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Index Concepts
5
2
=
22222
index/exponent
base
(5 times)
n
a

a

a

…

a
a =
(n times)
a to the power n”
“
記住:a zero
Law of positive integral indices
Class Activity 1
23 x 24 = (2 x 2 x 2) x (2 x 2 x 2 x 2) = 23+4 = 27
a2 x a3 = (a x a) x (a x a x a) = a2+3 = a5
b5 x b4 = (b x b x b x b x b) x (b x b x b x b) = b5+4 = b9
am  an = a  a  …  a  a  a  …  a
(m times)
(n times)
a
am  an = a (maxn…
times)
am  an = am + n
law 1
Class Activity 2
3
3 3 3 3 3
5 2
=
=3
2
3
3 3
5
a6 a  a  a  a  a  a
6 3
=
=3
3
a
aaa
a
aa
1
=
= 5 2
5
a
aaaaa 3
2
a
aaa
1
=
= 6 3
6
a
aaaaaa 3
3
a
aaaaa
5 1
a
=
=
1
a
a
5
a
aaaaa
53
=
=
a
3
a
aaa
5
1
aaaaa
a
1
= 7 5 = 2
=
7
aaaaaaa a
a
a
5
a
m
a

=
 1
n
a
 nm
a
mn
, where m > n and a  0
Law 2
, where m < n and a  0
EXAMPLE 1 – Simplify the following expressions
(a) a7.a3 = a7+3 = a10
n.a3 = an+3
a
(b)
(c)
(d)
a9
9-2
=
a
2
a
7
a
=
1
1
a
=
=
4
3
4
a
a
a
3
Class Activity
(23)2 = (2  2  2)  (2  2  2) = 232=
26
(a4)3 = (a  a  a  a)  (a  a  a  a) (a  a  a  a) = a43= a12
(b3)5=(bbb) (bbb) (bbb) (bbb) (bbb)=b35= a15
(a ) =
m n
a m  a m  a m  a m
n times of am
(a ) =
m n
(aa…a)  (aa…a) ……  (aa…a)
nm times of a
(a ) = a
m n
nm
Law 3
Example 2- Simplify the following expressions
(a)
(b)
(c)
5
2
(a ) =
52
a =
10
a
n
3
(a ) =
n3
a =
3n
a
4
m
4m
(a ) = a =a4m
Classwork- Simplify the following expressions
12
4 3
43
(1)
(x ) = x
= x
(2)
(xn)6=
xn6=
x6n
(3)
(x5)m=
x5m=
x5m
Class Activity
(2  3)4= (2  3) (2  3) (2  3) (2  3)
=(2  2  2  2)(3  3  3  3)
=24 34
(3  5)5=(3  5)(3  5)(3  5)(3  5) (3  5)
=(3  3  3  3  3)(5  5  5  5  5)
=35 55
(ab)3= (a  b) (a  b) (a  b)
=(a  a  a) (b  b  b)
=a3 b3
(ab)m= (a  b) (a  b)…. (a  b)
m times of ab
(ab)m= (a  a  …  a)  (b  b  …  b)
m times of a
(ab)m= am  bn
m times of b
Law 4
2
( )
3
4
=
2 2 2 2
  
3 3 3 3
=
2 2 2 2
3 3 3 3
4
=
3 3 3 3 3 3 3 3 3 3 3
( )=     =
5 5 5 5 5 5 5 5 5 5 5
5
2
4
3
5
3
=
5
5
a
a a a
aaa
a
=
=
( )=
 
3
b
b b b
bbb
b
a n a a a
a a  a  a  a a n
( ) =    =
b
b b b
b b  b  b  b = b n
a
3
3
n
a n a
( ) = n
b
b
(n times of ( ) n =
b
(where b  0)
Law 5
EXAMPLE 3: Simplify the following expressions
(a) (3a)3=
(b)
(c)
33a3= 27a3
a 4 = a4
( )
4
2
2
=
a4
16
2a23 ( 2a)2 3 23( a)2 3 8a6
=
=
( ) =
3
3
3
b
b
b
b
Classwork 1.3
(1) (2x)5=
25x5=
32x5
(2)
10
5)2 2 5 2
(
4
x
16
x
(3) 4 x 2
4
(
x
)
( ) = 2 2 = 22 = 4
y2 ( y ) ( y ) y
5
x 3= x3
( )
3
4
4
=
x3
64
EXAMPLE 4: Simplify the following expressions
3 3 3
3 3
3 2
a
(2ab)
2
a
b
8
a
b
a
=
=
=
=
2 4
2
4
2
4
4

3
2b
16a b
16a b
16a b
2b
2 n 1
2 n 1
2 n 1
3
3
1
3
3
= 2 n1 = 2 n  2 = ( 2 n  2 ) ( 2 n 1) =
n 1
(3 )
3
3
9
3
3
(a)
(b)
n2
(c)
3
 3 3n.32  3n 9.3n  3n 8.3n
=
=
=
=4
n
n
n
n
2.3
2.3
2.3
2 .3
n
Classwork 1.4
1.
3
(5 xy) 53 x 3 y 3 125 x 3 y 3
2
3 2
31
=
=
= 5 x . y = 5 x. y
2
2
2
25 x y 25 x y 25 x y
3
5
5
3 5
15
5
5
15 4
11
32 x y 2 x
(2 x y )
2x
2 .( x ) . y
=
=
=
= 3
2
2
2
4
2
4
8
8 5
2
4
2
2. (4 x y )
4 .( x ) .( y )
16 x y
y
y
5
3.
n 3
5
n
3.5
n 1
(120)5
5 .5  5 .5 (5  5)5
=
=
=
=
n
n
n
3.5
3.5
3.5
n
3
n
1
3
n
n
40
Exercise 1A
Level 1
(1b) a2.a3.a4=a2+3+4=a9
(2b)
1
a6
1
= 106 = 4
10
a
a
a
(3b) (x7)3 =x73 =x21
(4b) (4a2)3 =43. (a2)3=64. a23 =64. a6
Exercise 1A
Level 2
5
6
a
(5b)
=
3 4
2a .a
3a 5
=
34
a
(6b) (2b2)3(4b3)2=
3
3a 5
=
=
7
5
a
a7
3
2
a
23.(b2)3. 42(b3)2
=23.(b2)3. 42(b3)2
=8. b6. 16b6
=8. 16 b6. b6
=128 b6+6
=128 b12
(7b)
4
3
4 3
x y
x 4 y 3
x
x
x
y
( 2) ( ) = 2 4 3 =
=
=
8

3
5
8
3
(y ) x
y
x
y
y
y x
4
3
(8b)
2 3 4
4
2 4
3 4
4 8 12
abc
(ab c ) a (b ) (c )
=
=
abc
abc
abc
4 1 81 121
=a b c
=a b c
3 7 11
(9b)
3n
3 n 1
1
3n
1
2 .2 2 .2
2
= 3n = 2
=
n
3
n
2
2
8
 
(10b)
n 1
3
n 1
3
n
2.3
1
n
1
1
3
.
3

3
.
3
3
.(
3

3
)
=
=
n
n
2.3
2.3
n
1
n
1
(3  3 )
=
2
1
1
(3  3 ) 4
=
=
3
2
1
Mistakes students always make
23  54= (2  5) 7= 10 7
Wrong !
since these numbers has no common base.
3  46= 126
Wrong !
4
6
2
=
2
2
3
(x3)2=
x3+2=
3a.2a=5a
since these numbers has no common base.
x5
aaaaa
a
=
1
=
5
aaaaa
a
5
∵

a
55
=a
0
a =1
0
Thus, a = 1 (where a  0)
0
Law 6
3
a
1
1
=
=
5
5 3
2
a
a
a
From Law 2
3
a
3 5
2
=
a
=
a
5
a
∴
Thus
a
2
a
Any integer m and n
1
= 2
a
m
1
= m
a
Law 7
Example 5- Evaluate the following expressions
1
1
(a) (-2)-3 = 1
(2)
3
(b)
(c)
=
= 
8
8
1
1
2 -2
( ) = 22 = 4 = 9
( )
3
4
3
9
(3-3)2(20)
2
=
1 2 (1) = 1
1 1
=
( 3)
=
6
32
3
(3) 3 729
Classwork 1.5
1
1
1. (4) =
=
3
64
4
1
1
1
2.(5) 3 =
=

=
3
125
125
(5)
3
3
0
1
1
1
27
4 1


1
=
=
1 =
3
3.     =
3
64
4


4
64
3 2
 
3
27
3
 
3
3 2
6
1
(
2
)
1
2
4.
=
=
=
0
6
0
64
3
2
3
Example 6 - Simplify the following expressions and
express your answers in positive indices
(a) x 4 .x1
x
1
3
(b) 2 . p
2
=x
4 1 ( 3)
=2 p
1
2  ( 1)
= x = 1
 ( 3)
0
3 3
1 3 3
p
q
=
=
q
2 pq
1 3
p q
2 4 2

b
c
4
4
2





2
2
2
2
1
2



(c)(a2b 2c 1) 2=(a ) (b ) (c ) =a b c = 4
a
Classwork 1.6
3 2
(1)
(2)
(3)
(4)
aa
3 ( 2 )  ( 1)
3 2 1
2
=
a
=
a
=
a
1
a
1 3
1 3 1
3 x .x 3 x .x
1 2  31 2
=
=
3 x
2
2 2
(3 x)
3 x
1
1
3  4
=3 x = 3 4 =
4
3 x
27 x
1 2 1
(a b )
2
b
1 1
2 1
(a ) (b )
=
2
b
1 2
ab
= 2 =a
b
2
ab
(a b c ) = (a ) (b ) (c ) = a b c = 3
c
1  2 3 1
1 1
 2 1
3 1
1 2 3
Example 7: Simplify the following expressions and express your answers
in positive indices
(a)
2 3 2
1 2  2
(a b ) (a b )
4 6
= (a b )(a b )
6  4
4 2
= (a a )(b b )
[2( x
2
y)
] [
3
6
a
=a b
= a b = 10
b
1
3

y
1  6 3
6 3 1
=2 x y = 6
= 2x y
2x
4 2 6 4
(b)
2 4
]
6 10
Classwork 1.7
(1)
3 2 2
2 1 3
6 4 6 3
(x y ) (x y ) = x y x y = x x y y
12
66 43
12 1 x
= x y =x y =
y
(2)
 2 1
1 3  2
1 1 2  2 2  6
(2 xy ) (3x y ) = 2 x y .3 x y
1  2 1 2 2  6
=2 3 x x y y
12 2  6
x y
=
2
2.3
4
x
xy
=
=
4
18y
18
6 6 4 3
3
[( x
3
y)
] = [x y ]
2
2
6
2
2
=x y
12
4
4
12
x
= 4
y
2 025
2 3
ab
ab
(b ) (2a b) (8b ) = b 2 a b 8 b = 2 =
2 .8
32
3 0
1
2
5 1
0 2 2 2 1 5
Additional example
4
4
4 4
8
x .x
x
x
( x y ) ( x y ) = ( x y )( x y ) = 8 2 = 8 2 = 10
y .y
y
y
1
2 4
2
4
8
4
2
3 2 2
2
2
2
2
3 4
=
[3 (5 x y ) (2 y ) ]
1 (5 x y ) (2 y )
0
1
1 2
4
2
4
2
25 x y
=
3 4
(2 y )
25x y
=
12
16 y
4
25 x
= 122
y
25 x 4
= 10
y
Exercise 1B
(1b)
(3 ) = 1
2 0
2
1
(1c)
3 2
1
1
1 16
( ) =
= 2 =
=
3 2 3
9
4
9
( )
2
4
16
4
a a
 2 1( 3)
 2 1 3
0
=a
=a
= a =1
3
a
(4b) (a 2b 3 ) 1 = a 21b 31 = a 2b3
4
8a
 2 5
6 2
2  ( 5) 6 ( 2 )
3 4
(5a)
= 8b a =
(4a b )(2a b ) = 4 . 2 . b
a
b3
2 4
4
2
(6a)

( 4) x
16 x .2 x
2 0 2 1  2 2 1
=
(4 x y ) ( x y ) =
2
2
y
2
y
4 2
6
2
x
32 x
32 x
2
=
=
2
2
y
y
(3b)
1.4 SIMPLE EXPONENTIAL EQUATIONS
Equation involves unknowns in the exponents
(a0, a1, a-1)
a2x=ax
2x
a
=
1
x
a
a
2x  x
=a
2x-x=0
∴
x=0
0
Example 8: Solve the following exponential equations
(a) 6x=1
6x=60
∴ x=0
(b)
2x=8
2x=23
∴ x=3
Classwork 1.8
(2)
(1) 11x=1
11x=110
∴ x=0
3x=81
3x=34
∴ x=4
(3) 4x=256
4x=44
∴ x=4
2.13x=2
13x=1
13x=130
∴ x=0
(4)
Example 9: Solve the following exponential equations
2x+1=8x-2
(b)
4
2x=625
5
(a)
(22)2x+1=(23 )x-2
22(2x+1)=23(x-2)
52x=54
244x+2=23x-6
2x=4
∴ x=2
4x+2=3x-6
4x-3x=-6-2
∴ x=-8
Classwork 1.9
(1)
32x=81
(3)
(2)
32x=34
2x=4
∴ x=2
42x=256
42x=44
2x=4
∴
x=2
82x=16x+4
(23)2x=(24) x+4
26x=24(x+4)
26x=24x+16
6x=4x+16
6x-4x=16
2x=16
∴ x=8
(4) 94x-1=272x+4
(32)4x-1=(33) 2x+4
32(4x-1)=33(2x+4)
38x-2=26x+12
8x-2=6x+12
8x-6x=12+2
2x=14
∴ x=7
Example 10: Solve the following exponential equations
3
x2
 3  24 = 0
x
32.3x  3x  24 = 0
9.3  3  24 = 0
x
x
8.3  24 = 0
x
8.3 = 24
x
3x = 3
3 =3
x
∴
x =1
1
Classwork 1.10 - Solve the following exponential equations
(1)
(1)
2
x4
 3(2 )  104 = 0
x
3
x 3
3
x 1
 216 = 0
2 .2  3(2 )  104 = 0
32.3x 1  3x 1  216 = 0
16(2 )  3(2 )  104 = 0
9.3x 1  3x 1  216 = 0
8.3 x 1 = 216
x
4
x
x
x
13(2 ) = 104
x
3
2 =8
x
2 =2
x
∴
3
x=3
∴
x 1
= 27
3 x 1 = 33
x 1 = 3
x=2
Scientific Notation科學記數法
In scientific world, we often study objects which are very large or very small
e.g.
Mass of an element of hydrogen is 0.000 000 000 000 000 167 724 567 kg.
Velocity of light = 300 000 000 m/s
Difficult to read and write!!!
To overcome, we use a method called scientific notation to express these
numbers approximately
0.000 000 000 000 000 167 724 567 = 1.68  1016
corr. to 3 sig fig
300 000 000 = 3  108 kg
25870945830231 = 2.59  10 13
corr. to 3 sig fig
Positive number N can be expressed as N = d  10n, where n is an integer
and 1  d < 10.
Example 11- express the following numbers in scientific notation
(a) 0.000 001 = 1 0.000 001 = 1 10-6
(b) 18 000 = 1.8 10 000 = 1 104
(c) 2 000 000 = 2 1 000 000 = 2 106
(d) 0.000 216 = 2.16 0. 000 1 = 2.16 10-4
(e) 345  104 = 3.45  100  104 = 3.45  102  104 = 3.45  106
(f) 0.050 2  10-3 = 5.02  0.01  10-3 = 5.02  10-2  10-3 = 5.02  10-5
Classwork 1.1- express the following numbers in scientific notation
(1) 3 740 000 = 3.74 100 000 = 1 106
(2) 0.000 0007 89 = 7.89 0.000 0001 = 1 10-7
(3) 6 31  1011 = 6.311001011 = 6.31  1021011 = 6.311013
(4) 0.1 203710-5=1.20370.110-5=1.20 3710-110-5 =1.20 3710-5
(5) –0.0004 508103=-4.5080.0001104=-4.50810-4104=-4.50810-8
(6) 0.7 00810-4 =7.0080.110-4=7.00810-110-3=7.00810-4
Example 12- Evaluate the following,express your answers in scientific
notation and correct to 3 significant figures
(a) (3.52108)(7.911012) =(3.527.91 )(108 1012)
=27.84321020
=2.784321011020
=2.784321021
=2.781021
(b)
9
Corr. to 3 sig. Fig.
9
8 .16  10
8.16 10
6
6

=
=
1 .4191310 =1 .42 10
3
3
5.75 10
5.75 10
Corr. to 3 sig. Fig.
Classwork 1.12 ,express your answers in scientific notation and correct to 3 significant
figures
(1) (2.305107)(8.21012)=(2.3058.2)(107 105) =18.9011012=1.891011012
=1.891013
(2) (210-12)(71023) =(27)(10-121023) =141011 =1.41011011 = 1.41012
(3) (3.810-10)(810-11) =(3.88)(10-1010-11) =30.410-21 =3.0410110-21
= 3.0410-20
(4)
3 10 2
1
1
1
2
=
=
=
0
.
46875
10
4
.
69
10
10
4
.
69
10




3
6.4 10
(5) 8.009  10
3.5 10
(6)
6
4
5
= 2.2882  1010 = 2.29 1010
4.67  10
1
1

=
=
3.3357  10
3.34 10
4
1.4 10
Example 13 – Express the values of the following as integers or
decimal numbers
(a) 2105=2 100 000 =200 000
(b) 3.1410-4=3.14 0.000 1 = 0.000314
(c)(2.510-3)(3.610-4)=2.53.610(-3)+(-4) =910-7
=90.0000001
=0.0000009
Classwork 1.13
(1) 6.36106=6.36 1000000 = 6.360000
(2) 4.1210-5=4.120.0000 1 = 0.0000412
(3) (1.310-5)(5.7107)= 1.35.710-5+7 =7.41102 =7.41100
=741
2
(4) 9.75 10 =
7
= 1.3 10000000 = 13000000

1
.
3
10
9
7.5 10
Example 14- express the following numbers in scientific
notation and arrange them in ascending order
89, 98, 710, 107
Use calculator
89
98
710
107
=134217728
=1.34108
=43046721
=4.30107
=282475249
=2.82108
=10000000
=1107
10 Xy 7 =
8
Xy
9
=
9
Xy
8
=
7
Xy
10
=
107< 98 < 89 < 710
Compare powers of 10. The greater the power of 10, the greater the
number. If powers of 10 are equal, compare values of d of the numbers
Numeral Systems
In the algebraic expression
ax3+ bx2 + cx1 + dx0
x is an integer more than 1.
a, b, c and d are all non-negative integers less than x.
Denary System (十進制)- we use everyday
When x = 10, abcd is a denary number.
abcd = a  103  b  102  c  101  d  100
e.g. 3065(10)=3  103  0  102  6  101  5  100
3065(10)=3  1000  0  100  6  10  5  1
Place values
Place holder
Distinguish
values
3065
ten basic numerals:
0, 1, 2, 3,
4, 5, between
6, 7, 8, 9 365
(lessand
than
10)
The place value of each digit is ten times the place value of the digit on its right-hand side
Example 15
(a) Write down the place values of each digit of 50943
Digit
5
Place Value 10000
0
1000
9
100
4
10
3
1
(b) Express 50943 in the expanded form with base 10
50943(10)=510000  01000  9100  410 + 3  1
100
=5104
 0103  9102  4101 + 3
Classwork 1.15
(1) Write down the place value of each digit in the following numbers
(a) 24071
Digit
Place Value
2
10000
4
1000
0
100
7
10
9
1000
1
100
4
10
5
1
1
1
(b) 9145
Digit
Place Value
(2) Express the following as denary numbers
(a) 2541 = 2103  5102  4101 + 1 100
(b) 205041 = 2105  0104  5103 + 0 102 + 4 101 + 1 100
Challenge
What are the expansions of 12.302?
12.302 = 1  101  2  100  3  101  0  102  2  103
Example 16 Express the following as denary numbers.
(a)
(b)
2  102  3  101  7  100 =200 30  7 =237
9  105  5  103  6  101  7  100
=9  105  0  104  5  103  0  102  6  101  7  100
=905067
Classwork 1.16 Express the following as denary numbers.
(1) 6103  3101
=6103  0102  3101  0100
= 6030
(2) 410000  210  9
=4  104  0  103  0  102  2  101  9  100
=40029
(3) 5  10000  4 100  3  10
=5  104  0  103  4  102  3  101  0  100
=50430
Exercise 1E
(1) Write down the place value of each digit 2 in the following denary numbers
(a) 1232
100, 1
(b) 123423
10000, 10
(2) Express the following denary numbers in the expanded form with base 10
(a) 119= 1  102  1  101  9  100
(b) 2318= 2  103  3  102  1  101  8  100
(3) Express the following as denary numbers
(a) 6 10 + 8 1 = 6  101  8  100 =68
(c) 8 100000 + 4 100 + 1 = 8105  0104  0103  4102  0101 1100
=800401
(e) 11000 + 4 100 + 9  1 = 1103  4102  0101 9100 =1409
(4) Find the greatest and smallest denary numbers
(a) 2, 6, 4 Greatest 642, smallest 246 (b) 0,1,3,5 Greatest 5310, smallest 1035
(5) (a) 40014
10000, 1 10000/1=10000
(b) 241243 10000, 10 10000/10=1000
Denary System
$100
213(10) = 2  100
 1  10 $10  3  1
$1
0 to 9 (<10)
213(10) = 2  102
 1  101
 3  100
Binary System
23(10)= 1 
0
1  4
 1 2
1  1
0 to1 (<2)
=10111(2)
10111(2)= 1 24  0  23  1  22  1  21  1  20
Denary System
$100
213(10) = 2  100
 1  10 $10  3  1
$1
0 to 9 (<10)
213(10) = 2  102
 1  101
 3  100
Binary System
23(10)= 1
$16  0
$8
 1
$4
 1 $2
 1 $1
0 to1 (<2)
=10111(2)
10111(2)= 1 24  0  23  1  22  1  21  1  20
Denary System (十進制)
e.g. 3065(10)=3  103  0  102  6  101  5  100
0 to 9 (<10)
Binary System (二進制)- only computer understand
When x = 2, abcd is a binary number.
abcd(2)= a  23  b  22
 c  21  d  20
e.g. 1011(2)=1  23  0  22  1  21  1  20
1011(2)=1  8  0  4  1  2  1  1
Place values
Place holder
Distinguish
2 basic numerals:
0 to1 (< 2)values between 111 and 1011
The place value of each digit is two times the place value of the digit on its right-hand
side.
Example 17
(a) Write down the place value of each digit in 101 010(2)
Digit
1
Place Value 32
0
16
1
8
0
4
(b) Find the place value of the digit 0 in 11 011(2)
1
2
0
1
=22=4
Classwork 1.17
(a) Write down the place value of each digit in 111 01(2)
Digit
1
Place Value 16
1
8
1
4
0
2
(b) Find the place value of the digit 0 in 101 111(2)
0
1
=24=16
Example 18
Express the following binary numbers in the expanded form with base 2
(a) 1101(2)=1  23  1  22  0  21  1  20
(b) 1011011(2)=126  025  124  123  022  121  120
Classwork 1.18
Express the following binary numbers in the expanded form with base 2.
(1) 10001 (2)=1  24  0  23  0  22  0  21  1  20
(2) 100110 (2) =125  024  023  122  121  020
Example 19 Express the following as binary numbers.
(a) 1  8  1  4  0  2  0  1
 23  1  22  0  21  0  20
=1100(2)
=1
1  25  1  23  1  1
=1  25  0  24  1  23  0  22  0  21  1  20
(2)
=101001(2)
(b)
Classwork 1.19
(1) 1  16  1  4
=1  16  0  8  1  4  0  2  0  1
=1  24  0  23  1  22  0  21  0  20
=10100(2)
(2)1  24  1  22  1 =1  16  0  8  1  4  0  2  1  1
=10101(2)
Conversion of Binary number into Denary number
Example 20
(b) Convert 101010(2) into a denary number
101010(2)=1  25  0  24  1  23  0  22  1 21  0  20
=25  23  21
=42
Classwork 1.20
(1) (b) Convert 101 111(2) into a denary number
101111(2) =1  25  0  24  1  23  1  22  1 21  1  20
=25  23  22  21  20
=32+8+4+2+1
=47
(2) (b) Convert 11 110 010(2) into a denary number
11110010(2)=127  126  125  124  023  022 121  020
=128  64  32  16
 2
=242
Example 21 Convert 10011(2) into a denary number
10011 (2) = 1  24  0  23  0  22  1  21  1  20
=1  16  0  8  0  4  1  2  1  1
=16  2  1
=19
Classwork 1.21
(1) (b) Convert 111 000(2) into a denary number
111000(2) =1  25  1  24  1  23  0  22  0 21  0  20
=25  24  23
=32+8+4
=56
(2) (b) Convert 1010110(2) into a denary number
1010110(2) = 126  025  124  023  122 121  020
=64
 16
 4  2
=86
Conversion of Denary number into Binary number
Method 1- Fill in the number from left to right
24 2 3 22 21 20
16 8 4 2 1
30(10) = 1 1 1 1 0 = 11110(2)
Conversion of Denary number into Binary number
Method 2-
divide the denary number by 2 successively
write the answer in terms of remainders from bottom to top
2) 30
2)15
2) 7
2) 3
2)1
0
30(10) = 11110(2)
0
1
1
1
1
Example 22- Convert the following denary into binary number
2) 32
(a) 21
2) 21
(b) 32
2) 16
1
=10101(2) 2) 10
=100000(2)
2) 8
2) 5
0
2) 4
1
2) 2
2) 2
0
2) 1
2) 1
1
0
0
0
0
0
0
0
1
Classwork 1.22- Convert the following denary into binary number
2) 28
(1) 11
2) 11
(2) 28
0
2) 14
1
5
2)
=1011(2)
=11100(2)
2) 7
0
2) 2
1
1
2) 3
0
2) 1
1
1
2) 1
0
1
0
(3) 64
=1000000(2)
2) 64
2) 32
2) 16
2) 8
2) 4
2) 2
2) 1
0
0
0
0
0
0
0
1
0,1 (<2)
Binary System
23(10)= 1 $16  0 $8
 1 $4  1 $2  1 $1
12F(10)
=10111(2)
10111(2)= 1 24  0  23  1  22  1  21  1  20
Hexadecimal System (十六進制)
=303(10) = 1  $256  2  $16 + F 
=12F(16)
$1
0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F (<16)
12F(16)=1  162  2  161 + F  160
Example 23
(a) Write down the place value of each digit in 39A1(16)
Digit
3
Place Value 4096
163
9
256
162
A
16
161
1
1
160
(b) Find the place value of the digit in 4A067(16)=162=256
Classwork 1.23
(a) Write down the place value of each digit in 9FFC(16)
Digit
9
Place Value 4096
F
256
F
16
C
1
(a) Write down the place value of digit A in FC6A7(16) =16
Example 24
Express the following hexadecimal numbers in the expanded form with base 16
(a) 82A(16)=8  162  2  161 + A  160
(b) ABCD(16)=10  163  11  162 + 12  161 + 13  160
Classwork 1.24
Express the following hexadecimal numbers in the expanded form with base 16
(1) F3CF(16)=15  163  3  162 + 12  161 + 15  160
(2) 100FFF(16)=1165  0164 + 0163 + 15162 + 15161 + 15160
Example 25
Convert the following hexadecimal numbers into denary numbers
(a) 123(16)=1  162  2  161 + 3  160= 256 + 16 + 3 = 291(10)
(b) ABC (16)=10  162  11  161 + 12  160
=10  256  11  16 + 12
=2748
Classwork 1.25
(a) 2F3(16)=2 162  15161 + 3160= 512 +240 + 3 = 755(10)
(b) 3DD (16)=3162  13161 + 13160 =768  208 +13 =989(10)
(c) AFFF(16)=10163 15162 +15161 +15160 =40960 3840 +240+15 =989(10)
Example 26 – Convert the following denary numbers into hexadecimal
numbers
(a) 45(10) =2D(16)
(b) 345(10) =159(16)
)45
16) 2
0
)345
16) 21
16) 1
16
16
13 (D)
2
0
9
5
1
Classwork 1.26 – Convert the following denary numbers into
hexadecimal numbers
(1) 321(10)=141(16)
(2) 405(10) =195(16)
)321
16) 20
16) 1
)405
16) 25
16) 1
16
0
16
1
4
1
(3) 4110(10) =100E(16)
0
)4110
16) 256
16) 16
16) 1
5
9
1
16
0
14(E)
0
0
1