Matriculation QS015 2014 S.Y.Chuah June 18, 2014
Chapter 1 : Number Number System System 1.1 Real Numbers (a) Define Define Natural Natural Numbers Numbers N, Whole Number W, Integers Z, Prime Numbers, Rational Numbers Q, Irrational Numbers. (b) Represent Represent rational numbers and irrational irrational numbers in decimal forms. (c) Represen Representt the relations relationship hip of nu number mber sets in a real number number system system diagramdiagrammatically. (d) Represen Representt open, closed closed and half-open half-open interv intervals and their their represen representati tations ons on the number line. (e) Simplify Simplify union , intersection line.
∪
∩ of two or more intervals with the aid of number
1.2 Complex Numbers (a) Represent Represent a complex number number in Cartesian form. (b) Define the equality equality of two complex numbers. numbers. (c) Determine Determine the conjugate of a complex nu number, mber, ¯ Z . (d) Perform Perform algebraic operations on complex complex numbers. (e) Represen Representt a complex complex number number in polar form Z = r(cosθ + cosθ + i sinθ) sinθ) where r < 0 and π < θ < π. π.
−
1.3 Indices, Surds and Logarithms (a) State State the rules rules of indice indices. s. (b) Explain the meaning of surd and its conjugate and to carry out algebraic operations on surds. (c) State State the laws of logarithms logarithms.. (d) Cha Change nge the base base of logarithms 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 and irrational numbers.
0
Definiti Definition on 1.1 Natural Natural nu number mbers, s,N, are posit positiv ivee numbers umbers that that are used used for for count countin ing: g: 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, -2,-1,0,1,2, .
{···
···}
Definition 1.4 Prime numbers are natural numbers greater than 1 that can be divided by itself and 1 only. Primenumbers = Primenumbers = 2,3,5,7,11, 2,3,5,7,11, .
{
···}
Definition Definition 1.5 Rational numbers, numbers,Q, are numbers that can be written in the form pq where p and q are integers and q = 0. Q = pq p, q Z, q = 0
{
∈ ∈
}
In decimal form, rational numbers may be a terminating decimal , such as 34 = 0.75 or a 3 = 0.27272727 , in which a group of one or more digits repeating repeating decimal , such as 11 repears indefinitely. Examples of rational numbers are 3, 34 , 25 , 0.6, 1.212121 , 6, 20.
···
− −
···
¯ , are numbers that cannot be written in the form Definition 1.6 Irrational numbers,Q p where p and q are integers and q = 0. For example, π , e and 3. q
Figure 1: Real Number System
√
Figure 2: Venn Diagram represents different types of real numbers
Exercise 1:
(a) N
⊂W (b) Z ⊂ N
Determine whether each statement is true or false.
(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 in set notation as follows:
{}
≥ 2 can be expressed
{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 referring to 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]
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
(v) [ 6, 1)
−
(vi) [10,
∞)
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)
Hence, (1, 5)
∩ (3, 9)
∩ (3, 9) = (3, 5).
The union of two intervals is the set of real numbers that belong to one, or the other, or both of the intervals.
EXAMPLE 2: (1, 5)
Hence, (1, 5)
∪ (3, 9)
∪ (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
Exercise 3:
(a) ( 3, 5)
(ii) (A
∩ C ) ∪ B
Write each union or intersection as a single interval.
− − ∪ [0, 10) (b) (−∞, 10] ∩ (−5, 7)
(c) [0, 15]
∪ (−5, 1] (d) [2, ∞) ∩ [−2, 10)
1.2 Complex Numbers C x2 = 1 has no solution because square of real numbers cannot be negative. Therefore i is introduced to replace 1, i.e. i = 1. Hence, i2 = 1. Numbers which contain i is a complex number.
−
√ −
√ −
Let the complex number, z = a + bi, a, b known as the imaginary part.
−
∈ R, a is known as the real part and b is
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 = 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 = i√ b Example 1: Write in standard form, a + ib
√ −4 = √ 4 × √ −1 √ √ √ (B) 4 + −5 = 4 + ( 5 × −1) = √ √ (C) −7 + −27 = √ 2 − −48 − (D) = (A)
2
√ √ √ √ = √ −a × −b √ CAUTION!! a × b = ab but −a × −b √ √ 4 = √ 36 = 6 or √ 9√ 4 = 3 × 2 = 6 Thus√ 9 × √ But −9 × −4 = √ −9 × −4 = √ 36 = 6 √ √ So how to solve −9 × −4? [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 imaginary parts 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)
2
−x + 2yi = (2 − i)
Expand the right hand side of the equation, 2
−x + 2yi = (2 − i) = 4 − 4i + i = 3 − 4i
2
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, the result is another complex number.
Addition and substraction By usual rules of algebra,
± (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 the (a + bi)
lines 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 the form p + qi where p and q are real.
− bd and ad + bc. The product is therefore of
An important special case is (a + bi)
× (a − bi) = (aa − b(−b)) + (a(−b) + ba)i = (a2 + b2 ) + 0i = a 2 + b2
So with complex numbers, the sum of two squares, a2 + b2 can be factorised as (a + bi)(a bi).
−
Division First, we take
a + bi and consider two special cases. If d = 0, then c + di a + bi a + bi a b = = + i c + 0i c c c
And if c = 0, you can simplify the expression by multiplying numerator and denominator by i: a + bi a + bi (a + bi)i ai + bi2 = = = = . . . 0 + di di (di)i di2 In the general case
a + bi the trick is to multiply numerator and denominator by c c + di
− di.
Natural powers of i Natural powers of i take on particularly simple forms: i 2 i =
−1
i3 = i 2 i = i i4 = i 2 i2 = 1
· ·
−
i5 = i 4 i = i6 = i 4 i2 =
· ·
i7 = i8 =
In general, what are the possible values for in , n a natural number? Then evaluate each of 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 z .
− i, r = −2 + 3i, solve the following equations for the complex number
(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 + 4i iz + (3 + i)w = 1 + 5i
−
−
Problem 3
Simplify the following (A) 2i10
− 4i
Problem 4
Find
49
(B) (3i)3 + (i5 )10
√ 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 z but 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) = a
2
+ 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)
s t
∗
=
=
−
∗
−
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 complex number z = a + bi can either be represented by a translation of the plane, a units in the x-direction and b units in the y-direction (see diagram on the left) or as the point z with coordinates (a, b)(see diagram on the right).
The second of these representations is called an Argand diagram, named after JohnRobert Argand (1768 1822).
−
The axes are called the real axis (x-axis) and the imaginary axis (y-axis). These contain 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 point representing the complex number on an Argand Diagram.
| |
|z | = √ a
2
+ b2
The argument of z is the angle θ between the positive x-axis and the line from the origin to the point representing the complex number on an Argand diagram such that π < θ < π. It is denoted as arg(z ).
−
arg(z ) = θ = tan
1
−
b , π<θ<π a
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 θ then z can be written as z = r(cosθ + isinθ); π < θ < π, r or iθ z = re ; π < θ < π, r 0
−
−
≥0
≥
Example
Find the modulus and argument of the following complex numbers. Hence, find its polar form. (A) (5 +
√ 5i)(i − √ 5)
(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 be written 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
59049 = 95 = 9
×a×a×···×a
×9×9×9×9
2. For n = 0: a0 = 1 for a = 0
1320 = 1
00 is not defined 3. For n a negative integer: an =
1 a
n
−
for a = 0
7
3
−
=
1 73
Theorem 1 Properties of Integer Indices 1. am
×a
n
= a m+n
2. (am )n = a mn 3. (ab)m = a m bm 4. 5.
a b
m
am = m b m
a = an
am n , 1 , a = 0 an m −
−
EXAMPLE 1 Using Index Properties Simplify using index properties, and express answers using positive indices only. (A) (3a5 )(2a 3 ) = −
6x (B) 8x
2
−
5
−
=
Theorem 2 Further Index Properties For any a and b any real numbers and m, n and p any intergers (Excluding division by 0): 1. (am bn ) p = a pm b pn
n
a 3. b
−
m
−
2.
am bn
p
a pm = pn b
4.
bm = n a
a b
n
b a
−
=
n
Proof. (Try to complete the proof of Theorem 2)
EXAMPLE 2 Using Index Properties Simplify using index properties, and express answers using positive indices only. (A) (2a 3 b2) −
(B)
a3 b5
−
(D)
m 3 m3 n 2
−
−
2
−
(E) (x + y)
3
−
4x 3 y 5 (C) 6x 4 y 3 −
2
−
−
x (F) x
2
−
1
−
−y + y
2
−
1
−
m n
m n
Definition 2 b and b , Fractional Indices For m and n natural numbers and b any real number (except b cannot be negative when n is even): −
b
m n
1 n
= (b )m
b
−
m n
=
1 b
m n
EXAMPLE 3 Using Fractional Indices Simplify, and express answers using positive indices only. All letters represent positive real numbers. (A) 8
2 3
(C)
(B) ( 8)
−
5 3
(D)
4x x
u
1 2
1 3
1 2
1 2
− 2v
1 2
1
1
2
2
3u + v
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 the powers and solve the resulting equation. If they cannot be expressed in the same base, then take log of both sides of the equation(see example on log equations later).
EXAMPLE 5 Solve the following equations. (A) 91
x
= 93
(D) 3x2x
3
=8
(E) 42x
−
2
(B) 4x
−
3x
−
+2x
= 81
=8
16 2x (F) 2 = 8 −
x+1
(C) 27
=9
x
1.3.2 Surds A surd is a radical that is not evaluated, or cannot be precisely evaluated. The radicand is 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 and not convert it to the number? Definition 3 b For n a natural number greater than 1 and b a real number, we define b to be the principle nth root of b, that is
√ n
√ n
√ b = b n
1 n
Theorem 3 Properties of Surds For n a natural number greater than 1, and x and y positive real numbers: 1. 2. 3.
√ x = x √ xy = √ x × √ y √ x x = √ y y n
n
n
n
n
n
n
n
**Additional
√ x × √ x = x √ √ √ 2. x + x = 2 x √ √ a 3. a ÷ b = 1.
b
√ √ b) = a + b + 2√ ab √ √ √ √ 5. ( a + b)( a − b) = a − b 4. ( a +
2
EXAMPLE 6 Simplifying Surds Simplify (A)
8
(3x2 y)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 the surd. 2. No power of the radicand and the index of the surd have a common factor other than 1 3. No surd appears in a denominator. 4. No fraction appears within a surd.
Definition 5 Conjugate Surds The conjugate of a + b is a rational number.
√ √ − √ b as the product (√ a + √ b)(√ a − √ b) = a − b is a
Rationalising Operations
For example, we have this algebraic fractions 3 5
3 x 1
√
6
2
√ −
− √ 3
3
√ 2 − 1 √ 2+ 2
2a2 3b2
Here, to solve, we need to eliminate a surd from a denominator hence we refer this as rationalizing denominator. To rationalize the denominator, we do as below
Example 7 Simplifying the expression below which involves rationalizing surds. (A)
√ 35
(B)
(C)
3
(D)
2
2a 3b2
3 x 1
√ −
6
2
− √ 3
(E)
√ 6 + 2√ 3 √ √ + √ √ 2 5−3 2 2 6− 3
(F)
√ 2 − 1 √ 2 − 1 √ + √ 2+ 2 2− 2
1
2.1.2 Surds Equations There are equations involving surds. To solve surd equations, we have to look if the equations 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 equation If there is only one surd in the equation, put it on one side before starting the 4 STEPS solution.
√ 5x + 1 + 1 = x
CASE 2: TWO SURDS in the equation If there is only two surds in the equation, move one to the other side before starting the 4 STEPS solution.
√ 5x − 1 − √ x + 2 = 1
CASE 3: THREE SURDS in the equation If there is only three surds in the equation, make sure one of them is on one side before starting the 4 STEPS solution.
√ 8x + 17 − √ 2x = √ 2x + 9
1.3.3 Logarithm Definition 6 Definition of Logarithm For b > 0 and b = 1,
Logarithm form y = logb x y = log10 x y = loge x
is equivalent to is equivalent to is equivalent to
Index form x = b y x = 10y x = e y
** loge x = ln x , this is called a Natural Logarithm . A logarithm is an index or in other words, we can say that logarithm form is equivalent to index form so in order to solve any problem related to logarithm and index, logarithm form 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
5. logb M N = logb M + logb N
2. logb b = 1
6. logb
3. logb bx = x log bb = x
7. logb M P = P log b M
4. blog x = x
8. logb M = logb N iff M = N
b
M = logb M N
− log N b
EXAMPLE 8 Solve the following by using the properties of logarithm. (A) 2log10 5 + log10 70 + log10
45 35
− log
10
45 2
(B) Given log2 3 = 1.59 and log2 5 = 2.32, without using calculator, evaluate: (a) log2 0.6
(b) log2 30
3 (c) log2 3 5
(d) log2
√ 1.5 3
Change-of-Base Formula Let y = log b N where N and b are positive and N = 1.
y = logb N by = N logyb = logN loga by = loga N y loga b = loga N loga N y = 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)
logb M = logb M logb N
− log N
b
(b) logb (M + N ) = logb M + logb N
(c) (logb M ) p = p logb M
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
− log (x + 1) = 21 log 4 7
(B) 2x 8 = 3x 5x
·
(C) log9
·
x log9 x = 3 log9 3
7
