Contents
1 Algebra
1
1.1
Logic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
1
1.2
Propositional Functions. Quantifiers . . . . . . . . . . . . . . . . . . . . . . . . . . .
8
1.3
Methods of Proof . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
9
1.3.1
Proofs, Disproofs, Counter-examples . . . . . . . . . . . . . . . . . . . . . . .
11
1.4
Set Theory. Basic Ideas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
12
1.5
Set Theory. Further Properties . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
15
1.6
Number Systems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
20
1.7
Indices . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
1.8
Multiplication and Factorization . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
29
1.9
Quadratic Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
32
1.10 Inequalities . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
37
1.11 Absolute Values . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
44
1.12 Sigma Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
48
1.13 Mathematical Induction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
51
1
Chapter 1
Algebra
1.1
Logic
All of mathematics is based on the idea that one can proceed from one statement to another in a
logical way. Before we can start then we shall have to examine the logical foundations of mathematics.
Even this will not be done completely rigorously.
Propositions
Definition 1.1.1 A proposition is a statement that is either true or false.
Thus the following are propositions :
(i) 1 + 1 = 2 (true)
(ii) 1 + 1 = 3 (false)
The following are not propositions :
(iii) Hello! (This has no truth value).
(iv) x = 3. (We cannot say whether this is true or false without further information about x).
One can build compound propositions out of simple propositions in a number of ways. Suppose
then p and q are propositions.
Or
Definition 1.1.2 Suppose p and q are propositions. The proposition p or q is true if p is true, or
q is true or both are true; otherwise it is false. This is written p ∨ q.
Thus, when we say “p or q ” we include the possibility that both are true, and require that at least
one of p and q are true. In mathematics “or” is always used in this sense. This is often written in
common language as “and/or”.
Thus the following are true:
(i) 1 + 1 = 2 or 5 + 2 = 7
(ii) 1 + 1 = 2 or 5 + 2 = 4
1
2
CHAPTER 1. ALGEBRA
(iii) 1 + 1 = 3 or 5 + 2 = 7,
while
(iv) 1 + 1 = 3 or 5 + 2 = 4 is false.
Truth-Tables
This is most simply illustrated in a truth-table. Here we let T stand for “true” and F for “false”.
p
T
T
F
F
q
T
F
T
F
p∨q
T
T
T
F
This is to be read as follows :
The first line says that p ∨ q is true when both p and q are true, the second that p ∨ q is true when
p is true and q is false, and so on.
And
Definition 1.1.3 Let p and q be propositions. The proposition p and q is true if p and q are both
true; otherwise it is false. This is written p ∧ q.
The corresponding truth-table is
p
T
T
F
F
q
T
F
T
F
p∧q
T
F
F
F
For example,
(i) 1 + 1 = 2 and 5 + 2 = 7
is true while
(ii) 1 + 1 = 2 and 5 + 2 = 4
(iii) 1 + 1 = 3 and 5 + 2 = 7,
(iv) 1 + 1 = 3 and 5 + 2 = 4
are all false.
Negation
Definition 1.1.4 Let p be a proposition. Then the negation ¬p of p (read as “not p”) is the
proposition which is true when p is false and false when p is true.
3
1.1. LOGIC
The idea is very simple. For example, if p is the proposition “Lusaka is the capital of Zambia”, the
¬p is “Lusaka is not the capital of Zambia”.
The truth-table embodies the idea that if p is true then ¬p is false, while if p is false then ¬p is
true.
p
T
F
¬p
F
T
Thus if p is the proposition “1 6= 2” then ¬p is the proposition “1 = 2”.
Implication
A great many statements in mathematics are of the form “If p is true, it follows that q is true”.
More simply we may say “p implies q”. We wish to define this more formally.
Definition 1.1.5 Let p and q be propositions. Then p =⇒ q (read as “p implies q”) is the proposition which has the truth-table below.
p
T
T
F
F
q
T
F
T
F
p =⇒ q
T
F
T
T
The last two lines in the truth-table may seem strange. If we agree that they must have some
truth value, it is easy to see that both must be T .
Indeed, let p be the (false) proposition that 1 = 2, and let q be the true proposition 2 = 2. It is
easy to deduce q from p, so that p =⇒ q must be true. To see this, simply multiply both sides of
the equation 1 = 2 by zero, and then add 2. Thus the last entry on the third line of the truth-table
must be T .
On the other hand, let p be the (false) proposition that 1 = 2, and first let q be the false
proposition 2 = 3. Again one can deduce q from p by adding 1, so that again p =⇒ q must be true.
Hence the last entry on the bottom line of the truth-table must also be T .
The upshot is that p =⇒ q is true if p is true and q is true and also true whenever p is false
whatever the value of q. It is sometimes said that “a false proposition implies anything.”
For example the following are true :
(i) 1 + 1 = 2 =⇒ 2 + 2 = 4
(ii) 1 + 1 = 3 =⇒ 2 + 2 = 4
(iii) 1 + 1 = 3 =⇒ 2 + 2 = 5.
The important thing to realize is that if p is true and q is false, then p =⇒ q is false.
One cannot proceed from a true statement to a false one.
Moreover, it should be understood that the statements p =⇒ q and q =⇒ p are quite different.
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CHAPTER 1. ALGEBRA
Thus
x > 0 =⇒ x2 > 0
is true, while
x2 > 0 =⇒ x > 0
is false, since, for example, x might be −1.
We emphasize again the vital fact:
p =⇒ q is not the same as q =⇒ p
As another example consider the statements “This is an apple” and “This is a fruit”. Obviously
the statement
This is an apple =⇒ This is a fruit
is true, but the statement
This is a fruit =⇒ This is an apple
is false.
Converse
Definition 1.1.6 If p =⇒ q is a proposition, then its converse is the proposition q =⇒ p.
As we have seen above, p =⇒ q may be true, while its converse q =⇒ p may be false. For
example, if n is a positive integer
n is even
=⇒
2n is even
is certainly true, since whenever the left-hand side is true, so is the right. On the other hand, the
converse
2n is even =⇒ n is even
is false, since the left-hand side is true no matter what n is, but the right-hand is not – for instance
n might be 3.
However, there are many statements in mathematics which are true and whose converse is also
true. For example, suppose △ ABC is a triangle. Then the proposition
AC = BC
=⇒
∠A = ∠B
is true, since the base angles of an isoceles triangle are equal. The converse is
∠A = ∠B
=⇒
AC = BC
is also true, since if the base angles are equal, then the triangle is isoceles.
In this case, if p is AC = BC and q is ∠A = ∠B, then
p =⇒ q
and q =⇒ p.
It is convenient to use the following additional notation :
p ⇐⇒ q means p =⇒ q and q =⇒ p.
5
1.1. LOGIC
Iff
We say that p is true if and only if q is true. This is frequently abbreviated to p iff q. Let us
examine the relevant truth table, keeping in mind the truth-table for p =⇒ q.
p
T
T
F
F
q
T
F
T
F
p =⇒ q
T
F
T
T
q =⇒ p
T
T
F
T
p ⇐⇒ q
T
F
F
T
The last column is obtained from the fact that p ⇐⇒ q means p =⇒ q and q =⇒ p. Notice that
then that
p ⇐⇒ q is true provided p and q are both true or both false.
When p ⇐⇒ q, we sometimes say that p and q are logically equivalent.
The following results are then true whatever the truth values of the propositions in them.
(i) p ⇐⇒ (¬ ¬ p)
(ii) (¬(p ∨ q)) ⇐⇒ (¬ p ∧ ¬ q)
(iii) (¬(p ∧ q)) ⇐⇒ (¬ p ∨ ¬q)
De Morgan’s Laws
The first of these is fairly obvious. The next two are called De Morgan’s Laws. Note that when
the negation is taken into the bracket, “and” changes to “or” and vice-versa. We shall prove only
the first of De Morgan’s Laws. This will be done be examining the truth-values for each side.
p
T
T
F
F
q
T
F
T
F
p∨q
T
T
T
F
¬(p ∨ q)
F
F
F
T
↑
¬p
F
F
T
T
¬q
F
T
F
T
(¬ p ∧ ¬ q)
F
F
F
T
↑
Since the columns containing ¬(p ∨ q) and (¬ p ∧ ¬ q) are identical, the result follows.
The second of De Morgan’s Laws can be proved in the same way.
Truth-tables with three propositions
In the case that we have 3 propositions, the corresponding truth-table needs 8 rows.
Example 1.1.1
Let p, q and r be propositions. Show that
p ∧ (q ∨ r)
⇐⇒
(p ∧ q) ∨ (p ∧ r).
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CHAPTER 1. ALGEBRA
Solution
We arrange the truth-table systematically so that the first 4 entries of the p-column are all T and
the last 4 all F; the q-column is arranged in pairs of T’s and F’s, while the r-column has alternating
T’s and F’s. This covers all possibilities.
p
T
T
T
T
F
F
F
F
q
T
T
F
F
T
T
F
F
q∨r
T
T
T
F
T
T
T
F
r
T
F
T
F
T
F
T
F
p ∧ (q ∨ r)
T
T
T
F
F
F
F
F
p∧q
T
T
F
F
F
F
F
F
p∧r
T
F
T
F
F
F
F
F
(p ∧ q) ∨ (p ∧ r)
T
T
T
F
F
F
F
F
↑
↑
As before, one can see that the two expressions are equivalent.
Two other results are of importance.
Negation of Implication
We shall need to express the proposition ¬(p =⇒ q) in more basic terms. We know that p =⇒ q is
true if whenever p is true, then so is q. So we expect the negation of p =⇒ q should be that p is true
and yet q is false, that is we suspect that
(¬(p =⇒ q)) ⇐⇒ (p ∧ ¬ q).
This may be shown using a truth-table, as before.
p
T
T
F
F
q
T
F
T
F
p =⇒ q
T
F
T
T
¬(p =⇒ q)
F
T
F
F
↑
¬q
F
T
F
T
p ∧ (¬ q)
F
T
F
F
↑
Since the columns marked by the vertical arrows are identical, the result follows.
Contrapositive
Consider the statement
“If it is raining, then the ground will be wet.”
(1.1)
Now suppose the ground is not wet. Then clearly it cannot be raining! Thus we have shown
“If the ground is not wet, then it is not raining.”
(1.2)
If we let p be the statement “It is raining” and q be the statement “The ground is wet”, then (1.1)
is
p =⇒ q,
(1.3)
7
1.1. LOGIC
while (1.2) is the statement
(¬ q) =⇒ (¬ p)
(1.4)
It seems that (1.3) and (1.4) are logically equivalent. We now show this is indeed so.
p
T
T
F
F
q
T
F
T
F
p =⇒ q
T
F
T
T
¬q
F
T
F
T
¬p
F
F
T
T
(¬ q) =⇒ (¬ p)
T
F
T
T
↑
↑
This shows that p =⇒ q and (¬ q) =⇒ (¬ p) are logically equivalent. The expression
(¬ q) =⇒ (¬ p)
is known as the contrapositive of p =⇒ q. We emphasize again:
• The contrapositive of p =⇒ q is (¬ q) =⇒ (¬ p).
The contrapositive is logically equivalent to the original proposition.
• The converse of p =⇒ q is q =⇒ p.
The converse is NOT logically equivalent to the original proposition.
Exercises 1.1
1. Let p be the proposition 1 < 2, q be the proposition 2 + 3 = 6 and r be the proposition
2 × 3 = 6. Which of following are true
(a) p =⇒ q
(e) ¬ p =⇒ q
(i) q =⇒ p
(m) ¬ q =⇒ p
(q) p ⇐⇒ q
(b) p =⇒ r
(f) ¬ p =⇒ r
(j) q =⇒ r
(n) ¬ q =⇒ r
(r) p ⇐⇒ r
(c) p =⇒ ¬ q
(g) ¬ p =⇒ ¬ q
(k) q =⇒ ¬ p
(o) ¬ q =⇒ ¬ p
(s) ¬q ⇐⇒ r
(d) p =⇒ ¬ r
(h) ¬ p =⇒ ¬ r
(l) q =⇒ ¬ r
(p) ¬ q =⇒ ¬ r
(t) ¬ p ⇐⇒ ¬ r
2. Which of the following are true?
(a) (2 > 1) ∨ (2 = 1)
(b) (2 > 2) ∨ (2 = 2)
(c) (1 > 2) ∨ (1 = 2).
3. Let p, q and r be propositions. Prove the following using truth-tables:
(a) p ∧ (q ∧ r) ⇐⇒ (p ∧ q) ∧ r)
(c) ¬(p ∧ q) ⇐⇒ (¬p) ∨ (¬ q)
(b) p ∨ (q ∧ r) ⇐⇒ (p ∨ q) ∧ (p ∨ r)
(d) (p ∨ ¬ p) ∧ r ⇐⇒ r
4. State the converses and contrapositives of the following:
(a) If x > 0 then x2 > 0
(c) If xy = 1 then x = 1 or y = 1
(b) If xy = 0 then x = 0 or y = 0
(d) If x > 0 then x3 > 0.
Which out of (a), (b) (c), (d), their converses and their contrapositives are true and which are
false? Give reasons.
5. Negate the following:
(a) If Mary understands logic then Mary is clever.
(b) If John is lazy then John will fail.
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CHAPTER 1. ALGEBRA
(c) Either Mary is clever or John is lazy.
(d) Mary is clever and John is lazy.
Which out of (a) – (d) and their negations are true and which are false? Give reasons.
6. Let p and q be propositions. The exclusive or, p ∨ q is defined by the truth-table
p
T
T
F
F
q
T
F
T
F
p∨q
F
T
T
F
Thus p ∨ q is true if p is true or q is true, but not both.
Using this prove that if p, q and r are propositions then
(a) p∨(q∨r)
(b) p∨(q ∧ r)
1.2
⇐⇒
⇐⇒
(p∨q)∨r
(p ∧ q)∨(p ∧ r)
Propositional Functions. Quantifiers
Now consider the statement “x2 = 0”. We cannot assign a truth value to this since we do not
know anything about x. Such a statement, which includes some variable(s), which would become
a proposition if we knew something about the variable(s), is called a propositional function.
Propositional functions are generally denoted p(x), q(y), r(x, y) etc.
There are two simple methods to convert propositional functions into propositions. These are
by use of quantifiers. There are two basic quantifiers. The existential quantifier is symbolised
by ∃, read as “there exists”.
Thus the propositional function x2 = 0 can be turned into the proposition
(∃x)(x2 = 0).
In words this says there exists an x such that x2 = 0. This is true, since 02 = 0.
The second quantifier is the universal quantifier, symbolised by ∀, read as “for all” or “for
every”. Thus we can form the proposition
(∀x)(x2 = 0).
This says that for every x we have x2 = 0, which is clearly false.
The two quantifiers can be combined together. For example consider the following :
(∀x)(∃y)(y > x)
(∃y)(∀x)(y > x).
Note that the first of these is true, since it says that for any x one can find a bigger number, y.
On the other hand the second is false, since it says that one can find a number, y bigger than all
numbers, x. Thus the order in which the quantifiers are arranged is of importance.
The negations of the quantifiers follows similar rules to the negations of “and” and “or”. Suppose
then p(x) is a propositional function. Consider the proposition
¬(∃x)(p(x)).
9
1.3. METHODS OF PROOF
This means that it is not the case that there exists an x such that p(x) is true, or in other words
p(x) is always false. Thus
¬(∃x)(p(x)) ⇐⇒ (∀x)(¬p(x)).
Similarly, if it is not the case that p(x) is true for all x, then there must exist some x for which p(x)
is false, i.e.
¬(∀x)(p(x)) ⇐⇒ (∃x)(¬p(x)).
Note how taking the negation inside the quantifier changes the existential quantifier to the universal
and vice-versa. This process can be applied to more complex statements.
Example 1.2.1
Negate (∃x)(∀y)(p(x, y)).
Solution
We have
¬(∃x)(∀y)(p(x, y))
⇐⇒
⇐⇒
(∀x)(¬(∀y)(p(x, y)))
(∀x)(∃y)(¬p(x, y)).
Exercises 1.2
1. Negate the following (you may assume x and y are real numbers):
(a) (∀x)(∃y)(y = x2 )
(b) (∃y)(∀x)(y = x2 )
(c) (∀y)(∃x)(y = x2 )
(d) (∃x)(∀y)(y = x2 )
Which out of (a) – (d) and their negations are true and which are false? Give reasons.
1.3
Methods of Proof
The usual form of a theorem in mathematics is the following:
Suppose various hypotheses (H) hold. Then it follows that the conclusion (C) holds.
Fundamentally there are three types of proof. The first is a direct proof. The idea is to proceed
from the hypotheses to the conclusion. The proof of the following result is a direct proof.
Theorem 1.3.1 The sum of two even numbers is even.
Proof. Suppose m and n are even. Since every even number is a multiple of 2, then there exist k
and ℓ such that
m = 2k and n = 2ℓ.
Then
n + m = 2k + 2ℓ = 2(k + ℓ)
which is even.
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CHAPTER 1. ALGEBRA
The second method is known as an indirect proof. This really works only in the case that
there is a single hypothesis, H. As before, we wish to show H =⇒ C. In an indirect proof, we show
instead that ¬ C =⇒ ¬ H, that is we prove the contrapositive. We have already seen that these are
equivalent.
An example of an indirect proof is the following. Suppose we wish to show the following:
Theorem 1.3.2 If n2 is an even integer, then n is also even.
Proof. The proof is achieved by showing instead that
if n is not even then n2 is not even,
that is to say
n is odd =⇒ n2 is odd.
This is easily achieved, since if n is odd, then n = 2k + 1 for some integer k, in which case
n2 = (2k + 1)2 = 4k 2 + 4k + 1
which is obviously odd.
Closely allied to indirect proof is proof by contradiction (or reductio ad absurdum). Here we
assume the hypothesis H and suppose that the conclusion C is false. Then we show that this leads
to a contradiction. Since this cannot be, then the conclusion cannot be false, that is the conclusion
is true, and the proof is achieved. Here is an example.
Theorem 1.3.3 Suppose n is an integer. If n2 + 3 is odd, then n is even.
Proof. The hypothesis H is n2 + 3 is odd. We want to show that n is even.
Very well, suppose, for the sake of deriving a contradiction, that this is not the case. Then n is odd.
So
n = 2k + 1
for some integer k. But then
n2 + 3 = (2k + 1)2 + 3 = 4k 2 + 4k + 1 + 3 = 4k 2 + 4k + 4.
This is clearly even, which contradicts our hypothesis. So n cannot be odd, that is n is even.
√
The proof that 2 is irrational given in Chapter 1 is a typical proof by contradiction. Another
classic example is the following theorem. The proof is more than 2300 years old.
Theorem 1.3.4 (Euclid) There exist infinitely many prime numbers.
Proof. Suppose, for the sake of contradiction, that there are only finitely many prime numbers. Let
these be
p 1 , p 2 , . . . , pk .
Since any integer can be broken down into prime factors, then
any integer must be divisible by at least one of p1 , p2 , . . . , pk .
Now consider the number
n = p1 p2 · · · pk + 1.
(1.5)
11
1.3. METHODS OF PROOF
This is not divisible by p1 , since on dividing by p1 one is left with a remainder of 1. In the same
way it is not divisible by any of p1 , p2 , . . . , pk . This contradicts our previous statement (1.5). It
follows then that there are infinitely many primes.
Note that in a proof by contradiction, one must assume the conclusion is false. It is pointless to
assume that the hypothesis is false. If you wish to prove Pythagoras’s theorem, it would be madness
to start off “Suppose △ABC is not right-angled.”
The natural question to ask is which method should be used in which case. The answer, unfortunately, is that there is no recipe to tell us how to prove any particular theorem. If there were,
mathematicians would soon be out of work.
1.3.1
Proofs, Disproofs, Counter-examples
Consider a statement of the form
(∀x)p(x).
To show this is the case one must then show that
p(x) is true, no matter what the value of x!!!
It is simply not good enough to give an example, that is to say a value of x for which p(x) is true.
This is very different from the situation in the natural sciences. If a physicist, for example, observes
that when you drop an apple it falls downwards, and it falls downwards every time you try it, he
will conclude that it is always the case that when you drop an apple, then it falls downwards. This
is not acceptable to a mathematician, who demands a proof.
For example, if one is asked to prove that the sum of 2 odd integers is even no good to say
3 + 5 = 8, which is even.
That does not prove anything useful.
Sometimes a statement may look very plausible and yet be false. Consider the following sequence
of statements, all of which are true:
31 is a prime number
331 is a prime number
3331 is a prime number
One suspects that any number of the form
33 · · · 31
is a prime number. If one tries a few more, it turns out that
33331 is a prime number
333331 is a prime number
are all prime numbers.
Surely to goodness now you believe the suspicion. If we give it one more try we find
3333331 is a prime number.
12
CHAPTER 1. ALGEBRA
Only an absurdly obstinate person would not be a believer by now. Unfortunately, if we try just
one more, we find
33333331 = 1960784 × 17.
Our hypothesis has disintegrated!
If we start with a statement of the form (∀x)p(x), then to prove it true, we have to show it is
true for every x. On the other hand, if we want to show it is false, then we have only to produce a
single example where it is false. Such an example is called a counter-example.
For example, consider the statement
“Every even integer n such that n > 2 is the sum of 2 primes.”
It is easy to check this works for a few small values of n. Thus
4 = 2 + 2, 6 = 3 + 3, 8 = 5 + 3, 10 = 5 + 5, 12 = 7 + 5, etc.,
so it seems to work. But this is not a proof. Are we sure that
867514894
is the sum of 2 primes? Nor is it obvious how to prove the original statement. Mathematicians have
been trying for over 300 years with no success. Nobody said it would be easy.
In summary
Suppose one has a statement of the form (∀x)(p(x)).
If you wish to prove it true, you must show it for all x
If you wish to prove it false, you need produce only one example, called
a counter-example where it is false.
We emphasize one more time
One cannot prove a general statement to be true by giving an example.
Exercises 1.3
1. Prove or disprove the following statements – that is, provide a proof or a counter-example:
(a) If n2 is divisible by 3, then n is divisible by 3.
(b) If n2 is divisible by 9, then n is divisible by 9.
(c) If n is even and m is odd, then (m + n)2 is odd.
(d) If (m + n)2 is odd, then either m is odd and n is even, or n is odd and m is even.
1.4
Set Theory. Basic Ideas
Set, Element, Member
Definition 1.4.1 A set is a collection of objects. The objects which comprise the set are called
elements or members of the set.
13
1.4. SET THEORY. BASIC IDEAS
Thus the collection of all students in the world forms a set and any individual student is a
member of this set. Similarly, we can speak of the set of even numbers, the set of all tables, the set
of all triangles and so on.
We shall denote sets by capital letters A, B, C, X, Y, Z etc. and their elements by lower case
letters a, b, c, x, y, z etc.
Then if A is a set and x is a member of A, we write
x ∈ A.
If, on the other hand, x is not a member of A we write x 6∈ A.
For example, if E is the set of even integers, then 2 ∈ E but 17 6∈ E.
The simplest way of defining what a set contains is simply to list all of its elements between
braces. For example, the set of all vowels may be written as
{a, e, i, o, u}.
Please note that we use “curly brackets” {, }. Square brackets [, ] and round ones (, ) have quite
different meanings. Now we can say
a ∈ {a, e, i, o, u},
but
z 6∈ {a, e, i, o, u}.
In the same way we may write the set of positive even integers as
{2, 4, 6, 8, . . .}.
The three dots signify that we continue in the same fashion indefinitely.
This notation may also used as follows. Consider the set
{5, 10, 15, 20, . . . , 250}.
This denotes the set of all multiples of 5 between 5 and 250.
Now suppose we wish to write down the set of all prime numbers. Clearly we cannot list them
all, nor do they follow an obvious pattern. We circumvent this difficulty by writing this set as
{x | x is a prime number}.
Here one may read the vertical bar (sometimes a colon is used) as “such that”. Thus
{d | d is a day of the week}
is identical to
{Monday, Tuesday, Wednesday, Thursday, Friday, Saturday, Sunday}.
This leads us to the following definition.
Equality of Sets
Definition 1.4.2 Let A and B be sets. Then A is equal to B (written A = B) if every member of
A is a member of B, and every member of B is a member of A.
So A = B when A and B have exactly the same elements. If A is not equal to B, we write A 6= B.
14
CHAPTER 1. ALGEBRA
Example 1.4.1
Let A = {1, 2, 3, 4}, B = {2, 1, 4, 3} and C = {1, 2, 3}. Then A = B, but A 6= C.
Example 1.4.2
Let A = {1, 2, 3}, B = {3, 2, 1} and C = {1, 2, 2, 3, 3, 3}. Then A = B = C.
Note that the order in which the elements appear is not important. Furthermore, repetition of
elements is irrelevant.
Even when two sets are not equal there may be another relationship between them.
Subset
Definition 1.4.3 Let A and B be sets. Then A is a subset of B (or A is contained in B) if
every element of A is also an element of B.
This is written as
A⊆B
or
B ⊇ A.
We could also say that
A⊆B
if
x ∈ A =⇒ x ∈ B.
This concept can be conveniently illustrated by means of a Venn diagram as shown in Figure
1.4.1. Here A consists of all points within the smaller region and B of all points within the larger.
Evidently every member of A is a member of B.
B
A
Figure 1.4.1
Example 1.4.3
Let A = {3, 5, 7, 9}, B = {1, 2, 3, 4, 5, 6, 7, 8, 9} and C = {2, 4, 6, 8}.
Then A ⊆ B and C ⊆ B.
Do not confuse the symbols ⊆ and ∈.
Thus, while a ∈ {a, b, c} it is not true that a ⊆ {a, b, c}, since the left-hand side is not even a set.
Of course it is true that {a} ⊆ {a, b, c}.
It should also be noted that for any set A we have A ⊆ A. This follows immediately from Definition
1.4.3 (with A in place of B).
The following proposition contains some obvious facts.
15
1.5. SET THEORY. FURTHER PROPERTIES
Proposition 1.4.4 (a) A = B if and only if A ⊆ B and B ⊆ A.
(b) If A ⊆ B and B ⊆ C then A ⊆ C.
We may rewrite Definition 1.4.3 as:
A ⊆ B iff (x ∈ A =⇒ x ∈ B),
and Definition 1.4.2 as:
A = B iff (x ∈ A ⇐⇒ x ∈ B).
It is important not to confuse the symbols = and ⇐⇒ .
AN EQUALS SIGN MAY ONLY BE USED WHEN THE
EXPRESSIONS ON EITHER SIDE OF IT ARE EQUAL !!!
The following is WRONG !
x = 2 = x2 = 4,
for this would mean 2 = 4. What should be written is
x = 2 =⇒ x2 = 4.
Exercise 1.4
1. Which of the following statements are true and which false?
(a) 1 ∈ {1, 2, 5}
(b) 1 ⊆ {1, 2, 5}
(c) {2} ∈ {1, 2, 5}
(d) {2} ⊆ {1, 2, 5}
(e) {2, 3} ∈ {1, {2, 3}, 2, 3, 4}
(f) {2, 3} ⊆ {1, {2, 3}, 2, 3, 4}.
2. Let P, Q, R, S, T denote respectively the sets of parallelograms, quadrilaterals, rectangles,
squares and trapezia in the plane. Which of these are subsets of which others?
3. What is wrong with the following arguments? Write them out correctly.
(a) Solve 2x − 3 = 5.
Solution:
2x − 3 = 5
=
2x = 8
=
x = 4.
2
(b) Prove that x(x − 2) + 7(x + 1) = x + 5x + 7.
Solution:
x(x − 2) + 7(x + 1) =
2
x − 2x + 7x + 7 =
−2x + 7x + 7 =
5x =
1.5
x2 + 5x + 7
x2 + 5x + 7
5x + 7
5x,
proved.
Set Theory. Further Properties
If we are given two sets A and B, we may construct further sets from them in a variety of ways.
Union
Definition 1.5.1 If A and B are sets, then the union of A and B, denoted A ∪ B, is the set of
elements that belong to A or B or both.
16
CHAPTER 1. ALGEBRA
A
B
Figure 1.5.1
In the Venn diagram Figure 1.5.1, A is the left-hand region, B the right-hand region and A ∪ B is
the whole shaded portion. To put it another way
x∈ A∪B
⇐⇒
(x ∈ A ∨ x ∈ B).
(Recall that p ∨ q means p is true or q is true or both).
Example 1.5.1
Let A = {2, 5, 7} and B = {2, 4, 6, 8}. Then A ∪ B = {2, 4, 5, 6, 7, 8}.
Example 1.5.2
Let A = {x | x is an even integer} and B = {x | x is an odd integer}.
Then A ∪ B = {x | x is an integer}.
Example 1.5.3
If A = {3, 4, 5} and B = {4, 5, 6}, then A ∪ B = {3, 4, 5, 6}.
Note that we do not repeat the 4’s and 5’s.
The following result gives a few elementary properties of unions.
Proposition 1.5.2 If A, B and C are sets, then
(a) A ∪ A = A
(b) A ∪ B = B ∪ A
(c) A ∪ (B ∪ C) = (A ∪ B) ∪ C
(d) A ⊆ A ∪ B.
The proofs of these statements are left as exercises.
Intersection
Definition 1.5.3 If A and B are sets, then the intersection of A and B, denoted A ∩ B, is the
set of elements which belong to both A and B.
17
1.5. SET THEORY. FURTHER PROPERTIES
A
B
Figure 1.5.2
In the Venn diagram, Figure 1.5.2, A ∩ B is the shaded portion. Again we can say
x∈A∩B
⇐⇒
(x ∈ A ∧ x ∈ B).
This leads us to a problem. What would happen if A and B were as shown in Figure 1.5.3?
A
B
Figure 1.5.3
There is nothing which is in both A and B and so A ∩ B has no elements. However, we should like
A ∩ B to be a set. This leads us to our next definition.
Empty Set. Disjoint Sets
Definition 1.5.4 (a) The empty set is the set containing no elements. This is denoted ∅.
(b) If A ∩ B = ∅, we say A and B are disjoint.
Corresponding to Proposition 1.5.2, we have the following results.
Proposition 1.5.5 If A, B and C are sets, then
(a) A ∩ A = A
(b) A ∩ B = B ∩ A
(c) A ∩ (B ∩ C) = (A ∩ B) ∩ C
(d) A ∩ B ⊆ A
(e) ∅ ⊆ A.
Example 1.5.4
Let A = {1, 2, 4}, B = {3, 5, 6} and C = {2, 5, 6}.
Then A ∩ B = ∅, A ∩ C = {2} and B ∩ C = {5, 6}.
Finally, it is sometimes necessary to talk about the elements that are not in a set. For example,
if
one would probably think of
A = {x | x is an even integer}
B = {x | x is an odd integer}
as being the set of all elements not in A. However, it is equally true that Africa 6∈ A. This is not
what we want at all. It is therefore necessary to restrict ourselves to some (usually large) set in
which our whole discussion will take place.
18
CHAPTER 1. ALGEBRA
Universal Set
Definition 1.5.6 The universal set, E, is the set of all elements under discussion.
Complement of a Set
Definition 1.5.7 Let E be the universal set and suppose A ⊆ E. Then the complement of A in
E, denoted A′ , is the set of all elements in E which are not in A.
E
A
A′
Figure 1.5.4
In Figure 1.5.4, the outer box denotes the universal set E, and the shaded portion is A′ .
Proposition 1.5.8 Let A ⊆ E be any set. Then
(a) (A′ )′ = A
(b) A ∪ E = E;
(c) A ∪ ∅ = A;
(d) A ∩ A = ∅;
′
A∩E =A
A∩∅=∅
A ∪ A′ = E
(e) (∅)′ = E; (E)′ = ∅.
A number of algebraic relationships hold between sets. Proofs of these may be given using the
corresponding results from logic.
Example 1.5.5
Show that (A ∩ B)′ = A′ ∪ B ′ and (A ∪ B)′ = A′ ∩ B ′ . (These two relations are again known
as De Morgan’s Laws. )
Solution
Recall that, as we have already shown
¬(p ∧ q)
⇐⇒
(¬p ∨ ¬q).
(1.6)
If we let p be x ∈ A and q be x ∈ B, then
x ∈ (A ∩ B)′ ⇐⇒ ¬(x ∈ A ∩ B)
⇐⇒ ¬(x ∈ A ∧ x ∈ B)
⇐⇒ ¬(x ∈ A) ∨ ¬(x ∈ B)
⇐⇒ x ∈ A′ ∨ x ∈ B ′
⇐⇒ x ∈ (A′ ∪ B ′ )
(by(1.6))
19
1.5. SET THEORY. FURTHER PROPERTIES
Hence
(A ∩ B)′ = A′ ∪ B ′ .
It is left as an exercise to show that
(A ∪ B)′ = A′ ∩ B ′ .
Exercise 1.5
1. Which of the following statements are true, and which false ?
(a) {∅} = ∅
(b) ∅ ∈ {∅}
(c) For any set A, A ∩ ∅ = A
(d) If A, B, C are sets such that A ∪ B = A ∪ C, then B = C.
(e) If A, B, C are sets such that A ∪ B = A ∪ C for every set A, then B = C.
2. Let A = {1, 2, 5, 6}, B = {2, 3, 5, 7} and C = {1, 3, 6, 8} where E = {1, 2, 3, 4, 5, 6, 7, 8}. Find
(a) A ∪ B
(e) B ′ ∪ A
(b) B ∩ C
(f) (A′ ∩ B ′ ) ∪ C.
(c) (A ∪ B) ∩ (B ∩ C)
(d) A′
3. Let the universal set E = {1, 2, 3, 4, 5, a, b, c, d, e}, A = {1, 3, a, b} and B = {2, 4, 5, c, d}. Find
the following:
(a) A ∪ B
(e) (A ∪ B)′
(b) A ∩ B
(f) (B ′ )′
(g)
(c) A′
(A′ ∪ B)′
(d) A ∩ B ′
(h) all the subsets of A.
4. If X and Y are sets, simplify the following:
(a) X ∩ X
(d) X ∩ E
(g) (X ′ )′
(j) ∅′
(m) X ′ ∩ X
(p) X ∪ (X ∩ Y )
(s) (X ∪ Y ) ∩ (X ∩ Y )
(b) X ∪ X
(e) X ∪ X ′
(h) X ∩ ∅
(k) ((X ′ )′ )′
(n) X ∩ (X ∪ Y )
(q) (X ∪ Y ) ∪ X
(t) (X ∪ (X ∩ Y )) ∩ X.
(c) X ∪ ∅
(f) E ′
(i) X ∪ E
(l) (X ∩ E)′
(o) X ∪ (Y ∩ ∅)
(r) (X ∪ Y ) ∪ X ′
5. (a) Suppose S and T are sets. Under what conditions is S ∪ T = T ?
(b) Suppose X and Y are sets and X ∩ Y = X. What can you deduce about X and Y ?
(c) If A and B are sets and A ∩ B = A ∪ B, what can be deduced?
6. Prove Propositions 1.5.2, 1.5.5 and 1.5.8.
7. Using the facts that
p ∨ (q ∧ r)
⇐⇒
(p ∨ q) ∧ (p ∨ r)
p ∧ (q ∨ r)
⇐⇒
(p ∧ q) ∨ (p ∧ r)
and
(see example 1.1.1), show that if A, B and C are sets, then
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
and
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).
Use these identities, 1.5.2, 1.5.5, 1.5.8 and De Morgan’s Laws to simplify
(a) A ∩ (B ∪ A′ )
(d) (X ∩ Y )′ ∩ Y ′
(g) (X ∩ Y ′ )′ ∪ Y ′
(b) A′ ∪ (B ∩ A)
(e) X ∪ (X ′ ∩ Y )
(h) (X ′ ∪ Y ′ )′ ∩ X ′ .
(c) (A ∩ B) ∪ (A′ ∪ B)
(f) (A ∪ (B ∪ A′ )′ )′
20
CHAPTER 1. ALGEBRA
8. If A and B are sets, the difference set A − B is defined by
A − B = {x ∈ A | x 6∈ B},
that is the set of elements of A which are not in B (see Figure 1.5.5).
E
A
B
Figure 1.5.5
(a) Show that A − B = A ∩ B ′ . Use this identity, and other well-known identities, to simplify
the following:
(i) A − (B − A)
(ii) (A − B) − A
(iii) (A − B) ∩ (A − C).
(b) If A, B and C are sets such that A− B = C − B, does it follow that A = C? Give reasons.
9. How many subsets are there of the set {1, 2, 3}? (Remember that the empty set is a subset, as
is the whole set {1, 2, 3} itself.) How many subsets are there of {1, 2, 3, 4}? Guess a formula
which will tell you how many subsets there are of {1, 2, 3, . . . , n} for any positive integer n.
1.6
Number Systems
Counting is one of the most familiar concepts in all of mathematics. It is assumed that everybody
is conversant with the numbers 1, 2, 3, . . . .
Natural Numbers
The set of all such numbers is called the set of natural numbers, denoted N. Thus
N = {1, 2, 3, . . .}.
Now any two elements of N may be added or multiplied together to give another member of N.
Thus 2 ∈ N, 7 ∈ N and 2 + 7 = 9 ∈ N, 2 × 7 = 14 ∈ N.
However, it is not true that the difference between two members of N is a member of N.
Thus 5 ∈ N, 8 ∈ N, but 5 − 8 6∈ N.
Integers
It becomes necessary to enlarge N to the set of integers, Z. (Z is for the German “zahl” meaning
number.) Thus
Z = {. . . , −3, −2, −1, 0, 1, 2, 3, . . .}.
Note that N ⊆ Z.
Now if a, b ∈ Z then a + b ∈ Z, ab ∈ Z and −b ∈ Z. However it is not always true that a/b ∈ Z.
Thus for example 3 ∈ Z, 5 ∈ Z but 3/5 6∈ Z.
21
1.6. NUMBER SYSTEMS
Rational Numbers
Again we enlarge Z to the set of all rational numbers, Q. By definition Q is the set of all numbers
p/q where p, q ∈ Z and q 6= 0. i.e.
Q = {p/q | p, q ∈ Z and q 6= 0}.
Thus Q is just the set of “fractions”. Notice that Z ⊆ Q, for if n ∈ Z then n = n/1 ∈ Q. The
following are therefore rational numbers :
2 135
101
2
,
, − 23,
, −
, etc.
17 61
2
101
One might object to the restriction that q must not be zero, and try to define for example
1
=∞
0
(1.7)
where the right-hand side is “infinity”. This leads to immediate problems, for by (1.7) we have
1 = 0.∞
(1.8)
Since 0 = 0 + 0, we obtain by substitution
1
= (0 + 0)∞
= 0.∞ + 0.∞.
Then using (1.8) again, we have
1 = 1 + 1 = 2.
We thus see that division by zero cannot be allowed.
THOU SHALT NOT DIVIDE BY ZERO !!!
a
is meaningless.
0
Now it might be that we have enough numbers in Q to conduct all of arithmetic. This is not so,
however, as we shall now show.
Thus any expression of the form
First recall the following result, which we proved earlier
Lemma 1.6.1 If n ∈ N and n2 is even, so is n.
Now we show there are numbers which are not in Q.
Theorem 1.6.2
√
2 6∈ Q.
√
Proof. Suppose, for the sake of contradiction that 2 ∈ Q.
√
Now if 2 ∈ Q , then
√
p
2=
q
where p, q ∈ Z and q 6= 0.
Furthermore, we may assume that p and q are not both even, since if they were, we could cancel out
a factor of 2. Now on squaring we obtain
2q 2 = p2 .
22
CHAPTER 1. ALGEBRA
This means that p2 is even, and hence by Lemma 1.6.1, so is p. Let then
where k ∈ Z.
p = 2k
Substituting back gives us
2q 2 = 4k 2
so
q 2 = 2k 2 .
Thus q 2 is even and hence by Lemma 1.6.1 so is q.√Thus p and q are both even, contradicting our
earlier assumption. This contradiction shows that 2 6∈ Q.
√
This theorem was known to the ancient Greeks. It is not difficult to show
√ that 3 6∈ Q, that the
cube
root of 2 is irrational, and indeed if n is not a perfect square, that n 6∈ Q; thus for example,
√
6 6∈ Q. It was shown by Lambert in 1832 that π 6∈ Q. Furthermore, Cantor showed in 1874 that,
in a sense, very few numbers are rational.
Now let us look at the situation geometrically. Suppose we have a line extending indefinitely in
both directions (Figure 1.6.1).
−3
−2
−1
0
1
2
3
4
-
Figure 1.6.1
Choose a point on the line and label it 0. Choose any other point and label it 1 (conventionally to
the right of 0). Now lay off segments of length equal to the distance from 0 to 1 to the right of 1
and label these successively 2, 3, 4, . . . and similarly to the left of 0 labelling them −1, −2, −3, . . ..
Thus every element of Z corresponds to a point on the line. The reverse is clearly false, since there
are points between 0 and 1 which do not correspond to any element of Z.
A simple geometrical construction will then locate all points of the form p/q where p, q ∈ Z and
q 6= 0, i.e. all of Q. Then every rational corresponds to a point on the line. However Theorem
1.6.2 tells us that there are still points on the line√that do not correspond to elements of Q, one
example being the point whose distance from 0 is 2. Thus if we consider only points on the line
corresponding to members of Q, there are “gaps” left in the line.
Real Numbers
All of this means that we need to enlarge Q to the set of real numbers, R, in such a way that
to every real number there corresponds a point on the line, and to every point on the line there
corresponds a real number. The details of how this is done are fairly technical and we shall not
pursue the matter here. (Indeed, the very meaning of the word “line” is a thorny problem.)
In summary, each point on the line is uniquely associated with a real number and each real
number corresponds to a point on the line. For this reason we usually envisage the reals as points
on a line and talk of the “real line”.
Decimal Expansions
We can instead approach the idea of what the real numbers are by looking at decimal expansions.
As we know,
1/4 = 0.25
and
1/3 = 0.333 · · ·
23
1.6. NUMBER SYSTEMS
where the dots signify that the 3’s continue indefinitely. We shall call the first type, like 1/4 a
terminating decimal and the second type like 1/3 a recurring decimal. To see why 1/3 takes
this form, let
x = 0.333 · · ·
Then
10x = 3.333 · · ·
Subtracting gives
9x = 3
so that x = 1/3.
We can regard terminating decimals as a special kind of recurring decimals. For example,
1/5 = 0.2 = 0.2000 · · ·
Any recurring decimal may be written as a rational. For example consider
3.658658658 · · ·
Letting
x = 3.658658658 · · ·
we have
1000x = 3658.658658 · · ·
Hence, subtracting we have
999x = 3658
and so
3658
∈ Q.
999
In the same way every recurring decimal corresponds to a rational number. The converse, that is
every rational has a recurring decimal expansion, is also true.
x=
Now not every decimal expansion is recurring. For example,
0.101001000100001 · · ·
is not recurring. The real numbers, then, consist of all decimal expansions, recurring or not.
Exercise 1.6
1. Write out each of the following sets in full with each element listed:
√ √
√
(a) {−10, −9, −8, . . . , 10} ∩ N
(b) { 1, 2, . . . , 12} ∩ Q.
2. Write out in dot notation each of the following sets:
(a) {x ∈ Q | x = n/2 for some n ∈ N}
(b) {x ∈ Q | x = 2n/3 for some n ∈ N}.
√
√
3. Given the set A = {−1, −3, −3/17, π, 2, 0, 4} and the universal set E = R, write down
(a) A ∩ N
(b) A ∩ Z
(c) A ∩ Q.
4. Denote the rationals by Q and the irrationals by Q′ . Simplify
(a) N ∩ Q
(d) Z ∩ Q
(b) Z ∪ Q
(e) N ∪ Q
(c) Z ∩ Q′
(f) N ∩ Q′ .
5. The sets X and Y are defined by X = {3n + 1 | n ∈ Z}, Y = {3n + 2 | n ∈ Z}, and ′ denotes
complement in Z. Give one element of X, one of Y and one of (X ∪ Y )′ . Show that:
24
CHAPTER 1. ALGEBRA
(a) if x, y ∈ X then xy ∈ X
(b) if x, y ∈ Y then xy ∈ X
(c) if x, y ∈ X then x + y ∈ Y
(d) if x ∈ X and y ∈ Y , then (x + y) is divisible by 3.
Give an example to show that the statement “x, y ∈ X ′ implies x + y ∈ X ′ ” is false.
6. Let C = {4n + 1 | n ∈ Z}. Give three different elements of C.
(a) Show that if x, y ∈ X then xy ∈ C.
(b) Give an example to show that the statement “if x, y are integers and xy ∈ C, then
x, y ∈ C”, is false.
7. Suppose n is an integer. Give a precise condition for n to be even. Show that
(a) (−1)n = 1 if n is even
(b) (−1)n = −1 if n is odd.
Hence simplify (−1)(−1) , where n is any integer.
√
8. Let X = {a + b 2 | a, b ∈ Z}. Show that if x, y ∈ X then xy ∈ X.
n
9. Let X = {a/3 | a ∈ Z}. Suppose x, y ∈ X. Show that x + y ∈ X. Is xy ∈ X?
10. If x, y 6∈ Q, which of the following statements are true ?
(a) x2 ∈ Q
(b) x + y 6∈ Q
(c) xy 6∈ Q.
11. If x ∈ Q and y 6∈ Q, which of the following statements are true?
(a) x + y 6∈ Q
(c) x2 y 3 6∈ Q.
(b) xy 6∈ Q
12. If x, y ∈ Q, and y 6= 0, prove the following statements :
(a) x + y ∈ Q
(b) xy ∈ Q
(c) x/y ∈ Q.
13. Prove the following are irrational:
√
√
√
(a) 3
(b) 21/3
(c) 2 + 3
√
(d) 5 2 − 4/17.
14. Express the following recurring decimals as rational numbers in the form p/q, where p, q ∈ Z.
(a) 2.666 · · ·
1.7
(b) 4.363636 · · ·
(c) 0.999 · · ·
Indices
We denote for any a ∈ R, a.a by a2 , a.a.a by a3 , and in general a.a.
· · · .a} by an . Then if n, m ∈ N,
| {z
n times
am an = |a.a.{z
· · · .a} a.a.
· · · .a} = a.a.
· · · .a} = am+n
| {z
| {z
n times
We have shown that
m times
m+n times
am an = am+n for m, n ∈ N.
That is to say, when you multiply, you add indices. For example,
33 34 = 37 .
Note that the number which is being raised to a power must be the same throughout. Thus
23 34 6= 67 .
(1.9)
25
1.7. INDICES
(Nor is the left-hand side equal to anything else very interesting for that matter.)
Now suppose m, n ∈ N and m is greater than n. Then
am
an
=
=
=
a.a. · · · .a
where the top has m factors and the bottom n factors
a.a. · · · .a
a.a.a. · · · .a with (m − n) factors after n cancellations
am−n .
We have shown that
am
= am−n for n, m ∈ N, where m > n, a 6= 0.
an
(1.10)
Here it is necessary to specify that a 6= 0, otherwise we shall be dividing by zero. Equation (1.10)
says that when you divide, you subtract indices. Thus for example
27
= 22 .
25
Finally, if m, n ∈ N, then
(am )n = (a.a. · · · .a)(a.a. · · · .a) · · · (a.a. · · · .a),
where each bracket on the right-hand side has m factors and there are n such brackets. This gives
a total of mn factors. i.e.
(am )n = amn for n, m ∈ N.
(1.11)
For example
(54 )7 = 528 .
Now we have a meaning for a2 , a3 , a4 , . . ., and evidently it is sensible to define a1 = a. Can we
not attach some useful meaning to such expressions as a−3 , a1/2 or a0 ? (We are quite at liberty
to define a−3 to mean 3.98456a213 if we wish, but this would be a silly definition, and we should
prefer a definition which was useful.) It would be pleasant if (1.9), (1.10) and (1.11) held, not just
for m, n ∈ N, but for any m, n ∈ Z.
Let us first consider a−3 as an example. We wish to preserve the rule that when we multiply we
add indices. Now if this is to be so then
24 2−3 = 2.
But clearly
24
= 2,
23
and so we are led to define
1
.
23
More generally, if we assume (1.9) holds for any integers m and n, then
2−3 =
an+1 a−n = a.
Since
an+1
=a
an
we make the following definition.
Definition 1.7.1 If a ∈ R, a 6= 0 and n ∈ N, then a−n =
1
.
an
26
CHAPTER 1. ALGEBRA
Note that we must avoid division by zero.
Now consider the problem of defining a0 . If (1.10) is to hold for any integers m and n, then
putting m = n, we have
am
= am−m = a0 .
am
Clearly the left-hand side is 1 and so we make the following definition.
Definition 1.7.2 If a ∈ R, a 6= 0, then a0 = 1.
Note that the expression 00 is meaningless.
Finally, let us see if we can attach some meaning to 21/2 for example. Now if (1.11) holds even
when m is not an integer, then
(21/2 )2 = 21 = 2.
Since
√
( 2)2 = 2,
it seems natural to define
21/2 =
√
2.
More generally, if (1.11) holds for any m and n, then
(a1/n )n = a1 = a.
Since
√
( n a)n = a,
it would appear to be a good idea to define
√
a1/n = n a.
There
is a√slight problem in that a1/n will not exist if n is even and a is negative.
√
√ For example,
−2 and 4 −5 are not real numbers. There is no problem if n is odd; for example 3 −8 = −2, since
(−2)3 = −8.
Definition 1.7.3 If a ∈ R, n ∈ N, then a1/n =
1
, provided the right-hand side exists.
an
Remark.
√
If n is even and a > 0, then n a is defined to be the positive real solution to the
√
equation xn = a. So 4 is defined to be 2 and not ±2.
This should not be confused with the fact that if x2 = 4 then x = ±2. This follows
because x = 2 and x = −2 both satisfy the equation x2 = 4.
√
It follows that if m, n ∈ N, then am/n = (am )1/n = n am , provided the nth root of am exists.
Example 1.7.1
Simplify 23 8−2 45 .
27
1.7. INDICES
Solution
First notice that 8 = 23 and 4 = 22 . So
23 8−2 45 = 23 (23 )−2 (22 )5 = 23 2−6 210 = 27 = 128.
Example 1.7.2
Simplify
27 + 24 32
√
.
62 2
Solution
Similarly, 27 = 24 23 , 62 = 22 32 and
√
2 = 21/2 . We have
24 23 + 24 32
22 32 21/2
=
=
=
=
24 (23 + 32 )
22 32 21/2
4 −2 −1/2 −2 3
2 2 2
3 (2 + 32 )
23/2 3−2 17 √
√ 17
34 2
2 2
=
.
9
9
Example 1.7.3
Simplify
a2 (b3 c−2 )2
.
a−3 b2 c−4
Solution
We have
a2 (b3 )2 (c−2 )2 a3 b−2 c4
=
a2 b6 c−4 a3 b−2 c4
=
=
a 5 b 4 c0
a5 b 4 .
Example 1.7.4
Simplify
√
√
√
√
3 − 75 + 12 + 48.
Solution
We obtain
√
√
√
√
√
√
√
√
√
3 − 3.25 + 3.4 + 3.16 = 3 − 5 3 + 2 3 + 4 3 = 2 3.
√
√ √
√
This example
an important principle. Note that 5 3 = 25 3 = 75, (and we do
√ illustrates
√
NOT obtain 5 3 = 15).
Thus when taking a number inside a square root, one must square it.
28
CHAPTER 1. ALGEBRA
Similarly
√
√
12
12 √
= √ = 3.
2
4
Example 1.7.5
Solve for x the equation 2x = 1/64.
Solution
We note 26 = 64, and so
2x = 2−6 ,
and x = −6.
Example 1.7.6
2
Solve the equation 3−x =
1
.
81
Solution
Again
2
3−x = 3−4 ,
so that x2 = 4.
Hence x = 2 or x = −2.
Exercise 1.7
1. Simplify:
23 2−1/2
22
2.5−3 .52
(d)
1252/3
62 123
34 23
104 2−3
(e)
200−1/2 52
(a)
(b)
(g) a−3 a6 a−2/3
(h) [a2 b−3 /(a−3 b2 )]3
2. Simplify:
√ √
(a) 32 2
(d)
√
√
48 − 4 12
(g)
r
7
+
9
r
7
+
4
(b) (3−1 − 2−1 )−1
√
18
(e) √
9 2
r
7
36
3. Solve for x:
(a) 5x = 6252
x
(d) 3 = (3
−2
322 643
1285 16
9.26 − 32 .25.16
(f)
72 53 161/4
3/2 2
a b + a1/2 b3
(i)
a2 b−3
(c)
(h)
(c)
(f) (−27)−1/3
√
√
√
√
45 − 125 + 20 + 5.
(b) 2−x = 128
√ −1
3)
9x
3x
(c) 4−x = 2
2
x
x
(e) (1/16) = 4(64)
9x
(f) 81 =
.
9
29
1.8. MULTIPLICATION AND FACTORIZATION
1.8
Multiplication and Factorization
Multiplication and factorization are reverse processes. We shall deal with multiplication first.
In an expression such as ax + y it is to be understood that the multiplication is to be carried out
first and the addition after that. Thus 2.3 + 5 is equal to 11 (and not 16). If we wish to reverse the
order of operations we must use brackets. Thus in the expression a(x + y) we perform the addition
first, and then the multiplication. Since the inclusion of brackets changes the meaning, care should
be taken.
DO NOT LEAVE OUT BRACKETS WHICH SHOULD
BE THERE.
As is well known
a(x + y) = ax + ay
(1.12)
Now if two brackets are multiplied together we proceed as follows:
(a + b)(x + y) = (a + b)x + (a + b)y.
This is obtained by substituting (a + b) for a in (1.12). Then using (1.12) again, we have
(a + b)(x + y) = ax + bx + ay + by.
The effect is that every term in the first bracket is multiplied by every term in the second. Of
particular importance is the case where the two brackets are the same. Thus
(a + b)2 = (a + b)(a + b) = a2 + ab + ba + b2 = a2 + 2ab + b2 .
Please note that
(a + b)2 6= a2 + b2 .
It is then easy to see that (a − b)2 = a2 − 2ab + b2 , and in summary we have
(a ± b)2 = a2 ± 2ab + b2
Similarly,
(a + b)3
=
(a + b)(a + b)2
=
=
(a + b)(a2 + 2ab + b2 )
a3 + 2a2 b + ab2 + ba2 + 2b2 a + b3
=
a3 + 3a2 b + 3ab2 + b3 .
We could of course continue in this way indefinitely. Fortunately there is a simple rule which tells
us what the expansion of (a + b)n is for any n ∈ N.
Pascal’s Triangle
Figure 1.8.1 shows what is known as Pascal’s Triangle. It is constructed as follows:
Write down 1 in the first row and then two 1’s in the second row as shown. Each new row is started
off and ended with 1’s. The intervening numbers are obtained by adding the two numbers directly
above as shown by the arrows.
30
CHAPTER 1. ALGEBRA
1
1
1
2
1
1
1
5
1
4
R
3
R
1
3
6
10
1
4
10
1
5
1
Figure 1.8.1
In the third row we find the numbers 1 2 1. Note that
(a + b)2 = 1.a2 + 2.ab + 1.b2 .
In the fourth row we have 1 3 3 1 and as we have seen
(a + b)3 = 1.a3 + 3.a2 b + 3.ab2 + 1.b3 .
In the same way, using the next two rows, we have
(a + b)4 = a4 + 4a3 b + 6a2 b2 + 4ab3 + b4 ,
(a + b)5 = a5 + 5a4 b + 10a3 b2 + 10a2 b3 + 5ab4 + b5
and so on.
Of course, this is not a proof, but merely a rule of thumb, although a proof may be constructed
with some further knowledge. Please note that
(a + b)n 6= an + bn
(unless n = 1).
Thus
(a + b)7 6= a7 + b7
and (a + b)1/2 6= a1/2 + b1/2 .
Example 1.8.1
Expand (x + 4)5 .
Solution
Using Pascal’s Triangle
(x + 4)5
=
=
x5 + 5x4 .4 + 10x3 42 + 10x2 43 + 5x4 4 + 45
x5 + 20x4 + 160x3 + 640x2 + 1280x + 1024.
=
x4 + 4x3 (−3) + 6x2 (−3)2 + 4x(−3)3 + (−3)4
=
x4 − 12x3 + 54x2 − 108x + 81.
Example 1.8.2
Expand (x − 3)4 .
Solution
Again by Pascal’s Triangle
(x − 3)4
1.8. MULTIPLICATION AND FACTORIZATION
31
It is worth checking that the signs are correct when expanding an expression of the form (a − b)n .
This is easily done, for in the expansion one will obtain alternating positive and negative signs. For
example
(a − b)6 = a6 − 6a5 b + 15a4 b2 − 20a3 b3 + 15a2 b4 − 6ab5 + b6 ,
using one more row of Pascal’s Triangle.
Let us now turn to the question of factorization. We shall assume that factorizations of the types
found in the next four examples are familiar, but further examples of them may be found in the
exercises.
In particular a2 − b2 = (a − b)(a + b) (the difference of two squares), while a2 + b2 (the sum of
squares) has no (real) factors in general.
Example 1.8.3
x2 + 3ax = x(x + 3a).
Example 1.8.4
4x2 − 9y 4 = (2x − 3y 2 )(2x + 3y 2 ).
Example 1.8.5
2x2 − x − 15 = (x − 3)(2x + 5).
Example 1.8.6
x3 + 4x2 − 5x = x(x2 + 4x − 5) = x(x − 1)(x + 5).
Example 1.8.7
3ax − bx + 3ay − by = x(3a − b) + y(3a − b) = (x + y)(3a − b).
The following two factorizations, known respectively as the sum and difference of two cubes,
may not be so well known.
a3 + b3 = (a + b)(a2 − ab + b2 )
(1.13)
a3 − b3 = (a − b)(a2 + ab + b2 )
(1.14)
That these two formulae are correct may easily be checked by multiplying out the right-hand sides.
Example 1.8.8
Factorize 8x3 − 1.
Solution
Using (1.14) we have
8x3 − 1 = (2x − 1)(4x2 + 2x + 1).
32
CHAPTER 1. ALGEBRA
Example 1.8.9
Factorize 27a3 + 125b6.
Solution
Using (1.13), we obtain
27a3 + 125b6 = (3a + 5b2 )(9a2 − 15ab2 + 25b4 ).
Exercise 1.8
1. Expand:
(a) (x + 2y)2
(d) (x − y)3
(g) (x2 − y)3
(j) (x − y)4
(m) (x + 1/x)4
(p) (x + y)5
(b) (3x − 4y)2
(e) (x − 2y)3
(h) (x + y 2 )3
(k) (y − 3x)4
(n) (2x − 1/2)4
(q) (2x − y)5
(c) (x + y)3
(f) (2x − 3y)3
(i) (x + y)4
(l) (3y − 2x)4
(o) (2x − 1/x)4
(r) (a − 2b + c)2 .
2. Simplify where possible :
(a) (132 − 52 )1/2
(e) (x4 + y 4 )1/4
(b) (x2 − y 2 )1/2
x2 − 1
(f)
1−x
p
x2 + y 2
√
(h) ( x2 + 2)2 .
(c) (x1/3 + y 1/3 )3
4x
(g) 1 −
(1 + x)2
(d)
3. Factorise as far as possible :
(a) 9x2 − 4y 2
(b) 16y 4 − 1
(c) x5 − 36x3 y 2
(d) x5 − 9x
(e) x5 − 4x3
(f) a4 − b4
(g) 16x4 − y 4
(h) x2 + 5x − 14
(i) y 2 + 7y − 8
(j) 4x2 + 4x − 1
(k) 7x2 + 7x − 7
(l) x2 + 26x − 27
(m) 2x2 + x − 1
(n) 4y 2 − 5y + 1
(o) 4x3 − 4x2 − 3x
(p) y 4 − 3y 2 + 2
(q) x4 − x2 − 6
(r) x4 − 5x2 + 6
(s) x6 + 7x3 − 8
(t) x4 + x3 − x2 .
4. Factorise as far as possible:
(a) x3 − 27
(d) x6 − 64y 6
(g) x9 − y 9
(b) 8y 3 − a3
(e) a3 − b3
(h) x8 − 1
(c) 8x3 + 27
(f) x9 + y 9
(i) 1 − 3x + 3x2 − x3 .
5. By choosing suitable counter-examples, convince yourself that the following identities are all
WRONG:
√
√
√
√
√
(a) a + b = a + b
(b) (a + b)2 = a2 + b2
(c) ab = a b
(d) (a + b)−1 = a−1 + b−1
(g)
1.9
(e)
1 1
2
+ =
a
b
a+b
a
c
a+c
+ =
.
b
d
b+d
Quadratic Equations
Suppose we wish to solve for x the general quadratic equation
ax2 + bx + c = 0,
(f)
√
√
−a = − a
33
1.9. QUADRATIC EQUATIONS
where a 6= 0. Dividing by a, we have
b
c
x2 + x + = 0.
a
a
Now we note that
Hence
and so
x+
b 2
b
b2
= x2 + x + 2 .
2a
a
4a
c b 2
b2
b
− 2 =0
x2 + x + = x +
a
a
2a
4a
b 2
b2 − 4ac
=
.
2a
4a2
In the case that b2 − 4ac ≥ 0, taking square roots, we obtain the Quadratic Formula
r
√
b2 − 4ac
b2 − 4ac
b
=±
=
±
,
x+
2a
4a2
2a
x+
i.e.
b
x=− ±
2a
usually written in the familiar form below.
(1.15)
√
b2 − 4ac
,
2a
The Quadratic Formula
x=
−b ±
√
b2 − 4ac
2a
(1.16)
Further, in the case that b2 − 4ac < 0, then the left-hand side of (1.15) is negative and the
right-hand side is not. Hence, in this case there are no solutions.
Example 1.9.1
Solve the quadratic equations
(a) x2 + 4x − 7 = 0
(b) 4x2 − 12x + 9 = 0 (c) x2 + 2x + 8 = 0.
Solution
(a) Using(1.16) gives
x=
−4 ±
√
√
16 + 28
−4 ± 44
=
= −2 ± 11.
2
2
√
(b) Using the formula, we have
√
144 − 144
3
x=
= .
8
2
Notice that in this case there is only one (distinct) solution).
12 ±
(c) If we attempt to use the formula, we have
x=
−2 ±
√
4 − 32
.
2
The quantity under the square root is negative, and since no real number has a negative square
root, there are no solutions (at any rate in the set R of real numbers).
34
CHAPTER 1. ALGEBRA
From the preceding three examples we see that a quadratic equation may have two, one or no
solutions. Indeed it is the sign of the term under the square root in (1.16) which determines which of
these three possibilities happens. For if b2 − 4ac is negative there will be no solutions, if b2 − 4ac = 0
there will be the single solution given by x = −b/(2a) and if b2 − 4ac is positive there will be two
solutions given by (1.16).
Sometimes, however, the simplest method involves factorization. Suppose we wish to solve
x2 + 2x − 3 = 0.
We have
(x − 1)(x + 3) = 0.
Now if the product of two numbers is zero, then one of them must be zero. It follows that
x−1=0
or x + 3 = 0,
i.e. x = 1 or x = −3.
There is clearly a close connection between factorization and the solution of equations. We shall
return to this matter in a few moments, but a few words of warning are in order at this stage.
Although it is true that if ab = 0 then a = 0 or b = 0, it is NOT TRUE that if ab = 1
then a = 1 or b = 1. For example, we might have a = 1/2 and b = 2.
Now our previous example showed that there was a close connection between the factors of a
quadratic expression and the solution of the corresponding quadratic equation. We shall now explore
this idea in detail.
Suppose then x1 and x2 are solutions of ax2 + bx + c = 0. By (1.16) we have
√
√
−b + b2 − 4ac
−b − b2 − 4ac
x1 =
and x2 =
.
2a
2a
Then
√
√
b b2 − 4ac ih
b b2 − 4ac i
a(x − x1 )(x − x2 ) = a x +
−
x+
+
.
2a
2a
2a
2a
We may treat this as the difference of two squares, to obtain
h
a(x − x1 )(x − x2 ) = a
h
x+
b 2 b2 − 4ac i
−
= ax2 + bx + c.
2a
4a2
So x1 and x2 are solutions of ax2 + bx + c = 0 if and only if
ax2 + bx + c = a(x − x1 )(x − x2 ).
Of course, it is necessary for such solutions to exist if this statement is to make sense. As we have
seen this will happen provided b2 − 4ac ≥ 0. We have proved the following result.
Theorem 1.9.1 (a) If b2 − 4ac ≥ 0 then x1 and x2 are solutions of ax2 + bx + c = 0 if and only if
ax2 + bx + c ≡ a(x − x1 )(x − x2 ).
(b) If b2 − 4ac < 0 then ax2 + bx + c has no linear factors and ax2 + bx + c = 0 has no real solutions.
Example 1.9.2
Factorize (a) 2x2 − 3x − 6
(b) 5x2 − 2x + 6
√
(c) 25x2 − 5 3x + 3/4.
35
1.9. QUADRATIC EQUATIONS
Solution
(a) First we solve the quadratic equation
2x2 − 3x − 6 = 0.
Using (1.16), the solutions are given by
x=
3−
√
9 + 48
3
= ±
4
4
√
57
.
4
Hence by Theorem 1.9.1,
√ √ 57
57
3
3
x− +
.
2x2 − 3x − 6 = 2 x − −
4
4
4
4
(b) Here the term “b2 − 4ac” is 4 − 120 = −116. Since this is negative the expression cannot be
factorized.
√
(c) Again we solve the quadratic equation 25x2 − 5 3x + 3/4 = 0 to obtain
√
√
√
5 3 ± 75 − 75
3
x=
=
.
50
10
There is only one solution, but we may still apply Theorem 1.9.1 with x1 = x2 . Then
√
√
√
3
25x2 − 5 3x + = 25(x − 3/10)2 = (5x − 3/2)2 .
4
Some equations which are not quadratic can be reduced to quadratic equations by a substitution.
Example 1.9.3
Solve the equation x4 + 3x2 − 10 = 0.
Solution
This is actually a quadratic in x2 . Factorizing, we have
(x2 − 2)(x2 + 5) = 0.
√
Thus x2 = 2 or x2 = −5. The latter gives no solutions, so we obtain x = ± 2.
Example 1.9.4
Solve the equation 22x − 12.2x + 32 = 0.
Solution
Setting y = 2x we have
y 2 − 12y + 32 = 0,
so (y − 8)(y − 4) = 0, and y = 8 or y = 4. Thus
2x = 4
whence x = 2 or x = 3.
or 2x = 8
36
CHAPTER 1. ALGEBRA
The formula (1.16) was known to Islamic mathematicians in about 1000 A.D. The question of how
to solve the general cubic equation ax3 + bx2 + cx + d = 0 was solved by the Italian mathematicians
Cardano and Tartaglia in the early fifteenth century and within a short time Ferrari had solved
the general quartic equation (i.e. involving terms in x4 ). There the matter rested for about 300
years. Nobody was able to find a formula for solving the general fifth order equation. Finally the
Dane, Abel, showed that there was no such formula. The problem was concluded by the French
mathematician, Galois, who showed there was no general formula for the solution of any equations
of higher order than four. We shall return to this later, but only to deal with rather simple kinds of
higher order equation.
Exercise 1.9
1. Solve the following equations:
(a) x2 = 9
(d) x2 = 4x
(g) x5 = x3
(j) x2 − 4x = −4
(m) 3x2 − 6x − 3 = 0
(p) 2x2 + 2x − 1 = 0
(s) x4 − 2x2 − 1 = 0
2. Solve the following equations:
√
√
(a) x = x3
(d) x1/2 = −x3/2
(g) 3x−2 = x−1
(j) (2x − 3)4 = 1
(m) x(x + 1)3 = x(x + 1)
(p) 2.4x+1 + 4x − 1 = 0
(b) x2 = −16
(e) x = x5
(h) x2 − 2x = −2
(k) 3x2 − 6x + 2 = 0
(n) x2 − 2x + 1 = 0
(q) 3x3 + 3x2 = 6x
(t) y 6 + 9y 3 + 8 = 0.
(c) x2 = 2
(f) x3 = −27
(i) x2 + 2x = 2
(l) x2 − x − 2 = 0
(o) 2x2 + 3x − 1 = 0
(r) y 3 + 7y = 6y 2
√
√
(b) x = − x5
(e) 3x−1 = x
(h)(x − 2)(x − 1) = x(1 − x)(x − 2)
(k) (3x − 4)6 = 1
(n) x(x − 1)3 = x(x − 1)
(n) x(2x − 3)2014 = (3 − 2x)x.
(c) 4x−1/2 = 2
(f) x−1 + x−2 = 1
(i) 4 − x2 = (4 − x2 )2
(l) (x − 3)2 = x − 3
(o) 4x − 2x − 2 = 0
3. (a) Use the quadratic formula to solve the equation 2x2 − 3x − 1 = 0, and hence factorize
2x2 − 3x − 1.
(b) Factorize 7x2 + 35x + 14.
4. Find the real values of k for which the equation x2 + (k + 1)x + k 2 = 0 has
(a) real roots
(b) one root double the other.
5. Suppose the equation ax2 + bx + c = 0 has solutions x = α, x = β.
(a) Write down a factorization for ax2 + bx + c in terms of α and β.
(b) Show that α + β = −b/a , and αβ = c/a.
(c) Express α2 + β 2 in terms of a, b, and c. (Note that α2 + β 2 = (α + β)2 − 2αβ.)
(d) Express α3 + β 3 in terms of a, b and c.
6. Suppose the quadratic equation x2 + 2(k + 1)x + 2(k + 5) = 0 has roots α and β.
(a) Express α + β and αβ in terms of k.
(b) Show that (2α + αβ) and (2β + αβ) are the roots of a quadratic equation of the form
y 2 − 16y + a = 0 and express a in terms of k.
7. (a) Find the values of k for which the equation x2 + (k + 4)x + 5k = 0 has equal roots.
(b) The roots of the equation x2 + (k + 4)x + 5k = 0 are α and β. Given that k 6= 0, form
a quadratic equation, with the coefficients expressed in terms of k, whose roots are α/β
and β/α.
37
1.10. INEQUALITIES
√
8. Steve Illy believes that 1 − x2 = 1 − x for any x. Give an example to prove him wrong, and
find all values of x for which he is right.
9. Brenda Rackets believes that 1/2 − 2x−1 = 1/(2 − 2x)−1 = 2 − 2x.
1
1
Her brother Brian believes that 1/2 − 2x−1 = 1/2 − (2x)−1 = −
.
2 2x
What is the correct value of 1/2 − 2x−1 when x = 2? Show that Brian is wrong for all real
values of x, and find all values of x for which Brian is right.
10. Clive Lot believes that (2−1 − x−1 )−1 = 2 − x. Give an example to show that he is wrong,
and find all real values of x for which he is right.
1.10
Inequalities
We have seen that the real numbers R can be represented as a straight line. We shall call that part
of R which lies to the right of 0 (i.e. which includes 1) the positive reals and that part which lies to
the left of 0 the negative reals. The point 0 itself is neither positive nor negative. If x is positive,
we write x > 0 and if x is negative, x < 0.
Definition 1.10.1 We say that x is greater than y (or y is less than x) if x − y > 0, and write
this as x > y (or y < x).
Geometrically this means that x is to the right of y on R (Figure 1.10.1).
−3
−2
−1
x<0
0
1
2
3
x>0
Figure 1.10.1
Order Axioms
We shall assume the following:
(1). For every x ∈ R exactly one of the following is true: x > 0, x < 0, x = 0.
(2). If x, y ∈ R, x > 0 and y > 0, then x + y > 0 and xy > 0.
From these two properties (known as order axioms) it is possible to deduce all the other properties
listed below. We shall not present the proofs of these properties here, although their statements are
simple to understand (but see the exercises).
(3). If x > 0, then −x < 0 and x−1 > 0.
(4). If x > y and y > z then x > z.
(5). If x > y then x + c > y + c for any c ∈ R.
(6). If x > y and c > 0, then cx > cy.
(7). If x > y and c < 0, then cx < cy.
Properties 3 – 6 should cause no trouble, but property 7 demands care. Notice that 3 > 2 but
(−5).3 < (−5).2 since −15 < −10.
38
CHAPTER 1. ALGEBRA
Thus, if one multiplies an inequality by a negative number, the sign of the
inequality is reversed.
This means that one cannot multiply an inequality by an unknown if the sign of the unknown may
be positive or negative.
Thus, although 3 > 2 IT DOES NOT FOLLOW THAT 3x > 2x for every x (try
x = −1).
THIS ERROR IS MADE VERY FREQUENTLY.
AVOID IT !!!
We shall also use the following notation:
a ≤ b means that a > b or a = b. Similarly a ≥ b means that a < b or a = b.
It follows that both of the following are true:
3 ≤ 4,
3 ≤ 3.
In general, if a ∈ R, then {x | x > a} is the set of points to the right of a not including a itself,
while {x | x ≥ a} is the set of points to the right of a including a itself.
All of the properties 4 – 7 listed above hold if > (respectively < ) is replaced by ≥ (respectively ≤).
Frequently we are confronted with the situation where we know that a < x and x < b. Necessarily
then a < b, and we write this simply as a < x < b. This means that at the same time x must be
both greater than a and less than b, i.e. that x is between a and b.
For example 2 < x < 4 means 2 < x and x < 4, i.e. x is between 2 and 4. (It does NOT follow
that x = 3; it may be, for example, that x = 3.156.)
On the other hand 4 < x < 2 is meaningless, since this would mean that 4 < x and x < 2. It
would then follow from (4) above that 4 < 2, which is nonsense. Expressions such as 3 < x > 4
should never be used.
Interval Notation
Now suppose a < b. In the solution of inequalities upon which we are about to embark, the following
notations are useful.
(a, b) = {x | a < x < b}
[a, b] = {x | a ≤ x ≤ b}
(a, b] = {x | a < x ≤ b}
[a, b) = {x | a ≤ x < b}
(a, ∞) = {x | a < x}
[a, ∞) = {x |a ≤ x}
(−∞, a) = {x | x < a}
(−∞, a] = {x | x ≤ a}.
Such sets are called intervals, some of which are illustrated in Figure 1.10.2.
39
1.10. INEQUALITIES
(
a
)
b
a
b
(a, b)
[a, b]
)
b
a
a
[a, ∞)
[a, b)
Figure 1.10.2
In Figure 1.10.2, a round bracket means that the end point is not included, while a dot indicates
that it is. The idea behind the interval notation is to use a round bracket when the end-point is not
included and a square bracket when it is. Thus 3 ∈ [3, 5) but 3 6∈ (3, 5).
Note that the symbols −∞ and ∞ are used only as a quick way to write down intervals which
do not have finite end-points.
After the symbol ∞ (and similarly before the symbol −∞) one never has a square bracket,
because this would mean that ∞ was an element of the interval. This would be incorrect
because INFINITY IS NOT A REAL NUMBER.
We have taken great care to avoid any definition involving the use of infinite quantities except as a
notational device as above.
Notice further that the set [4, 7] is quite different from the set {4, 5, 6, 7}. The former
consists of all real numbers between 4 and 7 (inclusive) and contains such elements as
5.7, 6.133, 4.5 etc. The latter set has only four elements altogether.
Now let us turn to the problem of solving inequalities. We shall not develop any general theory,
but merely illustrate the procedures which may be used by giving examples.
Example 1.10.1
Solve the inequality 3x − 5 < 7.
Solution
Adding 5 to both sides (i.e. using property 5 above), we have
3x < 12.
Now dividing by 3 (i.e. multiplying by 1/3 and using property (6)) we obtain
x < 4,
or in interval notation x ∈ (−∞, 4). We could also say the solution set is {x | x < 4} = (−∞, 4).
Example 1.10.2
Solve 7 − 2x ≤ 4.
40
CHAPTER 1. ALGEBRA
Solution
On subtracting 7 we have
−2x ≤ −3.
Dividing by −2 (that is multiplying by −1/2) we obtain
x≥
3
,
2
i.e. x ∈ [3/2, ∞).
Note that we have multiplied by a negative and so reversed the sign of the inequality!
It is worth remarking at this stage that if ab > 0 (i.e. ab is positive) then either (i) a > 0
and b > 0 (i.e. both are positive) or (ii) a < 0 and b < 0 (i.e. both are negative). On
the other hand if ab < 0 then either (i) a > 0 and b < 0 or (ii) a < 0 and b > 0 (i.e. one
is positive and the other negative).
Example 1.10.3
Solve x2 + x − 6 < 0.
Solution
Factorizing we have
(x − 2)(x + 3) < 0.
Now it is clear that x − 2 = 0 when x = 2, and x + 3 = 0 when x = −3. Thus x − 2 > 0 when
x > 2 and x + 3 > 0 when x > −3.
x+3 − − 0 + + + + + + + + + + + + + + +
−4
−3
−2
−1
0
1
2
3
4
x−2 − − − − − − − − − − − − 0 + + + + +
Clearly then both factors are negative when x < −3 and both are positive when x > 2. The diagram
above illustrates the situation. We are interested in the case where the product is negative, i.e.
when one factor is positive and the other negative. This occurs between −3 and 2 and the solution
set is (−3, 2); that is, x is a solution if and only if −3 < x < 2.
The diagram below is a condensed version of the previous one.
−3
2
x+3
− 0
+
x−2
−
− 0 +
+ 0
− 0 +
(x + 3)(x − 2)
+
41
1.10. INEQUALITIES
We arrange the points where the various factors are zero along the top in ascending order. The
next two rows show the signs of the two factors in the intervals {x | x < −3}, {x | − 3 < x < 2}
and {x | x > 2}. The last line gives the sign of the product which is read off vertically using the
rules of elementary arithmetic that “a minus times a minus is a plus” and so on.
Example 1.10.4
Solve x2 − 3x − 4 ≥ 0.
−1
4
x+1
− 0
+
x−4
−
− 0 +
+ 0
− 0 +
(x + 1)(x − 4)
+
Solution
Factorizing, we have (x − 4)(x + 1) ≥ 0. This leads to the diagram above and the solution is
x ≤ −1 or x ≥ 4, that is the solution set is (−∞, −1] ∪ [4, ∞).
DO NOT ATTEMPT TO WRITE THIS AS −1 ≥ x ≥ 4 as this would imply
−1 ≥ 4.
Example 1.10.5
Solve x2 + 2x + 7 > 0.
Solution
Completing the square, we have (x + 1)2 + 6 > 0. This is always true as both terms are positive.
The solution is then all x ∈ R, i.e. the solution set is R.
Example 1.10.6
Solve x2 + 4x + 9 ≤ 0.
Completing the square we obtain (x + 2)2 + 5 ≤ 0. This is never true since the left-hand side
must be positive. The solution set is the empty set ∅.
Example 1.10.7
Solve
1
< 2.
x−1
Solution
Subtracting 2 gives us
1
− 2 < 0,
x−1
42
CHAPTER 1. ALGEBRA
that is
3 − 2x
< 0.
x−1
Multiplying by −1 (and changing the sign of the inequality!) gives
2x − 3
> 0.
x−1
Now the numerator is zero when x = 3/2 and the denominator when x = 1. We obtain the
diagram below (where u stands for undefined).
3/2
1
x−1
− 0
+
2x − 3
−
− 0 +
Quotient
+ u
− 0 +
+
The solution is then x < 1 or x > 3/2, that is the solution set is (−∞, 1) ∪ (3/2, ∞).
Note that we could not proceed as follows :
1
< 2 so
x−1
1 < 2(x − 1),
because we do not know, at the outset, whether x − 1 is positive or negative, and hence whether the
sign of the inequality should be reversed or not.
THIS IS A VERY BAD MISTAKE. AVOID IT !!!
Example 1.10.8
Solve
x
≤ 2.
2x + 3
Solution
As before, we have
so
i.e.
We may safely divide by 3 to reach
x
−2≤0
2x + 3
x − 4x − 6
≤ 0,
2x + 3
3(x + 2)
≥ 0.
2x + 3
x+2
≥ 0.
2x + 3
43
1.10. INEQUALITIES
−3/2
−2
x+2
− 0
+
2x + 3
−
− 0 +
Quotient
+ u
− 0 +
+
The corresponding diagram is shown above.
x+2
Thus
> 0 provided x < −2 or x > −3/2.
2x + 3
Also (x + 2)/(2x + 3) = 0 when x = −2.
The solution set is then (−∞, −2] ∪ (−3/2, ∞). Note that we do not include the end-point −3/2,
as this would entail division by zero.
Example 1.10.9
Solve 2x − 3 < 3x − 1 ≤ 5x.
Solution
This means 2x − 3 < 3x − 1 and 3x − 1 ≤ 5x,
i.e. −2 < x and −1 ≤ 2x.
So x > −2 and x ≥ −1/2 . This is clearly satisfied provided x ≥ −1/2.
Example 1.10.10
Solve x − 2 ≤ 2x + 1 < 5 − x.
Solution
This means x − 2 ≤ 2x + 1 and 2x + 1 < 5 − x.
Thus −3 ≤ x and 3x < 4, so x ≥ −3 and x < 4/3 , giving the solution set [−3, 4/3).
Exercise 1.10
1. State whether the following are true or false:
(a) 3 ∈ [2, 3)
(b) 3 ∈ [2, 4)
(c) −∞ ∈ (−∞, 3)
(d) 7 > 5
(e) 5 ≥ 7
(f) 7 ≥ 7
(g) [2, 5] = {2, 3, 4, 5}.
√ √
2. Sets A and B are given by A = (−1, 3], B = {0, −1, 3, 2, 4, 4, −3/4}. Find:
(a) A ∩ N
(b) B ∩ Z
3. Solve the inequalities:
(a) x + 2 > x
(d) −5x − 4 ≥ 2x + 7
(g) x2 ≤ −2x
(j) (x − 1)(x + 2) ≥ 0
(m) x2 < −4.
(c) B ∩ Q
(d) A ∩ B
(b) −3x < 12
(e) x2 < 4
(h) (x − 2)2 ≤ 0
(k) (x − 1)(x − 2) < 2
(e) X = {x ∈ B | x2 ∈ A}.
(c)−2x + 2 ≤ −6x − 8
(f) x2 ≥ 9
(i) x(x + 3) ≤ 0
(l) x2 ≥ −9
44
CHAPTER 1. ALGEBRA
4. Solve the inequalities:
(a) x3 ≤ 8
(b) x4 − 16 > 0
(c) x2 (x − 2)(x + 3) ≥ 0
(d) (x − 1)2 x < 0
(e) x(2x + 3)(3 − x) < 0
(f)
(g)
(h)
1
<2
x
1
1
(i) ≥
2
2−x
1
1
(l)
≥
3x + 2
5x + 2
(m) 4 < 3 − 2x < 5
(n) −x + 2 < x − 3 ≤ 7
(o) 2 + x ≤ 3 + x < 6 − 2x
(p) 2x ≥
(q) 8 ≤ 4−2x < 32
(r) 32x − 4.3x + 3 < 0.
1
1
< 3
x
x
1
−1
(j)
≤
2−x
5 − 3x
1
2
1
1
≥
x2
x
1
(k)
≤x
2−x
5. What is wrong with the following argument?
1
1
<
4
2
and so
1 2
2
<
1
.
2
Taking logs,
log
1 2
2
< log
1
2
.
Then, using elementary properties of logs,
2 log
1
2
< log
1
2
and by cancellation 2 < 1.
6. In this exercise you are meant to deduce properties 3 – 7 (see p.37) from properties 1 and 2.
Recall that x > y if and only if x − y > 0.
(a) Prove properties 5 and 6 from property 2.
(b) Prove property 3 from properties 1 and 2.
(c) Prove properties 4 and 7.
7. (a) Show that if 0 < x < 1, then 0 < x2 < x.
(b) Show that if x > 1, then x2 > x.
(c) Show that if 0 < x < y, then
1
1
> .
x
y
(d) Would (c) still be true if we only knew that x < y? Explain.
8. Suppose a, b 6= 0. Show that ab > 0 if and only if
1.11
a
> 0.
b
Absolute Values
It is often necessary to consider the magnitude of a number irrespective of its sign. Thus, if we wish
to say that an object has a mass of 5.32 kg correct to two decimal places, we mean that its mass
is 5.32 kg with an error of at most 0.005 kg. When we say this what we mean is that the error lies
between −0.005 kg and +0.005 kg. In saying that the error is at most 0.005 kg, we are effectively
dropping the minus sign if it is there.
45
1.11. ABSOLUTE VALUES
Absolute Value
Definition 1.11.1 The absolute value, |x|, of x ∈ R is
x if x ≥ 0
|x| =
−x if x < 0
This equation is to be read in the following way:
If x ≥ 0, then |x| = x, while if x < 0, then |x| = −x.
Thus for example, if x = 5, then |x| = 5, i.e. |5| = 5, while if x = −3, then |x| = −x = −(−3) = 3,
i.e. | − 3| = 3.
Hence |x| ≥ 0 for all x ∈ R. Do not be confused by the fact that |x| = −x if x < 0. This does
not mean that |x| < 0, since in this case |x| = −x and −x > 0.
Geometrically, |x| is just the distance from x to 0. Thus −3 is at a distance 3 = | − 3| from 0.
More generally |x − y| is the distance from x to y, irrespective of sign. Note then that
|x| = | − x|
for all x ∈ R.
Theorem 1.11.2 Let M > 0. If |x| < M , then −M < x < M .
Proof. Suppose firstly that x ≥ 0. Then |x| = x. Thus if |x| < M then 0 ≤ x < M and so clearly
−M < x < M .
Now suppose that x < 0. Then |x| = −x. Hence if |x| < M then −x < M . It follows (multiplying
by −1 and reversing the sign of the inequality) that x > −M . Since clearly M > x (the right-hand
side is negative) we have −M < x < M .
The converse of this result also holds.
Theorem 1.11.3 If M > 0 and −M < x < M , then |x| < M.
Proof. Suppose firstly that x ≥ 0. Then −M < x < M implies that 0 ≤ x < M . Also |x| = x and
we have |x| < M .
On the other hand, if x < 0, then our hypothesis is −M < x < 0. Thus 0 < −x < M . Moreover,
|x| = −x and so 0 < |x| < M , and finally |x| < M .
From these two results we obtain the following.
Suppose M > 0. Then |x| < M if and only if − M < x < M
(1.17)
This result remains valid if < is replaced by ≤ throughout. We may also interpret (1.17) geometrically. What we have shown is that |x| < M if and only if x ∈ (−M, M ).
(
−M
0
)
M
-
Figure 1.11.1
Thus, if |x| < M , then x lies in the shaded region shown in Figure 1.11.1. It follows that if |x| > M ,
then x is not in the shaded region and also x 6= ±M . We thus have the following result.
Suppose M > 0. Then |x| < M if and only if x < −M or x > M
(1.18)
46
CHAPTER 1. ALGEBRA
Example 1.11.1
Solve |x − 3| < 2.
(Geometrically this means find all points x, whose distance from 3 is less than 2.)
Solution
Using (1.17) with x − 3 in place of x, we have
−2 < x − 3 < 2,
so
1 < x < 5,
i.e. x ∈ (1, 5).
Example 1.11.2
Solve |5 − x| ≥ 4.
Solution
Using (1.18) (with ≥ in place of >), we have
5 − x ≥ 4 or 5 − x ≤ −4,
i.e.
−x ≥ −1
or
− x ≤ −9.
So x ≤ 1 or x ≥ 9. In interval notation, the solution set is (−∞, 1] ∪ [9, ∞).
Example 1.11.3
Solve
1
< 4.
|x − 2|
Solution
We may safely multiply through by |x − 2|, since this is never negative.
(This should be contrasted with the inequality 1/(x − 2) < 4 in which we cannot multiply through
by x − 2, since we do not know the sign of x − 2.) We obtain
1 < 4|x − 2|,
i.e
|x − 2| >
Thus
1
.
4
1
1
or x − 2 < − ,
4
4
i.e. x > 9/4 or x < 7/4. So the solution set is (−∞, 7/4) ∪ (9/4, ∞).
x−2>
Example 1.11.4
Solve |x − 2| < x.
47
1.11. ABSOLUTE VALUES
Solution
Using (1.17), we have −x < x − 2 < x, i.e.
−x < x − 2
and x − 2 < x.
Hence
2 < 2x
and
− 2 < 0.
The second inequality is always true, so we may ignore it.
So we have 2 < 2x , i.e. x > 1.
The following elementary result is useful in solving some further inequalities.
√
Theorem 1.11.4 If x ∈ R, then |x| = x2 .
√
x2 means the positive square root of x2 .
√
Now if x ≥ 0, then |x| = x, and clearly x = x2 .
Proof. First recall that
On the other hand, if x < 0, then |x| = −x. So
√
p
√
x2 = x.x = (−x)(−x) = −x = |x|.
Corollary 1.11.5 If x ∈ R, then |x|2 = x2 .
Proof. This follows immediately from Theorem 1.11.4 after squaring.
Example 1.11.5
Solve |x − 2| > |x + 1|.
Solution
Squaring and using Corollary 1.11.5, we have
(x − 2)2 > (x + 1)2 ,
i.e.
x2 − 4x + 4 > x2 + 2x + 1.
This yields −6x > −3, so x < 1/2.
In the solution of this inequality we have assumed that it is legitimate to square the original
inequality, i.e. we assumed that if 0 < a < b, then a2 < b2 . We shall now justify this.
Theorem 1.11.6 If 0 < a < b then a2 < b2 .
Proof. Since a, b > 0 then b + a > 0. Also since a < b, then b − a > 0. Hence
b2 − a2 = (b − a)(b + a) > 0,
i.e.
a2 < b 2 .
Note that for this theorem to be true it is necessary to assume that both a and b are positive.
Without this restriction the result is no longer true. For example −3 < 1, but (−3)2 > 12 .
48
CHAPTER 1. ALGEBRA
Exercise 1.11
1. Solve the following equations:
(a) |x| = 2
(d) |5 − 2x| = 1
(g) |x − 1| = |x2 + 1|
(b) |x| = −2
(e) | − 3 − 6x| = 2
(h) |x|2 − 9|x| + 8 = 0
(c) |3x + 4| = 1
(f) |x2 − 10| = 6
(i) |x2 + x − 6| = 6.
2. Solve the following inequalities:
(a) |x| ≤ 3
(d) |x| < −7
(g) |2 + 3x| > 2
(j) |4x − 3| < 6
1
1
(m)
<
|3x − 2|
|x − 6|
(b) |x| ≥ 5
(e) |3 − 2x| ≤ 3
(h) |3x − 2| < 4
(k) |x − 3| ≤ |x + 6|
1
(n)
≤ 3.
|2x − 1|
(c) |x| ≥ −2
(f) |4 − x| ≤ 2
(i) |1 − 2x| ≥ 2
(l) 1 < |2x − 1| ≤ 3
3. Use Theorem 1.11.4 to show
(a) |ab| = |a| |b| for all a, b ∈ R
(b) |a + b| ≤ |a| + |b| for all a, b ∈ R. Under what conditions does this inequality become an
equality?
(c)
|a| − |b| = |a − b| for all a, b ∈ R.
1
1
> . Is this result true if we are given only that a < b?
a
b
(b) Suppose a < b < 0. Show that a2 > b2 .
4. (a) Show that if 0 < a < b, then
1.12
Sigma Notation
It is often necessary for us to add together a large number of similar terms. It becomes inconvenient
to write them all down, so we employ a piece of notation that enables us to express these sums in a
concise way. Consider for example the series
2 + 4 + 6 + · · · + 100
(1.19)
where the three dots signify the obvious missing terms. This is just the sum of all even numbers
from 2 to 100. Now any even number may be written as 2k, where k ∈ Z. Thus we shall rewrite
(1.19) as
50
X
2k.
(1.20)
k=1
This is read as “the sum from k = 1 to 50 of 2k”. The letter
the first letter of the word “sum”.
P
, sigma, is just a Greek capital S –
When evaluating (1.20), first put k = 1, to obtain 2. Then set k = 2, which gives 4 and a total
so far of 2 + 4. Setting k = 3 yields 6 and adding again gives 2 + 4 + 6. Continuing in this way until
k = 50, we recover (1.19). Similarly
10
X
1
k=1
k2
=1+
1 1
1
+ + ···+
.
4 9
100
We need not sum from k = 1 or indeed use the letter k. Thus
12
X
r=3
(2r + 1) = 7 + 9 + 11 + · · · + 25.
49
1.12. SIGMA NOTATION
Furthermore, the upper (or lower) limit may also be a variable. So
n
X
k=1
k 2 = 1 2 + 2 2 + · · · + n2 .
The following three summations are very useful:
n
X
k=1
n
X
k=1
k 2 = 1 2 + 2 2 + 3 2 + · · · + n2 =
n
X
k=1
n(n + 1)
2
(1.21)
n(n + 1)(2n + 1)
6
(1.22)
n2 (n + 1)2
4
(1.23)
k = 1 + 2 + 3 + ···+ n =
k 3 = 1 3 + 2 3 + 3 3 + · · · + n3 =
Formula (1.21) can easily be proved, since it is just the sum of an arithmetic series. Formulae (1.22)
and (1.23) may be proved by the method of mathematical induction, which will be dealt with in the
next section.
Example 1.12.1
Expand
5
X
n=1
[2n − 3].
Solution
Setting n successively equal to 1, 2, 3, 4 and 5 we have
−1 + 1 + 3 + 5 + 7 = 15.
Example 1.12.2
Evaluate 12 + 22 + · · · + 1002 .
Solution
Using (1.22), we see that,
12 + 22 + · · · + 1002 =
100.101.201
= 338, 350.
6
Evaluating this sum by straightforward addition would take a very long time.
Example 1.12.3
Evaluate 502 + · · · + 1002 .
Solution
Now
502 + · · · + 1002
which may readily be evaluated.
= (12 + · · · + 1002 ) − (12 + · · · + 492 )
100.101.201 49.50.101
=
−
6
6
50
CHAPTER 1. ALGEBRA
Example 1.12.4
Express in sigma notation
4 − 7 + 10 − 13 + 16 − 19.
Solution
Each term is 3 more than the previous, and thus terms are of the form 3n + 1.
To obtain the changes of sign we multiply by (−1)n+1 , which gives +1 when n is odd and −1
when n is even. Since there are 6 terms, we sum from n = 1 to 6. Thus
4 − 7 + 10 − 13 + 16 − 19 =
6
X
(−1)n+1 (3n + 1).
n=1
Example 1.12.5
Evaluate
100 h
X
1
n
n=1
−
1 i
.
n+1
Solution
Expanding we have
h1
1
−
h 1
1i h1 1i h1 1i
1 i
+
−
+
−
+ ···+
−
.
2
2 3
3 4
100 101
All the terms except the first and last cancel out, and we obtain
1−
1
100
=
.
101
101
Exercise 1.12
1. Expand the following sums written in sigma notation :
(a)
15
X
k=3
(3k − 2)
(b)
20
X
3n
(c)
n=1
n
X
sin r
r=2
(d)
k
X
(−1)n n2 .
n=1
2. Evaluate exactly:
(a)
(d)
(g)
8
X
i
(−1) i
i=1
50
X
j=9
3
X
[(j + 1)2 − j 2 ]
(b)
(e)
7
X
2
k=2
201
X
k
(c)
(−1)k
k=−1
(f)
101
X
k
k=−99
3
X
j
j
j=1
(−j)j .
j=1
3. Express the following in sigma notation:
(a) −17 − 18 − 19 − 20 − · · · − 74
1
1
1
1
(c)
+
+
+ ···+
1.2 2.3 3.4
50.51
(e) 9 − 13 + 17 − 21 + · · · + 49
(g) 10 + 12 + · · · + 52
(b) 23 + 26 + 29 + 32 + · · · + 59
(d) 1 + 2 + 22 + 24 + 28 + · · · + 264
(f) 4 + 8 + 16 + 32 + · · · + 512
1
1
1
1
(h) −
+
−
+ ···+
15.19 17.21 19.23
59.63
51
1.13. MATHEMATICAL INDUCTION
4. Prove formula (1.21).
5. Check that formulae (1.22) and (1.23) hold for n = 1, 2, 3 and 4.
6. Use the formulae (1.21), (1.22) and (1.23) to find:
(a) 1 + 2 + 3 + · · · + 199
(b) 13 + 23 + 33 + · · · + 1013
(c) 213 + 223 + 233 + · · · + 1013
(d) 12 + 32 + 52 + · · · + 432
(e) 232 + 242 + 252 + · · · + 432 .
7. Evaluate exactly:
(a)
5
X
1
n=1
(b)
2n
10
X
1
n=1
2n
(c)
100
X
1
n=1
2n
.
What meaning would you think it sensible to attach to the sum
∞
X
1
n=1
1.13
2n
?
Mathematical Induction
We are now presented with the problem of proving the formulae we have just given for
n
X
n
X
k 2 and
k=1
3
k . We shall develop a method of proof, called mathematical induction, which will enable us
k=1
to prove these formulae as well as to establish many other useful facts.
Let us start with a slightly simpler problem. Suppose we are asked to derive and prove a formula
to evaluate
n
X
1
1
1
1
=
+
+ ··· +
.
i.(i
+
1)
1.2
2.3
n(n
+ 1)
i=1
1
1
1
1
2
Now when n = 1, we obtain
= ; when n = 2, we have
+
= ; when n = 3, we obtain
1.2
2
1.2 2.3
3
1
1
1
3
4
+
+
= and when n = 4, we have .
1.2 2.3 3.4
4
5
We are led to suspect that
n
X
i=1
1
n
=
i(i + 1)
n+1
for all n ∈ N.
(1.24)
The above argument does not constitute a proof however. We have only shown that (1.24) holds
for n = 1, 2, 3 and 4. We could proceed further and show, by direct computation, that (1.24)
holds for n = 1, 2, 3, . . . , 100 say, but even this would not guarantee that it held when for example
n = 235 867.
To illustrate the point further, consider the expression n2 +n+41. When n = 1, 2, 3, 4 we obtain
43, 47, 53 and 61 respectively, which are all prime numbers. We might suspect that n2 + n + 41 is
prime for any n ∈ N. If we try a few more, we obtain successively 71, 83, 97, 113, which are all
prime numbers. Surely then this is sufficient evidence!
In fact, n2 + n + 41 is a prime number for n = 1, 2, . . . , 39, but when n = 40 we have
402 + 40 + 41 = 40(40 + 1) + 41 = 40.41 + 41 = 412 ,
which is evidently not prime.
52
CHAPTER 1. ALGEBRA
We need to find a new method of proof. The idea is as follows. Suppose p(n) is a proposition
involving the natural number n. For example, we could have
5n + 3 is divisible by 4
or
n2 ≤ 2n+1
or (1.24), that is
n
X
1
n
=
.
i(i
+
1)
n
+
1
i=1
We wish to show these formulas hold for all n ∈ N, that is to say we wish to show p(n) is true for
all n.
The proof proceeds in two stages.
(I1) We show p(1) is true.
(I2) We show that if p(k) is true for some k ∈ N, then p(k + 1) is also true.
This will be enough to prove that p(n) is true for all n ∈ N . For by (I1), p(1) is true. Then by (I2)
with k = 1, p(2) is true. So by (I2) with k = 2, p(3) is true. So by (I2) with k = 3, p(4) is true, etc.
We may break this procedure down into three steps:
(1) Show p(1) is true.
(2) Assume p(k) is true.
(3) Show that it then follows that p(k + 1) is true.
Notice that (2) and (3) combined are the same as (I2). There is no gratuitous assumption in (2).
What we are showing is that if p(k) is true, then so is p(k + 1). This procedure is known as
mathematical induction.
n
X
1
Now let us return to the problem of evaluating
.
i.(i
+ 1)
i=1
By direct calculation we have been led to suspect that
n
X
n
1
=
.
i(i
+
1)
n
+
1
i=1
(1.25)
We shall prove this by induction.
(1) First we show it is true when n = 1.
In this case the left-hand side is just
(1) is achieved.
1
1
1
1
= , while the right-hand side is
= , so that
1.2
2
1+1
2
(2) Next suppose that for some k,
k
X
1
k
=
.
i(i
+
1)
k
+
1
i=1
(3) We must deduce (10.2) holds for n = k + 1. Now
k+1
X
k
X
1
1
1
=
+
,
i(i
+
1)
i(i
+
1)
(k
+
1)(k
+ 2)
i=1
i=1
(1.26)
53
1.13. MATHEMATICAL INDUCTION
since the left-hand side is the sum of the first k terms plus the (k + 1)th term. Then
k+1
X
1
i(i
+
1)
i=1
=
=
=
=
=
k
1
+
k + 1 (k + 1)(k + 2)
by (1.26)
k(k + 2) + 1
(k + 1)(k + 2)
k 2 + 2k + 1
(k + 1)(k + 2)
(k + 1)2
(k + 1)(k + 2)
k+1
,
k+2
which is just (1.25) with k + 1 instead of n. Thus (3) is completed.
Example 1.13.1
Show
n
X
i=
i=1
n(n + 1)
.
2
(1.27)
(We have already remarked that this can be proved directly, since the left-hand side is just an
arithmetic series. We shall show that (1.27) holds by induction as a further illustration of the
method.)
Solution
Firstly, when n = 1, the left-hand side is just 1, while the right-hand side is
(1.27) holds for n = 1.
Now assume
k
X
i=1
i=
k(k + 1)
.
2
1(1 + 1)
= 1. So
2
(1.28)
We must show (1.27) holds with n = k + 1. We have
k+1
X
i =
i=1
k
X
i + (k + 1)
i=1
=
=
=
k(k + 1)
+ (k + 1)
2
k 2 + 3k + 2
2
(k + 1)(k + 2)
,
2
by (1.28)
which is (1.27) when n = k + 1.
Example 1.13.2
Show
n
X
i=1
i2 =
n(n + 1)(2n + 1)
.
6
(1.29)
54
CHAPTER 1. ALGEBRA
Solution
When n = 1, the left-hand side is 12 = 1, and the right-hand side is 1.2.3/6 = 1.
Now assume
k
X
i2 =
=
k
X
i=1
k(k + 1)(2k + 1)
.
6
Then
k+1
X
i2
i=1
i2 + (k + 1)2
i=1
=
=
=
=
k(k + 1)(2k + 1)
+ (k + 1)2
6
h k(2k + 1) + 6(k + 1) i
(k + 1)
6
2k 2 + 7k + 6
(k + 1)
6
(k + 1)(k + 2)(2k + 3)
6
which is just (1.29) with n = k + 1.
Finally, let us give an example from number theory.
Example 1.13.3
Show 5n + 3 is divisible by 4.
Solution
When n = 1 we have 8 which is obviously divisible by 4.
Now assume 5k + 3 is divisible by 4. Then
5k+1 + 3
= 5.5k + 3
= (4 + 1)5k + 3
= 4.5k + (5k + 3).
By assumption, the term in brackets is divisible by 4, and so clearly is 4.5k . Hence 5k+1 + 3 is
divisible by 4.
Exercise 1.13
1. Prove by induction:
(a)
n
X
i(i + 1) =
(b)
n
X
i3 =
i=1
i=1
n
n(n + 1)(n + 2)
3
n2 (n + 1)2
4
(c) 2 > n, for any n ∈ N.
(d) 12 + 32 + 52 + · · · + (2n − 1)2 =
n(2n + 1)(2n − 1)
3
55
1.13. MATHEMATICAL INDUCTION
(e) 7n + 5 is divisible by 6, for any n ∈ N.
(f) 32n + 3 is divisible by 4, for any n ∈ N.
n
X
(g)
i2i = 2n+1 (n − 1) + 2.
i=1
(h)
n
X
1
(n + 1)(n + 4)
=
.
(i
+
1)(i
+
2)(i
+
3)
4(n
+ 2)(n + 3)
i=0
(i) 8n + 6 is divisible by 7, for any n ∈ N.
(j) 42n + 4 is divisible by 5, for any n ∈ N.
(k) 1 + 2z + 3z 2 + · · · + nz n−1 =
1 − (n + 1)z n + nz n+1
for any z ∈ R, z 6= 1.
(1 − z)2
1
1
1
1
n
+
+
+ ···+
=
.
1.2 2.3 3.4
n(n + 1)
n+1
1
1
1
1
+
+
+ ···+
.
Hence calculate
12.13 13.14 14.15
25.26
2. Prove that
3. Show that
n
X
1
n
=
.
(2i − 1)(2i + 1)
2n + 1
i=1
Hence find
50
X
(i)
1
(2i
−
1)(2i
+ 1)
i=1
(ii)
50
X
1
.
(2i
−
1)(2i
+ 1)
i=26
4. Suppose r ∈ N with r 6= 1. Show that
n
X
ri =
i=1
r(1 − rn )
,
1−r
for any n ∈ N. Hence find (in factors)
(i) 2 + 22 + 23 + · · · + 224
(ii) 2 − 22 + 23 − 24 + · · · − 224
(iii) 4 + 43 + 45 + 47 + · · · + 417 .
5. Prove that
n
X
(i + 1)(i + 2)(i + 3) =
i=1
1
(n + 1)(n + 2)(n + 3)(n + 4).
4
Deduce that if n is not divisible by 5, then
n
X
(i + 1)(i + 2)(i + 3)(i + 4)
i=1
is divisible by 30.