Matriculation QS015 2014

S.Y.Chuah

June 18, 2014

Chapter 1 : Number System

1.1 Real Numbers

(a) Define Natural Numbers N, Whole Number W, Integers Z, Prime Numbers,Rational Numbers Q, Irrational Numbers.

(b) Represent rational numbers and irrational numbers in decimal forms.

(c) Represent the relationship of number sets in a real number system diagram-matically.

(d) Represent open, closed and half-open intervals and their representations onthe number line.

(e) Simplify union , intersection of two or more intervals with the aid of numberline.

1.2 Complex Numbers

(a) Represent a complex number in Cartesian form.

(b) Define the equality of two complex numbers.

(c) Determine the conjugate of a complex number,Z.

(d) Perform algebraic operations on complex numbers.

(e) Represent a complex number in polar form Z = r(cos + isin) where r < 0and pi < < pi.

1.3 Indices, Surds and Logarithms

(a) State the rules of indices.

(b) Explain the meaning of surd and its conjugate and to carry out algebraicoperations on surds.

(c) State the laws of logarithms.

(d) Change the base of logarithms.

1

1.1 Real Numbers R

1.1.1 Sets of Real Numbers

Definition 1 (Real Numbers) The set of real numbers,R, comprises rational numbers andirrational numbers.

-

0

Definition 1.1 Natural numbers,N, are positive numbers that are used for counting:N = {1,2,3, }.

Definition 1.2 Whole numbers,W, are natural numbers including the number zero:W = {0,1,2,3, }.

Definition 1.3 Integers,Z, are whole numbers including their negatives:Z = { ,-2,-1,0,1,2, }.

Definition 1.4 Prime numbers are natural numbers greater than 1 that can be divided byitself and 1 only.Primenumbers = {2,3,5,7,11, }.

Definition 1.5 Rational numbers,Q, are numbers that can be written in the form pq

wherep and q are integers and q 6= 0.Q = {p

qp, q Z, q 6= 0}

In decimal form, rational numbers may be a terminating decimal, such as 34

= 0.75 or arepeating decimal, such as 3

11= 0.27272727 , in which a group of one or more digits

repears indefinitely. Examples of rational numbers are 3,34, 25, 0.6, 1.212121 , 6, 20.

Definition 1.6 Irrational numbers,Q, are numbers that cannot be written in the formpq

where p and q are integers and q 6= 0. For example, pi, e and 3.

Figure 1: Real Number System

Figure 2: Venn Diagram represents different types of real numbers

Exercise 1: Determine whether each statement is true or false.

(a) N W(b) Z N

(c)

3 Q(d) 8.2525 Q

(e) 0.21212212 . . . / Q(f) 0.23 Q

1.1.2 Intervals of Real Numbers

Intervals of real numbers can be illustrated using

1. Set notation denoted by {}. The solution to the inequality x 2 can be expressedin set notation as follows:

{x : x 2}

It is read as : The set of all x such that x is greater than or equal to 2.

2. Real number line denoted by

3. Interval notation denoted by [ ], ( ), [ ) or ( ].(a,b) - open interval[a, b] - closed interval(a,b], [a,b) - half-open interval(a,) - infinite interval

Exercise 2: Summary of Set notation and Interval notation

Problem 1

Note : The symbol is not numerical. When we write [a,), we are simply referringto the interval starting from a and continuing indefinitely to the right.

Problem 2

Graph all real numbers x such that

(i) (20,5)(ii) (2,)

(iii) (,7)(iv) [0, 6]

(v) [6, 1)(vi) [10,)

Problem 3

Graph each of the following on a number line.

(i) All integers x such that 3 < x < 3(ii) All whole numbers x such that x 4

(iii) All natural number x such that 2 x 3(iv) All real numbers x such that 1 x < 5(v) x : 3 < x 9, x primenumber(vi) x : 2 < x < 10, x R

1.1.3 Combining Intervals

Using the symbol of union () and intersection ().The intersection of two intervals is the set of real numbers that belong to both intervals.EXAMPLE 1:

(1, 5) (3, 9)

Hence, (1, 5) (3, 9) = (3, 5).

The union of two intervals is the set of real numbers that belong to one, or the other, orboth of the intervals.

EXAMPLE 2:

(1, 5) (3, 9)

Hence, (1, 5) (3, 9) = (1, 9).

EXAMPLE 3:Given

A={x : 2 < x 5, x R}B={x : 0 x < 6, x R}C={x : 3 x 4, x Z}

Find

(i) (A B) C (ii) (A C) B

Exercise 3: Write each union or intersection as a single interval.

(a) (3,5) [0, 10)(b) (, 10] (5, 7)

(c) [0, 15] (5, 1](d) [2,) [2, 10)

1.2 Complex Numbers Cx2 = 1 has no solution because square of real numbers cannot be negative. Therefore iis introduced to replace

1, i.e. i = 1. Hence, i2 = 1. Numbers which contain iis a complex number.

Let the complex number, z = a + bi, a, b R, a is known as the real part and b isknown as the imaginary part.

Re(a+ bi) = a, Im(a+ bi) = b

Names for Particular Kinds of Complex Numbers

Let a+ bi be a complex number, a and b are real numbers.If b 6= 0, then a+ bi is a complex number.If a = 0, then 0 + bi = bi is a pure complex number.If b = 0, then a+ 0i = a is a real number.If a = 0, b = 0 then 0 + 0i = 0 is called a complex zero number.

Square Roots of Negative Numbers

For any positive real number b,b = ib

Example 1: Write in standard form, a+ ib

(A)4 = 41

(B) 4 +5 = 4 + (51) =

(C)7 +27 =

(D)248

2=

CAUTION!!ab = ab but ab 6= ab

Thus

94 = 36 = 6 or 94 = 3 2 = 6But94 6= 94 6= 36 6= 6

So how to solve94? [See Multiplication of complex number]

1.2.1 The Equality(Uniqueness) of Complex Numbers

If two complex numbers are equal, their real parts are equal and their imaginaryparts are equal.

(a+ bi) = (c+ di) (a+ bi) (c+ di) = 0 (a c) + (b d)i = 0 a c = 0, b d = 0 a = c, b = d

EXAMPLE 3 Solve the following equations.(i) 2 + 3yi = (x 1) + 3i

By comparing the real and imaginary part,

2 = x 1 and 3y = 3 x = 3 and y = 1.

(ii) x+ 2yi = (2 i)2Expand the right hand side of the equation,

x+ 2yi = (2 i)2= 4 4i+ i2= 3 4i

By comparing the real and imaginary part,

x = 3 and 2y = 4 x = 3 and y = 2

1.2.2 Operations with complex numbers

When you add, substract, multiply or divide two complex numbers a+ bi and c+ di, theresult is another complex number.

Addition and substraction

By usual rules of algebra,

(a+ bi) (c+ di) = a+ bi c di= a c+ bi di= (a c) + (b d)i

Since a,b,c,d are real numbers, so are a c and b d. The expression at the end of thelines therefore has the form p+ qi where p and q are real.

Multiplication

By the usual rules for multiplying out brackets,

(a+ bi) (c+ di) = ac+ a(di) + (bi)c+ (bi)(di)= ac+ adi+ bci+ bdi2

= (ac bd) + (ad+ bc)i

Since, a,b,c,d are real numbers, so are ac bd and ad + bc. The product is therefore ofthe form p+ qi where p and q are real.

An important special case is

(a+ bi) (a bi) = (aa b(b)) + (a(b) + ba)i= (a2 + b2) + 0i

= a2 + b2

So with complex numbers, the sum of two squares, a2 + b2 can be factorised as(a+ bi)(a bi).

Division

First, we takea+ bi

c+ diand consider two special cases. If d = 0, then

a+ bi

c+ 0i=a+ bi

c=a

c+b

ci

And if c = 0, you can simplify the expression by multiplying numerator and denominatorby i:

a+ bi

0 + di=a+ bi

di=

(a+ bi)i

(di)i=ai+ bi2

di2= . . .

In the general casea+ bi

c+ dithe trick is to multiply numerator and denominator by c di.

Natural powers of i

Natural powers of i take on particularly simple forms:

ii2 = 1

i3 = i2 i = ii4 = i2 i2 = 1

i5 = i4 i =i6 = i4 i2 =

i7 =i8 =

In general, what are the possible values for in, n a natural number? Then evaluate eachof the following.

(A) i17 (B) i24 (C) i38 (D) i47

Exercise 4: Uniqueness and Operations of Complex Numbers

Problem 1

If p = 3+4i, q = 1 i, r = 2+3i, solve the following equations for the complex numberz.

(A) p+ z = q (B) qz = r

Problem 2

Solve these pairs f simultaneous equations for the complex numbers z and w.

(1 + i)z + (2 i)w = 3 + 4iiz + (3 + i)w = 1 + 5i

Problem 3

Simplify the following

(A) 2i10 4i49 (B) (3i)3 + (i5)10

Problem 4

Find

3 + 4i in the form of a+ bi where a, b R.

1.2.3 Complex Conjugates

If z = x+ yi, then its complex conjugate, denoted by z or z has the same real part as zbut an imaginary part of the opposite sign, written as z = x yi.

Theorem 1 Product of a Complex Number and Its Conjugate

(a+ bi)(a bi) = a2 + b2 [A real number ]Theorem 2 Sum of a Complex Number and Its Conjugate

(a+ bi) + (a bi) = 2a [A real number ]Theorem 3 Difference of a Complex Number and Its Conjugate

(a+ bi) (a bi) = 2bi [An imaginary number ]Proof. (Try to complete the proof of Theorem 1,2 and 3.)

EXAMPLE 1

(A) z = 3 5i (B) z = 1 3i

Conjugate complex numbers have important properties. Suppose for example, that s =a+ bi and t = c+ di are two complex numbers, so that s = a bi and t = c di, then

(a) (s t) =

(b) (st) =

(c)(st

)=

1.2.4 Complex Numbers in Polar Form

Geometrical representation of complex numbers

There are two ways of representing a complex number by using a plane. The complexnumber z = a+ bi can either be represented by a translation of the plane, a units in thex-direction and b units in the y-direction (see diagram on the left) or as the point z withcoordinates (a, b)(see diagram on the right).

The second of these representations is called an Argand diagram, named after John-Robert Argand (1768 1822).

The axes are called the real axis (x-axis) and the imaginary axis (y-axis). These con-tain all the points representing real numbers and imaginary numbers respectively.

Points representing the conjugates pairs a ib are .

The modulus of z, written as |z|, is the length of the line from the origin to the pointrepresenting the complex number on an Argand Diagram.

|z| = a2 + b2

The argument of z is the angle between the positive x-axis and the line from theorigin to the point representing the complex number on an Argand diagram such thatpi < < pi. It is denoted as arg(z).

arg(z) = = tan1b

a,pi < < pi

For example, given complex number z = 9 + 6i,

Exercise 5: Plot the complex numbers in the Argand diagram and calculate

the modulus and argument of each

(A) z = 2 + 2

3

(B) z = 4 + i(C) z = 2 3i(D) z = 1 i

Polar Form

If z is a complex number with modulus r and then z can be written as

z = r(cos + isin); pi < < pi, r 0or

z = rei; pi < < pi, r 0Example

Find the modulus and argument of the following complex numbers. Hence, find its polarform.

(A) (5 +

5i)(i5) (B) 2i

1.3 Indices, Surds and Logarithms

1.3.1 Indices

Large or small numbers are better expressed in terms of indices. A given number can bewritten as a base raised to the index, (base)index. For example,

Definition 1 an, n an integer and a a real number

1. For n a positive interger:

an = a a a a 59049 = 95 = 9 9 9 9 9

2. For n = 0:

a0 = 1 for a 6= 0 1320 = 1

00 is not defined

3. For n a negative integer:

an =1

anfor a 6= 0 73 = 1

73

Theorem 1 Properties of Integer Indices

1. am an = am+n

2. (am)n = amn

3. (ab)m = ambm

4.(ab

)m=am

bm

5.

am

an=

{amn,

1

anm, a = 0

EXAMPLE 1 Using Index PropertiesSimplify using index properties, and express answers using positive indices only.

(A) (3a5)(2a3) = (B)6x2

8x5=

Theorem 2 Further Index Properties For any a and b any real numbers and m, nand p any intergers (Excluding division by 0):

1. (ambn)p = apmbpn

2.

(am

bn

)p=apm

bpn

3.an

bm=bm

an

4.(ab

)n=

(b

a

)nProof. (Try to complete the proof of Theorem 2)

EXAMPLE 2 Using Index PropertiesSimplify using index properties, and express answers using positive indices only.

(A) (2a3b2)2

(B)

(a3

b5

)2

(C)4x3y5

6x4y3

(D)

(m3m3

n2

)

(E) (x+ y)3

(F)x2 y2x1 + y1

Definition 2 bmn and b

mn , Fractional Indices

For m and n natural numbers and b any real number (except b cannot be negative whenn is even):

bmn = (b

1n )m b

mn =

1

bmn

EXAMPLE 3 Using Fractional IndicesSimplify, and express answers using positive indices only. All letters represent positivereal numbers.

(A) 823

(B) (8) 53

(C)

(4x

13

x12

) 12

(D)(u

12 2v 12

)(3u

12 + v

12

)

2.1.1 Index Equations

[This topic appears in in Chapter 2]If both sides of an index equations can be expressed in the same base, then equate thepowers and solve the resulting equation.If they cannot be expressed in the same base, then take log of both sides of the equa-tion(see example on log equations later).

EXAMPLE 5 Solve the following equations.

(A) 91x = 93

(B) 4x3 = 8

(C) 27x+1 = 9

(D) 3x2x3x = 81

(E) 42x2+2x = 8

(F) 2x =162x

8

1.3.2 Surds

A surd is a radical that is not evaluated, or cannot be precisely evaluated. The radicandis often a constant, such as the square root of two:

We know that the square root of 2 is 1.4142.. But why do we leave it as a radical andnot convert it to the number?Definition 3 n

b For n a natural number greater than 1 and b a real number, we define

nb to be the principle nth root of b, that is

nb = b

1n

Theorem 3 Properties of Surds For n a natural number greater than 1, and x and ypositive real numbers:

1. nxn = x

2. nxy = n

x ny

3. nx

y=

nx

ny

**Additional

1.xx = x

2.x+x = 2

x

3.ab =

a

b

4. (a+b)2 = a+ b+ 2

ab

5. (a+b)(ab) = a b

EXAMPLE 6 Simplifying SurdsSimplify

(A) 8

(3x2y)8 = (B)

10

5 = (C) 3

x

64=

Definition 4 Simplified Surd Form

1. Non radicand contains a factor to a power greater than or equal to the index of thesurd.

2. No power of the radicand and the index of the surd have a common factor otherthan 1

3. No surd appears in a denominator.

4. No fraction appears within a surd.

Definition 5 Conjugate SurdsThe conjugate of a +

b isa b as the product (a +b)(a b) = a b is a

rational number.

Rationalising OperationsFor example, we have this algebraic fractions

35

3x 1

3

2a2

3b2

6

23

2 1

2 +

2

Here, to solve, we need to eliminate a surd from a denominator hence we refer this asrationalizing denominator.

To rationalize the denominator, we do as below

Example 7 Simplifying the expression below which involves rationalizing surds.

(A)35

(B)3

2a2

3b2

(C)3x 1

(D)6

23

(E)1

2

5 32 +

6 + 2

3

2

63

(F)

2 1

2 +

2+

2 1

22

2.1.2 Surds Equations

There are equations involving surds. To solve surd equations, we have to look if theequations have one, two or three surds in the equation.

Now, we will consider the three cases which have 4 equal steps to solve the equation.

4 STEPS TO SOLVE SURDS EQUATION:

1. Square both sides of the equation and isolate any remaining surds.

2. Square the equation again to remove any remaining surds.

3. Solve the resulting equation.

4. Check your answers

CASE 1: ONE SURD in the equationIf there is only one surd in the equation, put it on one side before starting the 4STEPS solution.

5x+ 1 + 1 = x

CASE 2: TWO SURDS in the equationIf there is only two surds in the equation, move one to the other sidebefore startingthe 4 STEPS solution.

5x 1x+ 2 = 1

CASE 3: THREE SURDS in the equationIf there is only three surds in the equation, make sure one of them is on onesidebefore starting the 4 STEPS solution.

8x+ 172x = 2x+ 9

1.3.3 Logarithm

Definition 6 Definition of Logarithm For b > 0 and b 6= 1,Logarithm form Index formy = logb x is equivalent to x = b

y

y = log10 x is equivalent to x = 10y

y = loge x is equivalent to x = ey

** loge x = lnx , this is called a Natural Logarithm.

A logarithm is an index or in other words, we can say that logarithm form is equivalent toindex form so in order to solve any problem related to logarithm and index, logarithmform and index form are interchangeable.

EXAMPLE 7 Solve these equations by interchanging logarithm form and index form.

(A) log10 x = 2 (B) ln(2 + x) = 1 (C) log2(x2 3x 2) = 3

Theorem 4 Properties of Logarithm

1. logb 1 = 0

2. logb b = 1

3. logb bx = x log bb = x

4. blogb x = x

5. logbMN = logbM + logbN

6. logbM

N= logbM logbN

7. logbMP = P logbM

8. logbM = logbN iff M = N

EXAMPLE 8 Solve the following by using the properties of logarithm.

(A) 2 log10 5 + log10 70 + log1045

35 log10

45

2

(B) Given log2 3 = 1.59 and log2 5 = 2.32, without using calculator, evaluate:

(a) log2 0.6 (b) log2 30 (c) log2 33

5(d) log2

3

1.5

Change-of-Base Formula Let y = logbN where N and b are positive and N 6= 1.

y = logbN

by = N

logyb = logNloga b

y = logaN

y loga b = logaN

y =logaN

loga b

EXAMPLE 9 Solve these by changing the base of logarithm.

(A) log8 16 (B) log27 81 (C) log3 7

EXAMPLE 10 Solve.

(A)log5 81

log5 27(B) (log2 2)

3 (C) log4(13 + 3)

CAUTION 3 common errors in logarithm

(a)logbM

logbN6= logbM logbN

(b) logb(M +N) 6= logbM + logbN(c) (logbM)

p 6= p logbM

2.1.3 Logarithm Equations

Logarithm equations can be solved by considering all the properties of logarithm carefuly.We will look at more examples involving logarithm equations.

EXAMPLE 11 Solve logarithm equations.

(A) log7 4x log7(x+ 1) =1

2log7 4

(B) 2x 8 = 3x 5x

(C) log9x

3=

log9 x

log9 3