7 × 7)
but how can we show that it is always true? The answer is to use
algebra.
If a|b then there is an integer p such that b = ap
If a|c then there is an integer q such that c = aq
So bx + cy = apx + aqy = a(px + qy)
a|(bx + cy) since px + qy ∈ℤ.
When
you
divide
two
numbers, the divisor divides
into the dividend. The
number of times it goes in is
the quotient and, if it does not
go exactly, there will be a
remainder.
Before going further, make sure you are familiar with the words
shown in the notes box on the left. The process of division can be
stated formally using the division algorithm (which is not really an
algorithm at all).
If a and b are integers (b ≠ 0), then there are unique
integers q and r such that a = qb + r, 0 ≤ r ≤ |b|.
For example, if a = 38 and b = 5, then 38 = 7 × 5 + 3. The quotient
is 5 and the remainder is 3. Note that the remainder is positive
and must be less than the divisor (when dividing by 5, you can only
get remainders up to 4). The modulus sign is there to allow for the
case of a negative divisor.
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IB Math Higher – Discrete Mathematics Option
For example if -4 is divided by 3 we get −4 = −2 × 3 + 2 so the
remainder is 2
Modular arithmetic appears on page 8 but note for now that if
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a = qb + r then it is true that a ≡ r (mod b). This is because a is the
same amount above a multiple of b as r is above 0.
Fundamental theorem of arithmetic
Prime numbers are those numbers which have two distinct factors,
1 and the number itself. Thus, 1 is not a prime number. The
fundamental theorem of arithmetic states that:
Any positive integer greater than 1 is either prime or can
be written as a unique product of prime numbers.
A key point is the uniqueness of the factorisations – each number
can only be written as a product of primes in one way.
Examples:
15 = 3 × 5
45 = 32 × 5
100 = 22 × 52
It is because of this theorem that prime numbers are thought of as
the “building blocks” of arithmetic.
If two numbers do not share any factors other than 1 then they are
known as relatively prime. Thus 4 and 15 are relatively prime,
whereas 4 and 14 are not (because they share the factor 2). Note,
for example, that though 9 is relatively prime to 14, and 14 is
relatively prime to 15, 9 is not relatively prime to 15.
Note that relatively prime
numbers are not necessarily
prime themselves. Prime
numbers, of course, do not
share any factors except 1 –
so two prime numbers will
always be relatively prime.
When writing a fraction in its simplest terms, we are in fact dividing
the numerator and denominator by common factors until it is no
longer possible to do so. When this occurs, the numerator and
denominator will be relatively prime.
gcd, lcm and Euclid’s algorithm
The greatest common divisor (gcd) of two integers is the largest
integer which divides into both. For example, gcd(15, 20) = 5. The
least common multiple is the smallest number which is a multiple
of both. So lcm(15, 20) = 60. It is possible to use the prime
factorisation of numbers to find the gcd and lcm.
a and b will be relatively
prime if gcd(a, b) = 1, since
there are no common factors.
Find gcd(56, 84) and lcm(56, 84)
3
2
56 = 2 × 7, 84 = 2 × 3 × 7
The gcd is found by multiplying the lower power of each factor which is
2
common to the two numbers. So gcd(56, 84) = 2 × 7 = 28
The lcm is found by multiplying the higher power of each factor so as to
3
include all factors. So lcm(56, 84) = 2 × 3 × 7 = 168.
The process is quite lengthy for large numbers, and Euclid’s
algorithm provides a quicker way of calculating the gcd.
If r is the remainder when a is divided by b then
gcd(a, b) = gcd(b, r)
This follows from the fact that
a=qb+r so any divisor of a
and b must also divide r
In practice, we divide the higher number by the lower number and
work out the remainder. We then repeat the process with the lower
IBDP Matematics HL – Discrete Mathematics Option
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