lOMoAR cPSD| 49511909
COS1501/XOS1501/MO001
1
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, lOMoAR cPSD| 49511909
Activity 1-11:
1. Factorising: If we need to factorise an expression of the form x2 + ax + b or (x2 – ax – b or x2 +
ax – b or x2 – ax + b) we need to find some c and some d, such that a = c + d and b = (c)(d)
and (x + c)(x + d) = x2 + ax + b.
(a) x2 + 6x + 9 = x x + (3x + 3x) + (3)(3)
= (x + 3)(x + 3)
(b) x2 – x – 2 = x x + (x – 2x) + (1)(-2)
= (x + 1)(x – 2)
(c) x2 – 5x + 6 = x x – 2x –3x + (-2)(-3)
= (x – 2)(x – 3)
(d) x2 + 4x – 12 = x x – 2x + 6x + (-2)(6)
= (x – 2)(x + 6)
2. Solve x2 − 4x + 4 = 0 by factorising.
Well, at school we learnt that: the “+” in front of the last term means that our factorised version
x2 − 4x + 4 will be either of the form (x + ?)(x + ?) or of the form (x − ?)(x −? ), and that the “−” in
front of the middle term tells us that our factorised version must be of the form (x − ?)(x − ?).
Experimenting with numbers that, when multiplied, give us 4, we soon find that our factorised version has
to be (x − 2)(x − 2), or (x − 2)2 if you prefer.
Our equation x2 − 4x + 4 = 0 may therefore be rewritten as (x − 2)2 = 0.
By Property 9, at least one of the factors on the left-hand side must be zero, and both factors are
(x - 2), so we get that x − 2 = 0 ie
that x = 2.
3. Complete the square to solve x2 – 4x = 12.
If x2 − 4x = 12
then x2 − 4x + 4 − 4 = 12 (by Property 8, since 4 − 4 = 0)
ie x2 − 4x + 4 = 12 + 4 (by Property 6 with k = 4)
ie (x − 2)2 = 16 (factorise)
ie x − 2 = 4 or x − 2 = −4 (taking square roots)
ie x = 6 or x = −2 (using Property 6 again, with k = 2).
4. Is 21 a prime number?
No. Refer to the definition of prime numbers on p 16. The
numbers 3 and 7 are factors of 21 (3 × 7 = 21).
5. What is the value of 5! (5 factorial)?
5! = 5 × 4 × 3 × 2 × 1 = 120.
2
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COS1501/XOS1501/MO001
Study unit 2
Activity 2-8
1. Define the words “even” and “odd” for positive integers.
Definitions:
- An integer n is even if n is a multiple of 2.
(We can say a positive integer n is even if n = 2k for some positive integer k. You can think of even positive
integers as numbers n of the form n = 2k, where k is some positive integer.) - An integer n is odd if n is not
even.
(Using the general form of an even positive integer, we can now say that n is odd if n = 2k + 1 for some
positive integer k. You can think of odd positive integers as numbers n of the form n = 2k + 1, where k is
some positive integer.)
2. Is it the case that m + (n k) = (m + n)(m + k) for all positive integers m, n and k?
Substitute a few values and see whether the idea is plausible.
Take m = 1, n = 2, and k = 3, then the left-hand side is m +
(n k) = 1 + (2 3) = 7 while the right-hand side becomes
(m + n)(m + k) = (1 + 2) (1 +3) = 3 4 = 12.
This is a counterexample to show that it is not the case that m + (n k) = (m + n) (m + k) for all positive
integers m, n and k.
3. Are there any even prime numbers besides 2?
No. Any even prime number other than 2 would have three factors: 1, because 1 is a factor of every
number; 2, because the number we are talking about is supposedly even; and the number itself, because
a number is always a factor of itself. However, primes cannot have so many factors, which means that 2
is the only even prime number.
4. If m and n are even positive integers, is m + n even?
If m and n are even positive integers, then each is a multiple of 2, in other words
m = 2k for some k Z+. and n = 2j for some j Z+. So
m + n = 2k + 2j
= 2(k + j)
which means m + n is also even.
5. If m and n are odd positive integers, is m n odd?
If m and n are odd positive integers, then both m and n can be written in the following general form:
m = 2k + 1 for some k in Z+.
and n = 2j + 1 for some j in Z+.
So m n= (2k + 1)(2j + 1)
= 4kj + 2k + 2j + 1
= 2(2kj + k + j) + 1
which means that m n is odd.
3
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An additional exercise:
If m and n are prime, is m + n and m − n prime?
No, not usually.
It can occasionally happen that m + n is also prime: take m = 3 and n = 2 then m + n = 5.
But for other values m + n may not be prime: take m = 3 and n = 7, for instance. 3 + 7 = 10, which is not a
prime number.
What about the difference between two prime numbers? The difference m − n will sometimes be prime
and sometimes not. E.g. 5 – 3 = 2, which is a prime number, but 23 – 3 = 20, which is not prime.
5 Learning unit 3 – Sets
Study Material
Study guide
You should cover study unit 3 in the study guide. Where the study unit refers to the CAI tutorial, you should
also work through the relevant part in the CAI tutorial. The theory, examples and exercises will help you
understand the concepts. If you have not received your CAI tutorial with your study material, please see
the Orientation section for downloading instructions.
Time allocated
You will need one week to master this learning unit.
Notes
Background
The previous study units covered different number systems that you will come across in the remaining
units in this study guide. In this study unit, you are introduced to set theory. Set theory includes topics
involving sets. It is essential for your computing studies that you understand set theory concepts because
sets can be considered as fundamental building blocks for mathematics.
Applications
You should be familiar with all the notations and properties of sets. We use set notation to define
different kinds of sets. In this study unit, we build new sets from old ones. The Venn diagrams in the next
study unit will visually help you to understand the definitions in this study unit.
Activities
Do the activities provided in study unit 3 to consolidate your knowledge of the work in this learning unit.
Study unit 3
Activity 3-3
1. In each of the following cases, describe the set more concisely, firstly, using list notation and then
using set-builder notation.
4
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COS1501/XOS1501/MO001
1
Downloaded by Vincent master
()
, lOMoAR cPSD| 49511909
Activity 1-11:
1. Factorising: If we need to factorise an expression of the form x2 + ax + b or (x2 – ax – b or x2 +
ax – b or x2 – ax + b) we need to find some c and some d, such that a = c + d and b = (c)(d)
and (x + c)(x + d) = x2 + ax + b.
(a) x2 + 6x + 9 = x x + (3x + 3x) + (3)(3)
= (x + 3)(x + 3)
(b) x2 – x – 2 = x x + (x – 2x) + (1)(-2)
= (x + 1)(x – 2)
(c) x2 – 5x + 6 = x x – 2x –3x + (-2)(-3)
= (x – 2)(x – 3)
(d) x2 + 4x – 12 = x x – 2x + 6x + (-2)(6)
= (x – 2)(x + 6)
2. Solve x2 − 4x + 4 = 0 by factorising.
Well, at school we learnt that: the “+” in front of the last term means that our factorised version
x2 − 4x + 4 will be either of the form (x + ?)(x + ?) or of the form (x − ?)(x −? ), and that the “−” in
front of the middle term tells us that our factorised version must be of the form (x − ?)(x − ?).
Experimenting with numbers that, when multiplied, give us 4, we soon find that our factorised version has
to be (x − 2)(x − 2), or (x − 2)2 if you prefer.
Our equation x2 − 4x + 4 = 0 may therefore be rewritten as (x − 2)2 = 0.
By Property 9, at least one of the factors on the left-hand side must be zero, and both factors are
(x - 2), so we get that x − 2 = 0 ie
that x = 2.
3. Complete the square to solve x2 – 4x = 12.
If x2 − 4x = 12
then x2 − 4x + 4 − 4 = 12 (by Property 8, since 4 − 4 = 0)
ie x2 − 4x + 4 = 12 + 4 (by Property 6 with k = 4)
ie (x − 2)2 = 16 (factorise)
ie x − 2 = 4 or x − 2 = −4 (taking square roots)
ie x = 6 or x = −2 (using Property 6 again, with k = 2).
4. Is 21 a prime number?
No. Refer to the definition of prime numbers on p 16. The
numbers 3 and 7 are factors of 21 (3 × 7 = 21).
5. What is the value of 5! (5 factorial)?
5! = 5 × 4 × 3 × 2 × 1 = 120.
2
Downloaded by Vincent master ()
, lOMoAR cPSD| 49511909
COS1501/XOS1501/MO001
Study unit 2
Activity 2-8
1. Define the words “even” and “odd” for positive integers.
Definitions:
- An integer n is even if n is a multiple of 2.
(We can say a positive integer n is even if n = 2k for some positive integer k. You can think of even positive
integers as numbers n of the form n = 2k, where k is some positive integer.) - An integer n is odd if n is not
even.
(Using the general form of an even positive integer, we can now say that n is odd if n = 2k + 1 for some
positive integer k. You can think of odd positive integers as numbers n of the form n = 2k + 1, where k is
some positive integer.)
2. Is it the case that m + (n k) = (m + n)(m + k) for all positive integers m, n and k?
Substitute a few values and see whether the idea is plausible.
Take m = 1, n = 2, and k = 3, then the left-hand side is m +
(n k) = 1 + (2 3) = 7 while the right-hand side becomes
(m + n)(m + k) = (1 + 2) (1 +3) = 3 4 = 12.
This is a counterexample to show that it is not the case that m + (n k) = (m + n) (m + k) for all positive
integers m, n and k.
3. Are there any even prime numbers besides 2?
No. Any even prime number other than 2 would have three factors: 1, because 1 is a factor of every
number; 2, because the number we are talking about is supposedly even; and the number itself, because
a number is always a factor of itself. However, primes cannot have so many factors, which means that 2
is the only even prime number.
4. If m and n are even positive integers, is m + n even?
If m and n are even positive integers, then each is a multiple of 2, in other words
m = 2k for some k Z+. and n = 2j for some j Z+. So
m + n = 2k + 2j
= 2(k + j)
which means m + n is also even.
5. If m and n are odd positive integers, is m n odd?
If m and n are odd positive integers, then both m and n can be written in the following general form:
m = 2k + 1 for some k in Z+.
and n = 2j + 1 for some j in Z+.
So m n= (2k + 1)(2j + 1)
= 4kj + 2k + 2j + 1
= 2(2kj + k + j) + 1
which means that m n is odd.
3
Downloaded by Vincent master
()
, lOMoAR cPSD| 49511909
An additional exercise:
If m and n are prime, is m + n and m − n prime?
No, not usually.
It can occasionally happen that m + n is also prime: take m = 3 and n = 2 then m + n = 5.
But for other values m + n may not be prime: take m = 3 and n = 7, for instance. 3 + 7 = 10, which is not a
prime number.
What about the difference between two prime numbers? The difference m − n will sometimes be prime
and sometimes not. E.g. 5 – 3 = 2, which is a prime number, but 23 – 3 = 20, which is not prime.
5 Learning unit 3 – Sets
Study Material
Study guide
You should cover study unit 3 in the study guide. Where the study unit refers to the CAI tutorial, you should
also work through the relevant part in the CAI tutorial. The theory, examples and exercises will help you
understand the concepts. If you have not received your CAI tutorial with your study material, please see
the Orientation section for downloading instructions.
Time allocated
You will need one week to master this learning unit.
Notes
Background
The previous study units covered different number systems that you will come across in the remaining
units in this study guide. In this study unit, you are introduced to set theory. Set theory includes topics
involving sets. It is essential for your computing studies that you understand set theory concepts because
sets can be considered as fundamental building blocks for mathematics.
Applications
You should be familiar with all the notations and properties of sets. We use set notation to define
different kinds of sets. In this study unit, we build new sets from old ones. The Venn diagrams in the next
study unit will visually help you to understand the definitions in this study unit.
Activities
Do the activities provided in study unit 3 to consolidate your knowledge of the work in this learning unit.
Study unit 3
Activity 3-3
1. In each of the following cases, describe the set more concisely, firstly, using list notation and then
using set-builder notation.
4
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