Indices_F3

1.1 Index Notation

CHAPTER 1 What is repeated multiplication in index form? LEARNING

The development of technology not only makes most of our daily STANDARD

WDVNV HDVLHU LW DOVR VDYHV H[SHQVHV LQ YDULRXV ¿HOGV )RU LQVWDQFH Represent repeated
multiplication in index form
the use of memory cards in digital cameras enable users to store and describe its meaning.

photographs in a large number and to delete or edit unsuitable

photographs before printing. DISCUSSION CORNER

Discuss the value of the
capacity of a pen drive.

BULLETIN

7KH QXFOHDU ¿VVLRQ RI
uranium U-320 follows the
pattern 30, 31, 32, …

In the early stage, memory cards were made with a capacity of 4MB. The capacity increases
over time to meet the demands of users. Do you know that the capacity of memory cards is
calculated using a special form that is 2n?

,Q )RUP \RX KDYH OHDUQW WKDW 3 = 4 × 4 × 4. The number 43 is written in index notation, 4
is the base and 3 is the index or exponent. The number is read as ‘4 to the power of 3’.

Hence, a number in index notation or in index form can be written as;

an Index
Base

You have also learnt that 42 = 4 × 4 and 43 î î )RU H[DPSOH

4×4=42 The value of index is 2

Repeated two times The value of index is the same as the number of times
4 is multiplied repeatedly.
4×4×4=43
The value of index is 3
Repeated three times
The value of index is the same as the number of times
4 is multiplied repeatedly.

Example 1

Write the following repeated multiplications in index form an. REMINDER

(a) 5 × 5 × 5 × 5 × 5 × 5 (b) 0.3 × 0.3 × 0.3 × 0.3 25  2 × 5 43  4 × 3
an  a × n
(c) ( –2) × (–2) × (–2) ( d) — 4 × —4 × — 4 × — 4 × —4
(e) m × m × m × m × m × m × m (f) n × n × n × n × n × n × n × n

2

Chapter 1 Indices

Solution: (b) 0.3 × 0.3 × 0.3 × 0.3 = (0.3)4 CHAPTER 1
(a) 5 × 5 × 5 × 5 × 5 × 5 = 56
repeated four times
repeated six times
( ) (d) — × — × — × — × — = — 5
(c) ( –2) × (–2) × (–2) = (–2)3 4 44 44 4
UHSHDWHG ¿YH WLPHV
repeated three times (f) n × n × n × n × n × n × n × n = n8

(e) m × m × m × m × m × m × m = m7 repeated eight times

repeated seven times

)URP WKH VROXWLRQ LQ ([DPSOH LW LV IRXQG WKDW WKH YDOXH RI LQGH[ LQ DQ LQGH[ IRUP LV WKH VDPH DV
the number of times the base is multiplied repeatedly. In general,

an = a × a × a × … × a ; a 
n factors

MIND TEST 1.1a
1. Complete the following table with base or index for the given numbers or algebraic terms.

53 (– 4)7 Base Index
5 7
( ) ( ) —2 m6– —3 4—2
7 n 6
9
n0 (0.2)9 x 4
8 2
( )x20 2 — 3 2

8

2. State the following repeated multiplications in index form an.

(a) 6 × 6 × 6 × 6 × 6 × 6 (b) 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5 × 0.5
( c) — 2 × — 2 × — 2 × — 2 (d) (–m) × (–m) × (–m) × (–m) × (–m)
H ²3 î ²3 î ²3
( ) ( ) ( ) ( ) ( ) ( ) I – –n × – n– × – – n × – n– × – – n × – n–

3. Convert the numbers or algebraic terms in index form into repeated multiplications.

( a) ( –3)3 ( b) ( 2.5)4 ( )( c) —3 5 ( )( d) – 2 —4 3

(e) k6 ( f) ( –p)7 ( )( g) —m 8 (h) (3n)5

3

CHAPTER 1 How do you convert a number into a number in index LEARNING
form? STANDARD

A number can be written in index form if a suitable base is selected. You Rewrite a number in index
can use repeated division method or repeated multiplication method to form and vice versa.
convert a number into a number in index form.

Example 2 FLASHBACK
Write 64 in index form using base of 2, base of 4 and base of 8.
Solution: 4 × 4 × 4 = 43

Repeated Division Method

(a) Base of 2 (b) Base of 4 (c) Base of 8
‡ LV GLYLGHG UHSHDWHGO\ ‡ LV GLYLGHG ‡ LV GLYLGHG

by 2. repeatedly by 4. repeatedly by 8.

2 ) 64 4 ) 64 8 ) 64
n=2 8) 8
2 ) 32 n=3 4) 4

n=6
2) 8 Hence, 64 = 82

2) 4

2) 2 Hence, 64 = 43

The division is
Hence, 64 = 26
continued until
LV REWDLQHG

Repeated Multiplication Method

(a) Base of 2 (b) Base of 4 (c) Base of 8
2×2×2×2×2×2 4×4×4 8 × 8 = 64

4 64 Hence, 64 = 82
8
Hence, 64 = 43 DISCUSSION CORNER

32 Which method is easier
64 to convert a number into
a number in index form?
Hence, 64 = 26 Is it the repeated division
or repeated multiplication
method? Discuss.

4

Example 3 Chapter 1 Indices CHAPTER 1

Write — 3 —2 – in index form using base of —2 . Repeated Multiplication Method
Solution: —25 × —25 × —25 × —25 × —52
—245–
Repeated Division Method —8 –
—6 2 –5
n=5 2 ) 32 n=5 — 3 —2 –
5 ) 625
2) 8 ( )Hence, — 3 —2 – = — 2 5
2) 4 5 ) 25
2) 2 5) 5


( )Hence, — 3 —2 – = — 2 5

MIND TEST 1.1b

1. Write each of the following numbers in index form using the stated base in brackets.

D [base of 3] E [base of 5] ( c) — 6 4– [ ]base of —45

base of – —
4
[ ( )] ( d) 0 .00032 [base of 0.2] H ± [base of (– 4)] ( f) —

How do you determine the value of the number in index form , an?

The value of an FDQ EH GHWHUPLQHG E\ UHSHDWHG PXOWLSOLFDWLRQ PHWKRG RU XVLQJ D VFLHQWL¿F
calculator.

Example 4

Calculate the values of the given numbers in index form. QU I Z

(a) 25 (b) (0.6)3 (m)4 = 16
What are the possible
2×2×2×2×2 0.6 × 0.6 × 0.6 values of m?

4×2 0.36 × 0.6

î
î 3

32 Hence, 0.63
Hence, 25 = 32

5

Example 5 SMART FINGER 1,234567.89 REMINDER
7 8 9÷
CHAPTER 1 (a) 54 = 625 4 5 6x = Negative or fractional base
(b) (–7)3 = –343 1 2 3- 4= must be placed within
AC 0 . + brackets when using a
( ) ( c) — 4 — – 5 ^4= )^ calculator to calculate
( (–) 7 ) ^ 3 ^6 values of given numbers.
=
( 2 ab/c 3 ) ^ DISCUSSION CORNER
( )(d) ²35 2 —6254– 2=
( 1 ab/c 3 ab/c 5 = Calculate questions (c),
= (d) and (e) in Example 5
( (–) 0 . 5 ) without using brackets.
(e) (– 0.5)6 Are the answers the
same? Discuss.
MIND TEST 1.1c

1. Calculate the value of each of the following numbers in index form.

(a) 94 (b) (– 4)5 (c) (2.5)3 (d) (– 3.2)3

( )( e) — 8 5 ( )( f) – — 6 4 ( )( g) ² 3 2 ( ) (h) – 2 — 3 3

1.2 Law of Indices

What is the relationship between multiplication of LEARNING
numbers in index form with the same base and repeated STANDARD
multiplication?
Relate the multiplication
Brainstorming 1 In pairs of numbers in index
form with the same
Aim: To identify the relationship between multiplication of base, to repeated
multiplications, and hence
numbers in index form with the same base and repeated make generalisation.

multiplication.

Steps:
1. Study example (a) and complete examples (b) and (c).
2. Discuss with your friend and state three other examples.
3. ([KLELW WKUHH H[DPSOHV LQ WKH PDWKHPDWLFV FRUQHU IRU RWKHU JURXSV WR JLYH IHHGEDFN

Multiplication of Repeated multiplication
numbers in index form
(a) 23 × 24 3 factors 4 factors 7 factors (overall)

(2 × 2 × 2) × (2 × 2 × 2 × 2) = 2 × 2 × 2 × 2 × 2 × 2 × 2 = 27

23 × 24 = 2 7 7=3+4
23 × 24 = 2 3 + 4

(b) 32 × 33 2 factors 3 factors 5 factors (overall)

(3 × 3) × (3 × 3 × 3) = 3 × 3 × 3 × 3 × 3 = 35

32 × 33 = 3

32 × 33 = 3

6

Chapter 1 Indices

Multiplication of Repeated multiplication
numbers in index form

(c) 54 × 52 4 factors 2 factors 6 factors (overall) CHAPTER 1

(5 × 5 × 5 × 5) × (5 × 5) = 5 × 5 × 5 × 5 × 5 × 5 = 56
54 × 52 = 5
54 × 52 = 5

Discussion:

What is your conclusion regarding the relationship between multiplication of numbers in index
form and repeated multiplication?

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23 × 24 = 23 + 4 DISCUSSION CORNER
32 × 33 = 32 + 3
54 × 52 = 54 + 2 Given,
am × an = bm × bn.

Is a = b? Discuss.

In general, am × an = a m + n

Example 6

Simplify each of the following.

(a) 72 × 73 (b) (0.2)2 × (0.2)4 × (0.2)5 ( c) 2 k2 × 4k3 ( d) 3 m4 × —6 m5 î m

Solution: (b) (0.2)2 × (0.2)4 × (0.2)5 REMINDER
= (0.2)2 + 4 + 5 a = a1
(a) 72 × 73 = (0.2)
= 72 + 3
= 75

( c) 2k2 × 4k3 (d) 3m4 × — m5 î m
= (2 × 4)(k2 × k3) 6

Operation of î ² î m4 × m5 × m )
WKH FRHI¿FLHQWV 6

= 8k2 + 3 = 6m SMART MIND
= 8k5 = 6m If ma × mb = m8, such

that a > 0 and b > 0,

what are the possible

MIND TEST 1.2a values of a and b?

1. Simplify each of the following. (b) (– 0.4)4 × (– 0.4)3 × (– 0.4)
(a) 32 × 3 × 34
( ) ( ) ( )(d) ± ²52 2 × ± ²52 3 × ± ²52 5
( ) ( ) ( )(c) —4 × —4 3 × —4 5
77 7 ( f) n 6 × — n2 × — n3 × n
( e) 4m2 × — m3 × (– 3)m4 25 4

2 (h) – — y5 × (– 6)y3 × — y4
(g) – x4 × — x × — x2 23

45

7

CHAPTER 1 How do you simplify a number or an algebraic term in index TIPS
form with different bases?
Group the numbers or
Example 7 algebraic terms with the
VDPH EDVH ¿UVW 7KHQ DGG
the indices for the terms
with the same base.

Simplify each of the following.

(a) m3 × n2 × m4 × n5 (b) (0.3)2 × (0.2)2 × 0.3 × (0.2)5 × (0.3)3
( c) p2 × m3 × p4 × n3 × m4 × n2 (d) –m4 × 2n5 × 3m × — 4 n2

Solution:

(a) m3 × n2 × m4 × n5 (b) (0.3)2 × (0.2)2 × 0.3 × (0.2)5 × (0.3)3
= (0.3)2 × (0.3) × (0.3)3 × (0.2)2 × (0.2)5
= m3 × m4 × n2 × n5 Group the terms = (0.3) × (0.2)(2 + 5)
= m3 + 4 × n2 + 5 with the same base. = (0.3)6 × (0.2)7

= m7 × n7 Add the indices for terms
= m7n7 with the same base.

( c) p2 × m3 × p4 × n3 × m4 × n2 (d) –m4 × 2n5 × 3m × —4 n2

= m3 × m4 × n3 × n2 × p2 × p4 ± î î î ² 4 m4 × m × n5 × n2
= m3 + 4 × n3 + 2 × p2 + 4
= m7 n5 p6 = – —23 m n5 + 2 REMINDER

= – —23 m5 n7 –an  ±a)n

Example:
± 2  ± 2

MIND TEST 1.2b ± 

1. State in the simplest index form. (b) (0.4)2 î 3 î î 5 î
(a) 54 × 93 × 5 × 92 ( d) –2k5 × p6 × — 4 p5 × 3k

F x5 × y3 × — 2 x × — 3 y4

What is the relationship between division of numbers in LEARNING
index form with the same base and repeated multiplication? STANDARD

Brainstorming 2 In pairs Relate the division of
numbers in index form
Aim: To identify the relationship between division of numbers in with the same base, to
index form with the same base and repeated multiplication. repeated multiplications,
and hence make
generalisation.

Steps:
1. Study example (a) and complete examples (b) and (c).
2. Discuss with your friend and state three other examples.
3. 3UHVHQW \RXU ¿QGLQJV

8

Chapter 1 Indices

Division of numbers Repeated multiplication
in index form
5 factors CHAPTER 1
(a) 45 ÷ 42
—4452 = —4 ×—4—×—4 ×—4––×–4– = 4 × 4 × 4 = 43
(b) 26 ÷ 22 4×4
3 factors (Remainder)
(c) (–3)5 ÷ (–3)3
2 factors

45 ÷ 42 = 4 3 3=5–2

45 ÷ 42 = 4 5–2

6 factors

—2262 = —2 ×—2—×—22 ×—× 22––×–2–—× 2– = 2 × 2 × 2 × 2 = 24

4 factors (Remainder)

2 factors

26 ÷ 22 = 2
26 ÷ 22 = 2

5 factors

—((––33—))35 = (—–3—) ×—((–—–33)—) ××–((––––33–))—××(–(–––33–))—×—(––3–) = (–3) × (–3) = (–3)2

2 factors (Remainder)

3 factors

(–3)5 ÷ (–3)3 = (–3)

(–3)5 ÷ (–3)3 = (–3)

Discussion

What is the relationship between division of numbers in index form and repeated
multiplication?

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW SMART MIND

45 ÷ 42 = 45 – 2 Given ma ± b = m7 and
26 ÷ 22 = 26 – 2 0 < a < 10. If a > b,
(–3)5 ÷ (–3)3 = (–3)5 – 3 state the possible values
of a and b.
In general, am ÷ an = am – n

Example 8

Simplify each of the following.

(a) 54 ÷ 52 (b) (–3)4 ÷ (–3)2 ÷ (–3) (c) m4n3 ÷ m2n
(d) 25x2y3 ÷ 5xy (e) m ÷ 4m5 ÷ m2 I ± p8 ÷ 2p5 ÷ 4p2

Solution: (c) m4n3 ÷ m2n
= m4n3 ÷ m2n
(a) 54 ÷ 52 (b) (–3)4 ÷ (–3)2 ÷ (–3) = m4 – 2 n ±
= (–3)4 ÷ (–3)2 ÷ (–3) = m2 n2
= 54 – 2 = (–3) ± ±
= 52 = (–3)

= –3

9

(d) 25x2y3 ÷ 5xy H m ÷ 4m5 ÷ m2 I ± p8 ÷ 2p5 ÷ 4p2
= — 4 (m ÷ m 5 ÷ m2) = ±—2 – (p8 ÷ p5) ÷ 4p2
CHAPTER 1 = 2 5 x2y3 ÷ 5x y
= 3(m ± ) ÷ m2 = –8p8–5 ÷ 4p2
= —255 x ± y ± = –8p3 ÷ 4p2
Operation of the = 3m5 – 2
= 5x y2 FRHI¿FLHQWV = – —84 (p3 ÷ p2)
= 3m3 = –2p3 – 2
= 5xy2 = –2p

= –2p

MIND TEST 1.2c

1. Simplify each of the following.

(a) 45 ÷ 44 (b) 7 ÷ 76 ÷ 72 (c) —mm84—nn6
(f) –25h4 ÷ 5h2 ÷ h
(d) —297x—x34y–y2–5 (e) m7 ÷ m2 ÷ m4

2. Copy and complete each of the following equations.

(a) 8 ÷ 84 ÷ 83 = 8 (b) m4n ÷ m n5 = m2n

(c) —m — n—m4 ×7—nm——n2= m5n (d) —27—x3y—6x×2—yx3—y – = 3x y5

3. If —224x—×× —332y = 6, determine the value of x + y.

What is the relationship between a number in index form LEARNING
raised to a power and repeated multiplication? STANDARD

Brainstorming 3 In pairs Relate the numbers in
index form raised to a
Aim: To identify the relationship between a number in index form power, to repeated
raised to a power and repeated multiplication. multiplication, and hence
make generalisation.

Steps:

1. Study example (a) and complete examples (b) and (c).

2. Discuss with your friend and state three other examples.
3. 3UHVHQW \RXU ¿QGLQJV

Index form raised Repeated multiplication in index form Conclusion
to a power
4 factors (32)4 = 32(4)
(a) (32)4 = 38

32 × 32 × 32 × 32

= 32 + 2 + 2 + 2 2 is added 4 times

4 times

= 32(4)

10

Chapter 1 Indices

Index form raised Repeated multiplication in index form Conclusion
to a power
3 factors (54)3 = 5 CHAPTER 1
(b) (54)3 =5
54 × 54 × 54
= 54 + 4 + 4 4 is added 3 times (43)6 = 4
=4
3 times

= 54(3)

(c) (43)6 6 factors

43 × 43 × 43 × 43 × 43 × 43

= 43 + 3 + 3 + 3 + 3 + 3 3 is added 6 times

6 times

= 43(6)

Discussion:

What is your conclusion regarding the index form raised to a power and repeated multiplication
in index form?

The conclusion in Brainstorming 3 can be checked using the following method.

Example (a) Example (b) Example (c)

(32)4 = 32 × 32 × 32 × 32 (54)3 = 54 × 54 × 54 (43)6 = 43 × 43 × 43 × 43 × 43 × 43
= 32 + 2 + 2 + 2 = 54 + 4 + 4 = 43 + 3 + 3 + 3 + 3 + 3
= 38 = 5 = 4

32(4) = 32 × 4 54(3) = 54 × 3 43(6) = 43 × 6
= 38 = 5 = 4

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW SMART MIND

(32)4 = 32(4) Given,
(54)3 = 54(3) mrt = 312
(43)6 = 43(6)
What are the possible
In general, (am)n = amn values of m, r and t
if r > t ?

Example 9

1. Simplify each of the following.

(a) (34)2 (b) (h3) (c) ((–y)6)3

2. Determine whether the following equations are true or false.

(a) (42)3 = (43)2 (b) (23)4 = (22)6 (c) (32)6 = (272)4

11

CHAPTER 1 Solution: (b) (h3) (c) ((–y)6)3
= (–y)6(3)
1. (a) (34)2 = h = (–y)
= h30
= 34(2) (c) (32)6 = (272)4
= 38 (b) (23)4 = (22)6
Same left right
2. (a) (42)3 = (43)2 left right
Same /HIW
left right /HIW (32)6 = 32(6) = 3
(23)4 = 23(4) = 2
/HIW 5LJKW
(42)3 = 42(3) = 46 5LJKW (272)4 = (33(2))4 Not the
(22)6 = 22(6) = 2
5LJKW same
(43)2 = 43(2) = 46 Hence, (23)4 = (22)6
is true. = 36(4)
Hence, (42)3= (43)2 = 324
is true. Hence, (32)6 = (272)4
is false.

MIND TEST 1.2d

1. Use law of indices to simplify each of the following statements.

D 5)2 (b) (3 )2 (c) (72)3 (d) ((– 4)3)7
(e) (k8)3 (f) (g2) (g) ((–m)4)3 (h) ((–c)7)3

2. Determine whether the following equations are true or false. (d) – (72)4 = (– 492)3

(a) (24)5 = (22) (b) (33)7 = (272)4 (c) (52)5 2)3

How do you use law of indices to perform operations of multiplication and division?

(am × bn)q (ambn)q = amq bnq
= (am)q × (bn)q
= amq × bnq ( )—abmn– q = —abmnq–q

(am ÷ bn)q
= (am)q ÷ (bn)q
= amq ÷ bnq

Example 10

1. Simplify each of the following.

(a) (73 × 54)3 (b) (24 × 53 î 2)5 (c) (p2q3r)4 (d) (5m4n3)2
(g) —(36m—m2n—3n3)–3 (h) —(2x—3y3—46)4x— × y—( 3 —xy2—)3
( )(e) —3252 4 ( )(f) —32xy–73 4

12

Chapter 1 Indices

Solution: (b) (24 × 53 î 2)5 FLASHBACK CHAPTER 1
= 24(5) × 53(5) î 2(5)
(a) (73 × 54)3 = 220 × 5 î am × an = am + n
= 73(3) × 54(3) am ÷ an = am – n
= 79 × 5 (am)n = amn

(c) (p2q3r)4 (d) (5m4n3)2 QU I Z
= p2(4) q3(4)r = 52m4(2)n3(2) mm = 256.
= p8q r4 = 25m8n6 What is the value of m?

( )(e) —2352 4 ( )(f) —23yx–37 4 DISCUSSION CORNER

= —3252((–44)) = —2344y–x7–3((–44)) Why is 1n = 1 for all
= —2328–0 = — –yx–2 – 8 values of n?
Discuss.

(g) —(3m—2n—3)–3 (h) —(2x—3y3—46)4x— × —y( 3 —xy2—)3
6m3n = — 24x—3 ( 4 — ) y 4 —3(46)—x× —3y3 x— — y–2(–3)–
= — —x —3y6 —x × —y2 7 —x3—y6
= —336m—m2(—33n)n — 3(3)
( ) = — — î— – x ± y ±
= 2—67mm—36n—n 9 36
x5 y
= — m6 – 3 n ±
2

= —2 m3 n8

MIND TEST 1.2e

1. Simplify each of the following.

(a) (2 × 34)2 E 3 × 95)3 F 3 ÷ 76)2 (d) (53 × 34)5

(e) (m3n4p2)5 (f) (2w2 x 3)4 ( )(g) —–b3—4a5– 6 ( )(h) —32ba–45– 3

2. Simplify each of the following. ( )(c) —4623– 3 ÷ —6432– (d) —((–(—–4)4—6))62—××—((––—55)22—)3
(h) —((bb—22dd34—))23
— — 3 ×–2 4—2 (b) —33—×64—(62—)3 (g) —((mm—52nn—37))2–3
( ) (a) 2


(e) —x2—yx6y—×2 —x3 (f) —(h(h—3kk2)—2)4

3. Simplify each of the following.

(a) — (2m—2n —4 )m—3 7×n— ( 3— m—n4)–2 ( b) (— 5x—y4 —) 2 x—×4y6—6x — y (c) 2—(4dd—53ee—65) ×—× ((—36dd—e3e2–)43)–2

13

How do you verify a0 = 1 and a–n = —a1n ; D  0? LEARNING
STANDARD
CHAPTER 1 Brainstorming 4 In pairs
Verify that a0 =
Aim: To determine the value of a number or an algebraic term with and a–n = –a1–n ; a  .
a zero index.

Steps:
1. Study and complete the following table.
2. What is your conclusion regarding zero index?

Division in Law of indices Solution Conclusion
index form 23 – 3 = 20 Repeated multiplication from the
solution
(a) 23 ÷ 23 —22 ×—× 22—××–22–
20

(d) m5 ÷ m5 m5 – 5 = m0 ²mm²×× ²mm ²×× mm²×ײmm²××±mm±± m0

(c) 54 ÷ 54

(d) (–7)2 ÷ (–7)2

(e) n6 ÷ n6

Discussion:
1. Are your answers similar to the answers of the other groups?
2. What is your conclusion regarding zero index?

)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW

20
m0

7KHUHIRUH D QXPEHU RU DQ DOJHEUDLF WHUP ZLWK D ]HUR LQGH[ ZLOO JLYH D YDOXH RI
In general, a0 a 

How do you verify a–n = –a1–n– ?

Brainstorming 5 In groups

A i m : To veri fy a–n = — a n.
Steps:

1. Study and complete the following table.

14

Chapter 1 Indices

Division in Solution Conclusion
index form from the
Law of indices Repeated multiplication solution CHAPTER 1
(a) 23 ÷ 25
23 – 5 = 2–2 —2 — × 2 — ×î— 2 —×î 2 – –× – –2 = – 2 – × – –2 = – 2 –2 2 – 2 = 2– –2
(b) m2 ÷ m5 m2 – 5 = m–3 –m––×—m—×m—m×—×m—m —× m — = —m —× m – – × – – m – = – m –3
(c) 32 ÷ 36 m – 3 = – m – 3–

(d) (– 4)3 ÷ (– 4)7

(e) p4 ÷ p8

Discussion Scan the QR Code or visit
1. Are your answers similar to the anwers of the other groups? http://bukutekskssm.my/
2. What is your conclusion? Mathematics/F3/Chapter1
AlternativeMethod.mp4
)URP %UDLQVWRUPLQJ LW LV IRXQG WKDW to watch a video that
describes alternative
2–2 = 2 — 2 method to verify a± = —a n.
m –3 = m— 3

In general, a–n = –a –n ; a  0 BULLETIN

Example 11 Negative index is a
number or an algebraic
term that has an index of
a negative value.

1. State each of the following terms in positive index form.

( a) a – 2 (b) x – 4 ( c) – 8– –5– TIPS

(d) –y ––9 – ( e) 2 m –3 ( f) — 5 n –8 Ƈ a–n ±a1±n
Ƈ an ±a1±–±n

( )(g) –2– ± ( )(h) –x– –7 ( ) ( )Ƈ ±ab± –n ±ab± n
3 y
=

2. State each of the following in negative index form. REMINDER
2a –n  2²1an±
(a) — 3 4 ( b) —m 5 (c) 75
SMART MIND
(d) n20 ( )(e) –45– 8 ( )(f) –mn–
( ) ± ² 4 ± = x y
3. Simplify each of the following.
What are the values of x
(a) 32 × 34 ÷ 38 (b) (—24—)2 —× (—35—)3 (c) —(4x—y2—)2 —× x—5y and y?
(28 × 36)2 (2x3y)5
15

CHAPTER 1 Solution: (b) x – 4 = –x1–4 (c) –8–1––5 = 85 (d) –y–1––9 = y9
1. (a) a–2 = a–12– (f ) —35 n– 8 = —53n–8

(e) 2m–3 = m—23 –2– –10 –3– 10 –x– –7 –y– 7

= =
( ) ( )(g) ( ) ( )(h)
32 yx

2. (a) 3—14 = 3– 4 (b) —m15 = m–5 (c) 75 = 7—1–5 (d) n20 = n—–12–0

( ) ( )(e)–4–8 –5– –8 ( ) ( )(f)–m–15 –n– –15

= =
54 nm

(b) —(2(4—2)28—×× (—3365)—)23 (c) —(4x—(y22x—)32y—×)5x—5y TIPS
= —2218—6××—331–512
3. (a) 32 × 34 ÷ 38 = 28 – 16 × 315 – 12 = —42x—2y—4 ×—x5—y1 y0 = 1
= 2–8 × 33 25x15y5 y1 = y
= 32 + 4 – 8 = —2383
= 3–2 = 1—6 x2 + 5 – 15 y4 + 1 – 5
= —1 32

32 = —1 x–8 y0
2

= —21x–8

MIND TEST 1.2f

1. State each of the following terms in positive index form.

(a) 5–3 (b) 8– 4 (c) x– 8 (d) y–16 (e) —a1– –4

(f) —201–––2 (g) 3n– 4 (h) –5n– 6 (i) —2 m–5 ( )(j) – —3 m– 4
7 8
( )(k) —2 –12 ( )(l) – —3 –14 ( )(m) —x –10
5 7 y ( )(n) —2x– – 4 ( )(o) —1 –5
3y 2x

2. State each of the following terms in negative index form.

(a) —1 (b) —1 (c) —1 (d) —1 (e) 102
54 83 m7 n9
( )(j) —yx 10
(f) (– 4)3 (g) m12 (h) n16 ( )(i) —47 9

3. Simplify each of the following.

(a) (—42(—)436—×)24–5 (b) —((22—3××—3342–))5–3 (c) —(23—)(–52—2×)5–(5–4–)2
(f) (—2m—–2—(n4)m—5 2×—n(43)—2m–4n–)–2
(d) —3m—2n9—4m×—3n(m5—n—3)––2 (e) —(2m—2n—2()9—–m3 —×3n()—32m–n–2–)–4

16

Chapter 1 Indices

How do you determine and state the relationship between LEARNING
fractional indices and roots and powers? STANDARD

Relationship between n¥a and a—n1 Determine and state the CHAPTER 1
relationship between
In Form 1, you have learnt about square and square root as well as cube fractional indices and
and cube root. Determine the value of x for roots and powers.

(a) x2 = 9 (b) x3 = 64 TIPS

Solution: Square roots are used (b) x3 = 64 Ƈ 9 = 32 Ƈ 64 = 43
(a) x2 = 9 to eliminate squares. 3¥x3 = 3¥ 3
x =4 Cube roots are used to
¥x2 ¥ 2 eliminate cubes.
x =3

Did you know that the values of x in examples (a) and (b) above can be determined by raising the
index to the power of its reciprocal?

(a) x2 = 9 The reciprocal (b) x3 = 64 BULLETIN
of 2 is —1 .
x2(—21 )= 9—21 x3(—31 )= 64(—31 ) 1 is the reciprocal of a.
x1 = 32(—21 ) 2 x1 = 43(—13 ) —a

x =3 The reciprocal x =4
of 3 is —13 .

From the two methods to determine the values of x in the examples SMART MIND
above, it is found that:
What is the solution for
2¥x = x–12 ¥– 4 ? Discuss.
3¥x = x–13

In general, n¥a = a–1n ; D  0

Example 12
1. Convert each of the following terms into the form a—1n .

(a) 2¥ E 3¥± F 5¥m (d) 7¥n

2. Convert each of the following terms into the form n¥a .
(a) 125—51 (b) 256—81 (c) (–1 000)—31 (d) n1—12

3. Calculate the value of each of the following terms.

(a) 5¥± E 6¥ F — 13 G ± —51

Solution: (b) 3¥± ± — 31 F 5¥m = m—1 (d) 7¥n = n—1
(b) 25681– = 8¥256 5 7
1. (a) 2¥36 = 36—12
2. (a) 125—15 = 5¥125 (c) (–1 000)—31 = 3¥(–1 000) (d) n—1 = 12¥n
12

17

CHAPTER 1 3. (a) 5¥–32 = (–32)—15 (b) 6¥729 = 729—16 (c) 512 13– = 83(3–1) (d) (–243)—15 = (–3)5(—15 )

= (–2)5(—51 ) = 36(—61 ) = 81 = (–3)1
=8 = –3
= (–2)1 = 31 TIPS

= –2 =3 <RX FDQ XVH D VFLHQWL¿F
calculator to check the
MIND TEST 1.2g answers.

1. Convert each of the following terms into the form a–1n . (d) 10¥n

(a) 3¥125 (b) 7¥2 187 (c) 5¥–1 024 (d) n–11–5

2. Convert each of the following terms into the form n¥a. (d) (–32 768)—51

(a) 4—21 (b) 32—51 (c) (–729)—31

3. Calculate the value of each of the following terms.

(a) 3¥343 (b) 5¥–7 776 (c) 262 144—61

What is the relationship between a—mn and (am)—n1 , (a—n1 )m, n¥am dan (n¥a)m?

You have learnt that:

amn = (am)n and n¥a1 = a—n1

From the two laws of indices above, we can convert a—mn into (am)—1n , (a—1n )m, n¥am and (n¥a)m.
Calculate the value of each of the following. Complete the table as shown in example (a).

a—mn (am)—n1 (a—n1 )m n¥am (n¥a)m

(a) 64—32 (=64420)—9136(—13 ) (64—31 )2 3¥642 (3¥64)2
= 163(—31 ) = 43(—31 )(2) = 3¥4 096 = 42
= 42 = 16
= 16

= 16 = 16

(b) 16—43

(c) 243—52

Are your answers in (b) and (c) the same when you use different index forms? Discuss.

From the activity above, it is found that:

a—mn = (am)—n1 = (a—1n )m
a—mn = n¥am = (n¥a)m

18

Chapter 1 Indices

Example 13

1. Convert each of the following into the form (am)–n1 and (a–n1)m. CHAPTER 1

(a) 81—32 (b) 27—23 (c) h—35

2. Convert each of the following into the form n¥am and (n¥a)m.

(a) 343—32 (b) 4 096—56 (c) m—25

Solution: (b) 27—32 = (272)—13 (c) h—35 = (h3)—51
1. (a) 81—32 = (813)—21 27—23 = (27—13 )2 h—53 = (h—15 )3

81—23 = (81—21 )3

2. (a) 343—23 = 3¥ 2 (b) 4 096—56 = 6¥ 5 (c) m—52 = 5¥m2
343—32 = (3¥343)2 4 096—56 = (6¥ 4 096)5 m—25 = (5¥m)2

MIND TEST 1.2h
1. Complete the following table.

—8116 —34 —2
a—mn 729—65 121—32 w—37 ( ) ( )x—25 —hk
3

(am)—1n

(a—n1 )m

n¥ am

(n¥ a )m

Example 14

1. Calculate the value of each of the following.

(a) 9—25 (b) 16—45

Solution: (b) 16—45 16—45 = (4¥16)5 = 25 = 32
1. (a) 9—25 16—45 = 4¥165 = 4¥1 048 576 = 32
Method 1
Method 1 9—25 = (¥9)5 = (3)5 = 243 Method 2
Method 2 9—52 = ¥95 = ¥59 049 = 243

19

MIND TEST 1.2i

CHAPTER 1 1. Calculate the value of each of the following.

(a) 27—23 (b) 32—25 (c) 128—27 (d) 256—38

(e) 64—34 (f) 1 024—25 (g) 1 296—43 (h) 49—23
(l) 10 000—34
(i) 2 401—41 (j) 121—23 (k) 2 197—32

2. Complete the following diagrams with correct values.

(a)

Example 16 Chapter 1 Indices

1. Calculate the value of each of the following. (c) (—24—3—54—×—5——32 )–2 CHAPTER 1
4¥ × ¥ 4
(a) ——49——21 ×—1—25—– —13—– (b) —16——43 —× 8—1–——14–
4¥ î 5¥ (26 × 34)—21 (c) (—24—3—54—×—5—32—)2
4¥ × ¥ 4
Solution: (b) —16——43 —× 8—1–——41– = —284—13——4154—(2×—) ×2–55–—24—–23–(–2)–
(26 × 34)—12 = —343(——415()——58×) ×—525–(—342–)
(a) ——49——21 ×—1—25—– ——13 = —3318—×× 5—534
4¥ î 5¥ = —224—6(3—(—4321—))××—3344—2(–(—21—–14) –) = 38 – 1 × 53 – 4
= —(77—24(—)21——14) ××—5(—53(5–)———3115 )– = —23—×—3––1 = 37 × 5–1
= —357
= —7711—××—55–11– 23 × 32 = —2 1—587–
= 71–1 × 5–1 –1 = 23 – 3 × 3–1 – 2 = 437 —25
= 70 × 5–2
= 1 × —512 = 20 × 3–3
= —1 = 1 × —1

25 33
= —1

27

MIND TEST 1.2j

1. Simplify each of the following. (b) —(m—n(2m)—36×n—3()¥——23m—n)–4 (c) —¥ — x3—yz—2 ×—4—x2z–
(a) —3¥—c2d(—c3–e—3d×—2ce—)13—2d 2—e—32– ¥ x5yz8

2. Calculate the value of each of the following. (c) ——(2—6 ×—3—4 ×—52—)—23——
4¥ î ¥ î 3¥
(a) —¥ —– 4—× 1—14 (b) —(5—–3 ×—3—6)——13 ×—4¥— ––
49 × 121 (125 × 729 × 64)– —13 (f) —64——32 ×—3¥—1—25—× —(2—× ——51 )––3
42 × 4¥
(d) —9—¥5—12—×—3¥—34—3 ×—¥–1–2–1––– (e) —(24—×—36—)—21—× —3¥ — î– —¥ –
(64)—13 × (81)—34 × (14 641)—14 16—43 × 27—31

3. Given m = 2 and n = –3, calculate the value of 64—m3 × 512(– —1n ) ÷ 81—2nm .
4. Given a = —21 and b = —32 , calculate the value of 144a ÷ 64b × 256—ab .

21

How do you solve problems involving laws of indices? LEARNING
STANDARD
CHAPTER 1
Solve problems involving
laws of indices.

Example 17 FLASHBACK
&DOFXODWH WKH YDOXH RI ¥ î —23 ÷ 6 without using a calculator.
Common prime factors
of 6 and 12 are 2 and 3.

Understanding the Planning a strategy Implementing the strategy
problem
Calculate the value of Convert each base ¥ î —32 ÷ 6
numbers in index form into prime factors and = 3—12 × (2 × 2 × 3)—23 ÷ (2 × 3)
with different bases. calculate the value by = 3—12 × 2—23 × 2—23 × 3—23 ÷ (21 × 31)
applying laws of indices. = 3—12 + —32 – 1 × 2—32 + —32 – 1
Making a conclusion
¥ î —32 ÷ 6 = 12 = 31 × 22

= 12

Example 18 REMINDER
Calculate the value of x for the equation 3x × 9x + 5 ÷ 34 = 1.
Ƈ ,I am = an
Understanding the Planning a strategy then, m = n
problem
The question is an equation. Ƈ ,I am = bm
Calculate the value of Hence, the value on the left side then, a = b
variable x which is part of the equation is the same as
of the indices. the value on the right side of the Checking Answers
equation. Convert all the terms
into index form with base of 3. You can check the answer
by substituting the value of
x into the original equation.

3x × 9x + 5 ÷ 34 = 1

Left Right

Substitute x = –2 into left
side of the equation.

Implementing the strategy Making a conclusion 3–2 × 9–2 + 5 ÷ 34

3x × 9x + 5 ÷ 34 = 1 If 3x × 9x + 5 ÷ 34 = 1, = 3–2 × 93 ÷ 34
3x × 32(x + 5) ÷ 34 = 30 then, x = –2
3x + 6 = 0 = 3–2 × 32(3) ÷ 34
3x + 2(x + 5) – 4 = 30 3x = –6
3x + 2x + 10 – 4 = 30 x = —–36– = 3–2 + 6 – 4
x = –2
33x + 6 = 30 = 30 The same value
am = an =1 as the value on
m=n the right side
of the equation.

22

Chapter 1 Indices

Example 19 Checking Answers
Calculate the possible values of x for the equation 3x2 × 32x = 315.
Substitute the values of x CHAPTER 1
Understanding Planning a Implementing the strategy into the original equation.
the problem strategy 3x2 × 32x = 315
3x2 × 32x = 315 If am = an,
Calculate All the 3x2 + 2x = 315 then, m = n. Left Right
the value of bases
x which is involved in x2 + 2x = 15 Solve the Substitute x = 3
part of the the equation quadratic
indices. are the Left: Right:
same.
x2 + 2x – 15 = 0 equation using 3(3)2 × 32(3) 315
factorisation
= 39 × 36

(x – 3)(x + 5) = 0 method. = 39 + 6

x – 3 = 0 or x + 5 = 0 = 315 The same value

x=0+3 x=0–5 Substitute x = –5

Making a conclusion x=3 x = –5 Left: Right:

The possible values of x for 3(–5)2 × 32(–5) 315
the equation 3x2 × 32x= 315
are 3 and –5. = 325 × 3–10

= 325 + (–10)

= 315 The same value

Example 20 FLASHBACK

Solve the following simultaneous equations. Simultaneous linear
equations in two
25m × 5n = 58 and 2m × —21n = 2 variables can be solved
using substitution
Solution: 2m × —21n = 2 method or elimination
25m × 5n = 58 method.
52(m) × 5n = 58
Checking Answers

52m + n = 58 2m × 2–n = 21 Substitute m = 3 and n = 2
into original simultaneous
2m + n = 8 1 2m + (–n) = 21 equations.

m–n=1 2 25m × 5n = 58

Equation 1 and 2 can be solved by substitution method. Left Right
From 1 :
2m + n = 8 Left: Right:

n = 8 – 2m 3 25m × 5n 58
= 52(m) × 5n
= 52(3) × 52

= 56 + 2

Substitute 3 into 2 Substitute m = 3 into 1 = 58 The same value
m–n=1
2m + n = 8 2m × —1 = 2
m – (8 – 2m) = 1 2(3) + n = 8 2n
m – 8 + 2m = 1 You can also
m + 2m = 1 + 8 6+n=8 substitute m = 3 Left Right
3m = 9 into equation
m = —9 Left: Right:
3
m=3 n = 8 – 6 2 or 3 . 2m × —21n 2

n=2 = 23 × —212

Hence, m = 3 and n = 2. = 23 × 2–2
= 23 + (–2)
= 21

= 2 The same value

23

Example 21

CHAPTER 1 My equation is
3(9x) = 27y.

My equation is
16(4x) = 16 y.

The values of the variables x
and y can be determined if you
can solve both the equations.

Chong and Navin performed an experiment to determine the relationship between variable x and
variable y. The equation Chong obtained was 16(4x) = 16 y, while the equation Navin got was
3(9x) = 27y. Calculate the values of x and y which satisfy the experiment Chong and Navin have

performed.

Solution:

16(4x) = 16y 3(9x) = 27y You can also substitute
42(4x) = 42(y) 3(32x) = 33(y) y = 3 into equation
42 + x = 42y 31 + 2x = 33y 2 or 3 .
2 + x = 2y 1 + 2x = 3y
1 2

Equations 1 and 2 can be solved by elimination Substitute y = 3 into equation 1

method. Multiply equation 1 1 : 2 + x = 2y
1 × 2 : 4 + 2x = 4y 2 + x = 2(3)
by 2 to equate the x =6–2
x =4
3 FRHI¿FLHQWV RI YDULDEOH x.
Hence, x = 4, y = 3
2 : 1 + 2x = 3y

3– 2:

3+0=y

y= 3

Dynamic Challenge

Test Yourself

1. State whether each of the following operations which involves the laws of indices is true or
false. If it is false, state the correct answer.

(a) a5 = a × a × a × a × a (b) 52 = 10 (c) 30 = 0
(d) (2x3)5 = 2x15
(g) 32—25 = (2¥32)5 (e) m0n0 = 1 (f) 2a– 4 = —21a–4
(i) (5m—14 )– 4 = 6—m25–
( ) ( )(h)—mn –4 —mn 4

=

24

Chapter 1 Indices

2. Copy and complete the following diagram with suitable values.

5Ƒ × 55 53(Ƒ) CHAPTER 1

( )—51Ƒ 3 512 ÷ 5Ƒ

5—1Ƒ 59 (¥25)Ƒ

—56 —5×25–Ƒ– ( )—1 Ƒ
(5Ƒ)—32 5
(Ƒ¥125)Ƒ

3. Copy and complete the following diagram.

Operations that ( )20 —3 –2
involve laws as –3—1– 4 as 5 as 72 × 5–3 as (5–1 × ¥25)3
of indices

Value

Skills Enhancement

1. Simplify each of the following.

(a) (mn4)3 ÷ m4n5 (b) 3x × —16 y4 × (xy)3 F ¥xy × 3¥xy2 × 6¥xy5

2. Calculate the value of each of the following. (c) (256)—83 × 2–3
(f) (125)—32 × (25)– —32 ÷ (625)– —41
(a) 64—13 × 5–3 (b) 7–1 × 125—23
(c) axa8 = 1
(d) 24 × 16– —43 (e) ¥49 × 3–2 ÷ (¥81)–1 (f) 2x = —12610–x
(i) 25x ÷ 125 = —51x
3. Calculate the value of x for each of the following equations.

(a) 26 ÷ 2x = 8 (b) 3– 4 × 81 = 3x

(d) 4 × 8x + 1 = 22x (e) (ax)2 × a5 = a3x

(g) 36 ÷ 3x = 81(x – 1) (h) (m2)x × m(x + 1) = m–2

25

Self Mastery

CHAPTER 1 1. Calculate the value of each of the following without using a calculator.

(a) 4—31 × 50—23 × 10—53 (b) 5—52 × 20—32 ÷ 10–2 (c) 60—12 × 125—23 ÷ ¥15

2. Calculate the value of x for each of the following equations.

(a) 64x—12 = 27x– —52 (b) 3x—32 = —247 x– —34 (c) 25x– —23 – —35 x—31 = 0

3. Calculate the possible values of x for each of the following equations.

(a) ax2 ÷ a5x = a6 (b) 2x2 × 26x = 27 (c) 5x2 ÷ 53x = 625

4. Solve the following simultaneous equations. (b) 4(4x) = 8y + 2 and 9x × 27y = 1
(a) 81(x + 1) × 9x = 35 and 82x × 4(22y) = 128

5. In an experiment performed by Susan, it was found
WKDW WKH WHPSHUDWXUH RI D PHWDO URVH IURP Û& WR TÛ&
according to equation T = 25(1.2)m when the metal

was heated for m seconds. Calculate the difference in
WHPSHUDWXUH EHWZHHQ WKH ¿IWK VHFRQG DQG WKH VL[WK
second, to the nearest degree Celsius.

6. Encik Azmi bought a locally made car for RM55 000.

After 6 years, Encik Azmi wishes to sell the car. Based

on the explanation from the used car dealers, the price RM55 000

of Encik Azmi’s car will be calculated using the formula

( )RM55000—8 n In this situation, n is the number of years
9
.

after the car is bought. What is the market value of Encik

Azmi’s car? State your answer correct to the nearest RM.

7. Mrs Kiran Kaur saved RM50 000 on 1 March 2019
in a local bank with an interest of 3.5% per annum.
After t years, Mrs Kiran Kaur’s total savings, in RM,
is 50 000 (1.035)t. Calculate her total savings on
1 March 2025, if Mrs Kiran Kaur does not withdraw
her savings.

26

Chapter 1 Indices

P ROJ EC T

Materials: One sheet of A4 paper, a pair of scissors, a long ruler, a pencil. CHAPTER 1

Instructions: (a) Carry out the project in small groups.
(b) Cut the A4 paper into the biggest possible square.

Steps:

1. Draw the axes of symmetry (vertical and horizontal only) as shown in Diagram 1.

2. Calculate the number of squares formed. Write your answers in the space provided in
Sheet A.

3. Draw the vertical and horizontal axes of symmetry for each square as shown in
Diagram 2.

4. Calculate the number of squares formed. Write your answers in the space provided in
Sheet A.

5. Repeat step 3 and step 4 as many times as possible. 8

11

2 7
2 6

3

Diagram 1 45
Diagram 2

6. Compare your answers with those of other groups. Scan the QR Code or
7. What can you say about the patterns in the column ‘Index form’ visit http://bukutekskssm.
my/Mathematics/F3/
in Sheet A? Chapter1SheetA.pdf
8. Discuss the patterns you identify. to download Sheet A.

Sheet A

Number of axis Index form Number of Index form
of symmetry square
– 20
0 21 1 22
2 4
8 16

27

CHAPTER 1 an Index CONCEPT MAP 54 = 5 × 5 × 5 × 5
m × m × m × m × m = m5
Base Indices
an = a × a × a × … × a

n factors

Multiplication Division Power
am × an = am + n am ÷ an = am – n (am)n = amn (am × an)p = amp × anp
36 ÷ 34 = 36 – 4 (34)2= 38 (3a4)3 = 27a12
23 × 25 = 23 + 5

Fractional index Negative index Zero index
a0 = 1 ; a  0
a—n1 = n¥a 8–13 = 3¥8 a–n = —a1n ; a  0 20 = 1
a—mn = (am)—1n = (a—n1 )m 8–32 = (82)–31 = (8–13)2 5–3 = —513 m0 = 1
8–23 = 3¥ 2 = (3¥8)2
a—mn = n¥am = (n¥a)m

SELF-REFLECT

At the end of this chapter, I can:

1. Represent repeated multiplication in index form and describe its meaning.

2. Rewrite a number in index form and vice versa.

3. Relate the multiplication of numbers in index form with the same base, to repeated
multiplications, and hence make generalisation.

4. Relate the division of numbers in index form with the same base, to repeated
multiplications, and hence make generalisation.

5. Relate the numbers in index form raised to a power, to repeated multiplication, and hence
make generalisation.

6. Verify that a0 = 1 and a–n = a—1n ; a 

7. Determine and state the relationship between fractional indices and roots and powers.

8. Perform operations involving laws of indices.

9. Solve problems involving laws of indices.

28

Chapter 1 Indices

EXPLORING MATHEMATICS

Do you still remember the Pascal’s Triangle that you learnt in the Chapter 1 Patterns and CHAPTER 1
Sequences in Form 2?

The Pascal’s Triangle, invented by a French mathematician, Blaise Pascal, has a lot of unique
properties. Let us explore two unique properties found in the Pascal’s Triangle.

Activity 1 Sum Index form

1 1 20
2 21
11 4 22
121

1331

14641

1 5 10 10 5 1

1 6 15 20 15 6 1

1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 1

Instructions: Sheet 1 Sheet 1(a)

1. Carry out the activity in pairs.

2. Construct the Pascal’s Triangle as in Sheet 1.

3. Calculate the sum of the numbers in each row. Write the sum in index form with base of 2.

4. Complete Sheet 1(a). Discuss the patterns of answers obtained with your friends.

5. Present your results. TIPS

Activity 2 115 = 161 051
11n Value
110 1 1 5 10 10 5 1
111 11
112 121 1 +1 +1—11
113 1 331
114 11 161051
115
116 121
117
118 1331
119
1110 14641

Sheet 2(a) 1 5 10 10 5 1
1 6 15 20 15 6 1

1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1

1 9 36 84 126 126 84 36 9 1

1 10 45 120 210 252 210 120 45 10 1

Sheet 2

Instructions:
1. Carry out the activity in small groups.
2. Construct the Pascal’s Triangle as in Sheet 2.
3. Take note of the numbers in each row. Each number is the value of index with base of 11.
4. Complete Sheet 2(a) with the value of index with base of 11 without using a calculator.
5. Present your results.
6. Are your answers the same as those of other groups?

29