Chapter 6 exponents and surds

Exponents and surds
6Contents:
A Exponent or index notation [1.4, 1.9]
B Exponent or index laws [1.9, 2.4]
C Zero and negative indices [1.9, 2.4]
D Standard form [1.9]
E Surds [1.10]
F Properties of surds [1.10]
G Multiplication of surds [1.10]
H Division by surds [1.10]
Opening problem #endboxedheading
Amadeo Avogadro (1776-1856) established that one gram of
hydrogen contains 6:02 £ 1023
atoms.
Things to think about:
² How can we write this number as an ordinary number?
² How many atoms would be in one tonne of hydrogen gas?
² Can you find the mass of 1030
atoms of hydrogen?
We have seen previously that 2 £ 2 £ 2 £ 2 £ 2, can be written as 25
.
25
reads “two to the power of five” or “two with index five”.
In this case 2 is the base and 5 is the exponent, power or index.
We say that 25
is written in exponent or index notation.
EXPONENT OR INDEX NOTATION [1.4, 1.9]A
25
exponent,
orpower
index
base
The use of also called or , allows us to write products of factors and also to
write very large or very small numbers quickly.
exponents powers indices,
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Y:HAESEIGCSE01IG01_06123IGCSE01_06.CDR Friday, 14 November 2008 10:52:42 AM PETER

124 Exponents and surds (Chapter 6)
Example 1 Self Tutor
Find the integer equal to: a 34
b 24
£ 32
£ 7
a 34
= 3 £ 3 £ 3 £ 3
= 9 £ 9
= 81
b 24
£ 32
£ 7
= 2 £ 2 £ 2 £ 2 £ 3 £ 3 £ 7
= 16 £ 9 £ 7
= 1008
Write as a product of prime factors in index form:
a 144 b 4312
a 2 144
2 72
2 36
2 18
3 9
3 3
1
) 144 = 24
£ 32
b 2 4312
2 2156
2 1078
7 539
7 77
11 11
1
) 4312 = 23
£ 72
£ 11
EXERCISE 6A.1
1 Find the integer equal to:
a 23
b 33
c 25
d 53
e 22
£ 33
£ 5 f 23
£ 3 £ 72
g 32
£ 52
£ 13 h 24
£ 52
£ 11
2 By dividing continuously by the primes 2, 3, 5, 7, ..... , write as a product of prime factors in index
form:
a 50 b 98 c 108 d 360
e 1128 f 784 g 952 h 6500
3 The following numbers can be written in the form 2n
. Find n.
a 32 b 256 c 4096
4 The following numbers can be written in the form 3n
. Find n.
a 27 b 729 c 59 049
5 By considering 31
, 32
, 33
, 34
, 35
.... and looking for a pattern, find the last digit of 333
.
6 What is the last digit of 777
?
7 Find n if:
a 54
= n b n3
= 343 c 11n
= 161 051 d (0:6)n
= 0:046 656
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Y:HAESEIGCSE01IG01_06124IGCSE01_06.CDR Wednesday, 17 September 2008 9:33:15 AM PETER

Exponents and surds (Chapter 6) 125
Historical note
#endboxedheading
1 = 13
3 + 5 = 8 = 23
7 + 9 + 11 = 27 = 33
etc.
Nicomachus of Gerasa lived around 100 AD. He discovered
an interesting number pattern involving cubes and sums of odd
numbers:
NEGATIVE BASES
So far we have only considered positive bases raised to a power.
We will now briefly look at negative bases. Consider the statements below:
(¡1)1
= ¡1
(¡1)2
= ¡1 £ ¡1 = 1
(¡1)3
= ¡1 £ ¡1 £ ¡1 = ¡1
(¡1)4
= ¡1 £ ¡1 £ ¡1 £ ¡1 = 1
(¡2)1
= ¡2
(¡2)2
= ¡2 £ ¡2 = 4
(¡2)3
= ¡2 £ ¡2 £ ¡2 = ¡8
(¡2)4
= ¡2 £ ¡2 £ ¡2 £ ¡2 = 16
From the pattern above it can be seen that:
² a negative base raised to an odd power is negative
² a negative base raised to an even power is positive.
Evaluate:
a (¡2)4
b ¡24
c (¡2)5
d ¡(¡2)5
a (¡2)4
= 16
b ¡24
= ¡1 £ 24
= ¡16
c (¡2)5
= ¡32
d ¡(¡2)5
= ¡1 £ (¡2)5
= ¡1 £ ¡32
= 32
CALCULATOR USE
Power keys
x2 squares the number in the display.
^ raises the number in the display to whatever
power is required. On some calculators this
key is yx , ax or xy .
Notice the effect of
the brackets in
these examples.
Not all calculators will
use these key sequences.
If you have problems,
refer to the calculator
instructions on page .12
Different calculators have different keys for entering powers, but in
general they perform raising to powers in a similar manner.
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Y:HAESEIGCSE01IG01_06125IGCSE01_06.CDR Monday, 15 September 2008 2:30:09 PM PETER

Find, using your calculator: a 65
b (¡5)4
c ¡74
Answer
a Press: 6 ^ 5 ENTER 7776
b Press: ( (¡) 5 ) ^ 4 ENTER 625
c Press: (¡) 7 ^ 4 ENTER ¡2401
EXERCISE 6A.2
1 Simplify:
a (¡1)4
b (¡1)5
c (¡1)10
d (¡1)15
e (¡1)8
f ¡18
g ¡(¡1)8
h (¡3)3
i ¡33
j ¡(¡3)3
k ¡(¡6)2
l ¡(¡4)3
2 Simplify:
a 23
£ 32
£ (¡1)5
b (¡1)4
£ 33
£ 22
c (¡2)3
£ (¡3)4
3 Use your calculator to find the value of the following, recording the entire display:
a 28
b (¡5)4
c ¡35
d 74
e 83
f (¡7)6
g ¡76
h 1:0512
i ¡0:62311
j (¡2:11)17
Notice that: ² 23
£ 24
= 2 £ 2 £ 2 £ 2 £ 2 £ 2 £ 2 = 27
²
25
22
=
2 £ 2 £ 2 £ 2 £ 2
2 £ 2
= 23
² (23
)2
= 2 £ 2 £ 2 £ 2 £ 2 £ 2 = 26
² (3 £ 5)2
= 3 £ 5 £ 3 £ 5 = 3 £ 3 £ 5 £ 5 = 32
52
²
µ
2
5
¶3
=
2
5
£
2
5
£
2
5
=
2 £ 2 £ 2
5 £ 5 £ 5
=
23
53
These examples can be generalised to the exponent or index laws:
² am
£ an
= am+n
To multiply numbers with the same base, keep the base
and add the indices.
²
am
an
= am¡n
, a 6= 0 To divide numbers with the same base, keep the base
and subtract the indices.
² (am
)n
= am£n
When raising a power to a power, keep the base and
multiply the indices.
² (ab)n
= an
bn
The power of a product is the product of the powers.
²
³a
b
´n
=
an
bn
, b 6= 0 The power of a quotient is the quotient of the powers.
EXPONENT OR INDEX LAWS [1.9, 2.4]B
1
1
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Y:HAESEIGCSE01IG01_06126IGCSE01_06.CDR Thursday, 18 September 2008 12:14:38 PM PETER

Simplify using the laws of indices:
a 23
£ 22
b x4
£ x5
a 23
£ 22
= 23+2
= 25
= 32
b x4
£ x5
= x4+5
= x9
Simplify using the index laws: a
35
33
b
p7
p3
a
35
33
= 35¡3
= 32
= 9
b
p7
p3
= p7¡3
= p4
Simplify using the index laws:
a (23
)2
b (x4
)5
a (23
)2
= 23£2
= 26
= 64
b (x4
)5
= x4£5
= x20
Remove the brackets of:
a (3a)2
b
µ
2x
y
¶3
a (3a)2
= 32
£ a2
= 9a2
b
µ
2x
y
¶3
=
23
£ x3
y3
=
8x3
y3
To multiply, keep
the base and add
the indices.
To divide, keep the
base and subtract
the indices.
To raise a power to
a power, keep the
base and multiply
the indices.
Each factor within the
brackets has to be
raised to the power
outside them.
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Express the following in simplest form, without brackets:
a (3a3
b)4
b
µ
x2
2y
¶3
a (3a3
b)4
= 34
£ (a3
)4
£ b4
= 81 £ a3£4
£ b4
= 81a12
b4
b
µ
x2
2y
¶3
=
(x2
)3
23 £ y3
=
x2£3
8 £ y3
=
x6
8y3
EXERCISE 6B
1 Simplify using the index laws:
a 23
£ 21
b 22
£ 22
c 35
£ 34
d 52
£ 53
e x2
£ x4
f a3
£ a g n4
£ n6
h b3
£ b5
a
24
23
b
35
32
c
57
53
d
49
45
e
x6
x3
f
y7
y4
g a8
¥ a7
h b9
¥ b5
a (22
)3
b (34
)3
c (23
)6
d (102
)5
e (x3
)2
f (x5
)3
g (a5
)4
h (b6
)4
a a5
£ a2
b n3
£ n5
c a7
¥ a3
d a5
£ a
e b9
¥ b4
f (a3
)6
g an
£ a5
h (b2
)4
i b6
¥ b3
j m4
£ m3
£ m7
k (a3
)3
£ a l (g2
)4
£ g3
5 Remove the brackets of:
a (ab)3
b (ac)4
c (bc)5
d (abc)3
e (2a)4
f (5b)2
g (3n)4
h (2bc)3
i
µ
2
p
¶3
j
³a
b
´3
k
³m
n
´4
l
µ
2c
d
¶5
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6 Express the following in simplest form, without brackets:
a (2b4
)3
b
µ
3
x2y
¶2
c (5a4
b)2
d
µ
m3
2n2
¶4
e
µ
3a3
b5
¶3
f (2m3
n2
)5
g
µ
4a4
b2
¶2
h (5x2
y3
)3
Consider
23
23
which is obviously 1.
Using the exponent law for division,
23
23
= 23¡3
= 20
We therefore conclude that 20
= 1.
In general, we can state the zero index law: a0
= 1 for all a 6= 0.
Now consider
24
27
which is
2 £ 2 £ 2 £ 2
2 £ 2 £ 2 £ 2 £ 2 £ 2 £ 2
=
1
23
Using the exponent law of division,
24
27
= 24¡7
= 2¡3
Consequently, 2¡3
=
1
23
, which means that 2¡3
and 23
are reciprocals of each other.
ZERO AND NEGATIVE INDICES [1.9, 2.4]C
1
1
In general, we can state the negative index law:
If a is any non-zero number and n is an integer, then a¡n
=
1
an
.
This means that an
and a¡n
are reciprocals of one another.
In particular notice that a¡1
=
1
a
.
Using the negative index law,
¡2
3
¢¡4
=
1
¡2
3
¢4
= 1 ¥
24
34
= 1 £
34
24
=
¡3
2
¢4
So, in general we can see that:
³a
b
´¡n
=
µ
b
a
¶n
provided a 6= 0, b 6= 0.
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Y:HAESEIGCSE01IG01_06129IGCSE01_06.CDR Thursday, 18 September 2008 12:18:04 PM PETER

Simplify, giving answers in simplest rational form:
a 70
b 3¡2
c 30
¡ 3¡1
d
¡5
3
¢¡2
a 70
= 1 b 3¡2
=
1
32
= 1
9
c 30
¡ 3¡1
= 1 ¡ 1
3 = 2
3 d
¡5
3
¢¡2
=
¡3
5
¢2
= 9
25
EXERCISE 6C
1 Simplify, giving answers in simplest rational form:
a 30
b 6¡1
c 4¡1
d 50
e 32
f 3¡2
g 53
h 5¡3
i 72
j 7¡2
k 103
l 10¡3
2 Simplify, giving answers in simplest rational form:
a (1
2 )0
b
54
54
c 2t0
d (2t)0
e 70
f 3 £ 40
g
53
55
h
26
210
i (1
4 )¡1
j (3
8 )¡1
k (2
3 )¡1
l (1
5 )¡1
m 20
+ 21
n 50
¡ 5¡1
o 30
+ 31
¡ 3¡1
p (1
3 )¡2
q (2
3 )¡3
r (11
2 )¡3
s (4
5 )¡2
t (21
2 )¡2
3 Write the following without brackets or negative indices:
a (3b)¡1
b 3b¡1
c 7a¡1
d (7a)¡1
e
µ
1
t
¶¡2
f (4t)¡2
g (5t)¡2
h (5t¡2
)¡1
i xy¡1
j (xy)¡1
k xy¡3
l (xy)¡3
m (3pq)¡1
n 3(pq)¡1
o 3pq¡1
p
(xy)3
y¡2
4 Write as powers of 2, 3 or 5:
a 25 b 1
25 c 27 d 1
27
e 16 f 1
16 g 2
3 h 3
5
i 9
125 j 32
81 k 22
5 l 93
8
5 Write as powers of 10:
a 1000 b 0:01 c 100 000 d 0:000 001
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10 000 = 104
1000 = 103
100 = 102
10 = 101
1 = 100
1
10 = 10¡1
1
100 = 10¡2
1
1000 = 10¡3
Consider the pattern alongside. Notice that each time we divide by
10, the exponent or power of 10 decreases by one.
We can use this pattern to simplify the writing of very large and very small numbers.
For example, 5 000 000
= 5 £ 1 000 000
= 5 £ 106
and 0:000 003
=
3
1 000 000
=
3
1
£
1
1 000 000
= 3 £ 10¡6
STANDARD FORM
Standard form (or scientific notation) involves writing any given number as a number between 1
and 10, multiplied by an integer power of 10,
i.e., a £ 10n
where 1 6 a < 10 and n 2 Z :
Write in standard form: a 37 600 b 0:000 86
a 37 600 = 3:76 £ 10 000
= 3:76 £ 104
fshift decimal point 4 places to the left and £10 000g
b 0:000 86 = 8:6 ¥ 104
= 8:6 £ 10¡4
fshift decimal point 4 places to the right and ¥10 000g
Write as an ordinary number:
a 3:2 £ 102
b 5:76 £ 10¡5
a 3:2 £ 102
= 3:20 £ 100
= 320
b 5:76 £ 10¡5
= 000005:76 ¥ 105
= 0:000 057 6
STANDARD FORM [1.9]D
÷10
÷10
÷10
-1
-1
-1
÷10
÷10
÷10
÷10
-1
-1
-1
-1
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y:HAESEIGCSE01IG01_06131IGCSE01_06.CDR Friday, 10 October 2008 9:08:28 AM PETER

Simplify the following, giving your answer in standard form:
a (5 £ 104
) £ (4 £ 105
) b (8 £ 105
) ¥ (2 £ 103
)
a (5 £ 104
) £ (4 £ 105
)
= 5 £ 4 £ 104
£ 105
= 20 £ 104+5
= 2 £ 101
£ 109
= 2 £ 1010
b (8 £ 105
) ¥ (2 £ 103
)
=
8 £ 105
2 £ 103
= 8
2 £ 105¡3
= 4 £ 102
To help write numbers in standard form:
² If the original number is > 10, the power of 10 is positive (+).
² If the original number is < 1, the power of 10 is negative (¡).
² If the original number is between 1 and 10, leave it as it is and multiply it by 100
.
EXERCISE 6D.1
1 Write the following as powers of 10:
a 100 b 1000 c 10 d 100 000
e 0:1 f 0:01 g 0:0001 h 100 000 000
2 Express the following in standard form:
a 387 b 38 700 c 3:87 d 0:0387
e 0:003 87 f 20:5 g 205 h 0:205
i 20 500 j 20 500 000 k 0:000 205
3 Express the following in standard form:
a The circumference of the Earth is approximately 40 075 kilometres.
b The distance from the Earth to the Sun is 149 500 000 000 m.
c Bacteria are single cell organisms, some of which have a diameter of 0:0004 mm.
d There are typically 40 million bacteria in a gram of soil.
e The probability that your six numbers will be selected
for Lotto on Saturday night is 0:000 000 141 62.
f Superfine sheep have wool fibres as low as 0:01 mm in
diameter.
4 Write as an ordinary decimal number:
a 3 £ 102
b 2 £ 103
c 3:6 £ 104
d 9:2 £ 105
e 5:6 £ 106
f 3:4 £ 101
g 7:85 £ 106
h 9 £ 108
a 3 £ 10¡2
b 2 £ 10¡3
c 4:7 £ 10¡4
d 6:3 £ 10¡5
e 1:7 £ 100
f 9:5 £ 10¡4
g 3:49 £ 10¡1
h 7 £ 10¡6
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Scientific and graphics calculators are able to display very large and very small numbers in standard form.
Likewise, if you perform 0:0024 ¥ 10 000 000 your calculator might display or ,
which actually represent 2:4 £ 10¡10
.
If you perform your calculator might display or or ,
all of which actually represent .
2300000 400000
9 2 10
¡ ¡ ¡ ¡ ¡
¡ ¡
£
£: 11
9.2 11¡ ¡E9.211
2.4-10
9.2 +11¡E¡
2.4 -10¡ ¡E
a The wavelength of visible light is 9 £ 10¡7
m.
b In 2007, the world population was approximately 6:606 £ 109
.
c The diameter of our galaxy, the Milky Way, is 1 £ 105
light years.
d The smallest viruses are 1 £ 10¡5
mm in size.
e 1 atomic mass unit is approximately 1:66 £ 10¡27
kg.
7 Write in standard form:
a 18:17 £ 106
b 0:934 £ 1011
c 0:041 £ 10¡2
8 Simplify the following, giving your answer in standard form:
a (8 £ 103
) £ (2 £ 104
) b (8 £ 103
) £ (4 £ 105
)
c (5 £ 104
) £ (3 £ 105
) d (2 £ 103
)3
e (6 £ 103
)2
f (7 £ 10¡2
)2
g (9 £ 104
) ¥ (3 £ 103
) h (8 £ 105
) ¥ (4 £ 106
)
STANDARD FORM ON A CALCULATOR
You will find instructions for graphics calculators on page 16.
EXERCISE 6D.2
1 Write each of the following as it would appear on the display of your calculator:
a 4 650 000 b 0:000 051 2 c 5:99 £ 10¡4
d 3:761 £ 1010
e 49 500 000 f 0:000 008 44
2 Calculate each of the following, giving your answers in standard form. The decimal part should be
correct to 2 decimal places:
a 0:06 £ 0:002 ¥ 4000 b 426 £ 760 £ 42 000 c 627 000 £ 74 000
d 320 £ 600 £ 51 400 e 0:004 28 ¥ 120 000 f 0:026 £ 0:00 42 £ 0:08
Use your calculator to find:
a (1:42 £ 104
) £ (2:56 £ 108
) b (4:75 £ 10¡4
) ¥ (2:5 £ 107
)
Instructions are given for the Casio fx-6890G: Answer:
a 1:42 EXP 4 £ 2:56 EXP 8 EXE 3:6352 £ 1012
b 4:75 EXP (¡) 4 ¥ 2:5 EXP 7 EXE 1:9 £ 10¡11
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3 Find, in standard form, with decimal part correct to 2 places:
a (5:31 £ 104
) £ (4:8 £ 103
) b (2:75 £ 10¡3
)2
c
8:24 £ 10¡6
3 £ 104
d (7:2 £ 10¡5
) ¥ (2:4 £ 10¡6
) e
1
4:1 £ 104
f (3:2 £ 103
)2
4 For the following give answers in standard form correct to 3 significant figures:
a How many millimetres are there in 479:8 kilometres?
b How many seconds are there in one year?
c How many seconds are there in a millennium?
d How many kilograms are there in 0:5 milligrams?
5 If a missile travels at 3600 km/h, how far will it travel in:
a 1 day b 1 week c 2 years?
6 Light travels at a speed of 3 £ 108
metres per second. How far will light travel in:
a 1 minute b 1 day c 1 year?
Give your answers in standard form with decimal part correct to 2 decimal places.
Assume that 1 year = 365 days.
For the remainder of this chapter we consider surds and radicals, which are numbers that are written using
the radical or square root sign
p
:
Surds and radicals occur frequently in mathematics, often as solutions to equations involving squared terms.
We will see a typical example of this in Chapter 8 when we study Pythagoras’ theorem.
RATIONAL AND IRRATIONAL RADICALS
Some radicals are rational, but most are irrational.
For example, some rational radicals include:
p
1 =
p
12 = 1 or 1
1p
4 =
p
22 = 2 or 2
1
q
1
4 =
q¡1
2
¢2
= 1
2
Two examples of irrational radicals are
p
2 ¼ 1:414 214
and
p
3 ¼ 1:732 051:
Strictly speaking, a surd is an irrational radical. However, in this and many other courses, the term surd
is used to describe any radical. It is reasonable to do so because the properties of surds and radicals are the
same.
SURDS [1.10]E
Give your answers in standard form with decimal part
correct to places. Assume that year days.2 1 = 365
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Historical note #endboxedheading
The name surd and the radical sign p both had a rather absurd past. Many
centuries after Pythagoras, when the Golden Age of the Greeks was past,
the writings of the Greeks were preserved, translated, and extended by Arab
mathematicians.
The Arabs thought of a square number as growing out of its roots. The roots
had to be extracted. The Latin word for “root” is radix, from which we get
the words radical and radish! The printed symbol for radix was first R, then r,
which was copied by hand as
p
.
The word surd actually came about because of an error of translation by the
Arab mathematician Al-Khwarizmi in the 9th century AD. The Greek word
a-logos means “irrational” but also means “deaf”. So, the Greek a-logos was
interpreted as “deaf” which in Latin is surdus. Hence to this day, irrational
radicals like
p
2 are called surds.
BASIC OPERATIONS WITH SURDS
We have seen square roots and cube roots in previous courses. We can use their properties to help with
some simplifications.
Simplify:
a (
p
5)2
b
µ
1
p
5
¶2
a (
p
5)2
=
p
5 £
p
5
= 5
b
µ
1
p
5
¶2
=
1
p
5
£
1
p
5
= 1
5
Simplify:
a (2
p
5)3
b ¡2
p
5 £ 3
p
5
a (2
p
5)3
= 2
p
5 £ 2
p
5 £ 2
p
5
= 2 £ 2 £ 2 £
p
5 £
p
5 £
p
5
= 8 £ 5 £
p
5
= 40
p
5
b ¡2
p
5 £ 3
p
5
= ¡2 £ 3 £
p
5 £
p
5
= ¡6 £ 5
= ¡30
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ADDING AND SUBTRACTING SURDS
‘Like surds’ can be added and subtracted in the same way as ‘like terms’ in algebra.
Consider 2
p
3 + 4
p
3, which has the same form as 2x + 4x.
If we interpret this as 2 ‘lots’ of
p
3 plus 4 ‘lots’ of
p
3, we have 6 ‘lots’ of
p
3.
So, 2
p
3 + 4
p
3 = 6
p
3, and we can compare this with 2x + 4x = 6x.
Simplify:
a 3
p
2 + 4
p
2 b 5
p
3 ¡ 6
p
3
a 3
p
2 + 4
p
2
= 7
p
2
fCompare: 3x + 4x = 7xg
b 5
p
3 ¡ 6
p
3
= ¡1
p
3
= ¡
p
3
fCompare: 5x ¡ 6x = ¡xg
EXERCISE 6E
1 Simplify:
a (
p
7)2
b (
p
13)2
c (
p
15)2
d (
p
24)2
e
µ
1
p
3
¶2
f
µ
1
p
11
¶2
g
µ
1
p
17
¶2
h
µ
1
p
23
¶2
2 Simplify:
a ( 3
p
2)3
b ( 3
p
¡5)3
c
³
1
3p
5
´3
3 Simplify:
a 3
p
2 £ 4
p
2 b ¡2
p
3 £ 5
p
3 c 3
p
5 £ (¡2
p
5)
d ¡2
p
2 £ (¡3
p
2) e (3
p
2)2
f (3
p
2)3
g (2
p
3)2
h (2
p
3)3
i (2
p
2)4
4 Simplify:
a
p
2 +
p
2 b
p
2 ¡
p
2 c 3
p
2 ¡ 2
p
2
d 2
p
3 ¡
p
3 e 5
p
7 + 2
p
7 f 3
p
5 ¡ 6
p
5
g 3
p
2 + 4
p
2 ¡
p
2 h 6
p
2 ¡ 9
p
2 i
p
5 + 7
p
5
j 3
p
2 ¡ 5
p
2 ¡
p
2 k 3
p
3 ¡
p
3 + 2
p
3 l 3
p
5 + 7
p
5 ¡ 10
5 Simplify:
a 3
p
2 + 2
p
3 ¡
p
2 + 5
p
3 b 7
p
2 ¡ 4
p
3 ¡ 2
p
2 + 3
p
3
c ¡6
p
2 ¡ 2
p
3 ¡
p
2 + 6
p
3 d 2
p
5 + 4
p
2 + 9
p
5 ¡ 9
p
2
e 3
p
2 ¡ 5
p
7 ¡
p
2 ¡ 5
p
7 f 3
p
2 + 4
p
11 + 6 ¡
p
2 ¡
p
11 ¡ 3
g 6
p
6 ¡ 2
p
2 ¡
p
2 ¡ 5
p
6 + 4 h 5
p
3 ¡ 6
p
7 ¡ 5 + 4
p
3 +
p
7 ¡ 8
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Discovery Properties of surds
#endboxedheading
Notice that
p
4 £ 9 =
p
36 = 6 and
p
4 £
p
9 = 2 £ 3 = 6, which suggests that
p
4£
p
9 =
p
4 £ 9:
Also,
r
36
4
=
p
9 = 3 and
p
36
p
4
=
6
2
= 3, which suggests that
p
36
p
4
=
r
36
4
.
What to do:
Test the following possible properties or rules for surds by substituting different values of a and b. Use
your calculator to evaluate the results.
1
p
a £
p
b =
p
ab for all a > 0, b > 0.
2
qa
b
=
p
a
p
b
for all a > 0, b > 0.
3
p
a + b =
p
a +
p
b for all a > 0, b > 0.
4
p
a ¡ b =
p
a ¡
p
b for all a > 0, b > 0.
You should have discovered the following properties of surds:
²
p
a £
p
b =
p
a £ b for a > 0, b > 0
²
p
a
p
b
=
r
a
b
for a > 0, b > 0
However, in general it is not true that
p
a + b =
p
a +
p
b or that
p
a ¡ b =
p
a ¡
p
b:
Write in simplest form:
a
p
3 £
p
2 b 2
p
5 £ 3
p
2
a
p
3 £
p
2
=
p
3 £ 2
=
p
6
b 2
p
5 £ 3
p
2
= 2 £ 3 £
p
5 £
p
2
= 6 £
p
5 £ 2
= 6
p
10
PROPERTIES OF SURDS [1.10]F
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Simplify: a
p
32
p
2
b
p
12
2
p
3
a
p
32
p
2
=
q
32
2
=
p
16
= 4
b
p
12
2
p
3
= 1
2
q
12
3 fusing
p
a
p
b
=
r
a
b
g
= 1
2
p
4
= 1
2 £ 2
= 1
Write
p
32 in the form k
p
2.
p
32
=
p
16 £ 2
=
p
16 £
p
2 fusing
p
ab =
p
a £
p
bg
= 4
p
2
SIMPLEST SURD FORM
A surd is in simplest form when the number under the radical sign is the smallest
integer possible.
Write
p
28 in simplest surd form.
p
28
=
p
4 £ 7 f4 is the largest perfect square factor of 28g
=
p
4 £
p
7
= 2
p
7
EXERCISE 6F
1 Simplify:
a
p
2 £
p
5 b
p
3 £
p
7 c
p
3 £
p
11
d
p
7 £
p
7 e
p
3 £ 2
p
3 f 2
p
2 £
p
5
Look for the
largest perfect
square factor.
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2 Simplify:
a 3
p
3 £ 2
p
2 b 2
p
3 £ 3
p
5 c
p
2 £
p
3 £
p
5
d
p
3 £
p
2 £ 2
p
2 e ¡3
p
2 £ (
p
2)3
f (3
p
2)3
£ (
p
3)3
3 Simplify:
a
p
8
p
2
b
p
2
p
8
c
p
18
p
2
d
p
2
p
18
e
p
20
p
5
f
p
5
p
20
g
p
27
p
3
h
p
18
p
3
i
p
3
p
30
j
p
50
p
2
k
2
p
6
p
24
l
5
p
75
p
3
4 Write the following in the form k
p
2:
a
p
8 b
p
18 c
p
50 d
p
98
e
p
200 f
p
288 g
p
20 000 h
q
1
2
p
3:
a
p
12 b
p
27 c
p
75 d
q
1
3
p
5:
a
p
20 b
p
45 c
p
125 d
q
1
5
7 a Find:
i
p
16 +
p
9 ii
p
16 + 9 iii
p
25 ¡
p
9 iv
p
25 ¡ 9
b Copy and complete: In general,
p
a + b 6= :::::: and
p
a ¡ b 6= ::::::
8 Write the following in simplest surd form:
a
p
24 b
p
50 c
p
54 d
p
40 e
p
56 f
p
63
g
p
52 h
p
44 i
p
60 j
p
90 k
p
96 l
p
68
m
p
175 n
p
162 o
p
128 p
p
700
9 Write the following in simplest surd form:
a
q
5
9 b
q
18
4 c
q
12
16 d
q
75
36
The rules for expanding brackets involving surds are identical to those for ordinary algebra.
We can thus use:
a(b + c) = ab + ac
(a + b)(c + d) = ac + ad + bc + bd
(a + b)2
= a2
+ 2ab + b2
(a ¡ b)2
= a2
¡ 2ab + b2
(a + b)(a ¡ b) = a2
¡ b2
MULTIPLICATION OF SURDS [1.10]G
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Expand and simplify:
a
p
2(
p
2 +
p
3) b
p
3(6 ¡ 2
p
3)
a
p
2(
p
2 +
p
3)
=
p
2 £
p
2 +
p
2 £
p
3
= 2 +
p
6
b
p
3(6 ¡ 2
p
3)
= (
p
3)(6 + ¡2
p
3)
= (
p
3)(6) + (
p
3)(¡2
p
3)
= 6
p
3 + ¡6
= 6
p
3 ¡ 6
a ¡
p
2(
p
2 + 3) b ¡
p
3(7 ¡ 2
p
3)
a ¡
p
2(
p
2 + 3)
= ¡
p
2 £
p
2 + ¡
p
2 £ 3
= ¡2 ¡ 3
p
2
b ¡
p
3(7 ¡ 2
p
3)
= (¡
p
3)(7 ¡ 2
p
3)
= (¡
p
3)(7) + (¡
p
3)(¡2
p
3)
= ¡7
p
3 + 6
Expand and simplify: (3 ¡
p
2)(4 + 2
p
2)
(3 ¡
p
2)(4 + 2
p
2)
= (3 ¡
p
2)(4) + (3 ¡
p
2)(2
p
2)
= 12 ¡ 4
p
2 + 6
p
2 ¡ 4
= 8 + 2
p
2
a (
p
3 + 2)2
b (
p
3 ¡
p
7)2
a (
p
3 + 2)2
= (
p
3)2
+ 2 £
p
3 £ 2 + 22
= 3 + 4
p
3 + 4
= 7 + 4
p
3
b (
p
3 ¡
p
7)2
= (
p
3)2
¡ 2 £
p
3 £
p
7 + (
p
7)2
= 3 ¡ 2
p
21 + 7
= 10 ¡ 2
p
21
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a (3 +
p
2)(3 ¡
p
2) b (2
p
3 ¡ 5)(2
p
3 + 5)
a (3 +
p
2)(3 ¡
p
2)
= 32
¡ (
p
2)2
= 9 ¡ 2
= 7
b (2
p
3 ¡ 5)(2
p
3 + 5)
= (2
p
3)2
¡ 52
= (4 £ 3) ¡ 25
= 12 ¡ 25
= ¡13
EXERCISE 6G
1 Expand and simplify:
a
p
2(
p
5 +
p
2) b
p
2(3 ¡
p
2) c
p
3(
p
3 + 1) d
p
3(1 ¡
p
3)
e
p
7(7 ¡
p
7) f
p
5(2 ¡
p
5) g
p
11(2
p
11 ¡ 1) h
p
6(1 ¡ 2
p
6)
i
p
3(
p
3 +
p
2 ¡ 1) j 2
p
3(
p
3 ¡
p
5) k 2
p
5(3 ¡
p
5) l 3
p
5(2
p
5 +
p
2)
a ¡
p
2(3 ¡
p
2) b ¡
p
2(
p
2 +
p
3) c ¡
p
2(4 ¡
p
2) d ¡
p
3(1 +
p
3)
e ¡
p
3(
p
3 + 2) f ¡
p
5(2 +
p
5) g ¡(
p
2 + 3) h ¡
p
5(
p
5 ¡ 4)
i ¡(3 ¡
p
7) j ¡
p
11(2 ¡
p
11) k ¡(
p
3 ¡
p
7) l ¡2
p
2(1 ¡
p
2)
m ¡3
p
3(5 ¡
p
3) n ¡7
p
2(
p
2 +
p
6) o (¡
p
2)3
(3 ¡
p
2)
a (1 +
p
2)(2 +
p
2) b (2 +
p
3)(2 +
p
3) c (
p
3 + 2)(
p
3 ¡ 1)
d (4 ¡
p
2)(3 +
p
2) e (1 +
p
3)(1 ¡
p
3) f (5 +
p
7)(2 ¡
p
7)
g (
p
5 + 2)(
p
5 ¡ 3) h (
p
7 ¡
p
3)(
p
7 +
p
3) i (2
p
2 +
p
3)(2
p
2 ¡
p
3)
j (4 ¡
p
2)(3 ¡
p
2)
a (1 +
p
2)2
b (2 ¡
p
3)2
c (
p
3 + 2)2
d (1 +
p
5)2
e (
p
2 ¡
p
3)2
f (5 ¡
p
2)2
g (
p
2 +
p
7)2
h (4 ¡
p
6)2
i (
p
6 ¡
p
2)2
j (
p
5 + 2
p
2)2
k (
p
5 ¡ 2
p
2)2
l (6 +
p
8)2
m (5
p
2 ¡ 1)2
n (3 ¡ 2
p
2)2
o (1 + 3
p
2)2
a (4 +
p
3)(4 ¡
p
3) b (5 ¡
p
2)(5 +
p
2) c (
p
5 ¡ 2)(
p
5 + 2)
d (
p
7 + 4)(
p
7 ¡ 4) e (3
p
2 + 2)(3
p
2 ¡ 2) f (2
p
5 ¡ 1)(2
p
5 + 1)
g (5 ¡ 3
p
3)(5 + 3
p
3) h (2 ¡ 4
p
2)(2 + 4
p
2) i (1 + 5
p
7)(1 ¡ 5
p
7)
a (
p
3 +
p
2)(
p
3 ¡
p
2) b (
p
7 +
p
11)(
p
7 ¡
p
11) c (
p
x ¡
p
y)(
p
y +
p
x)
Did you notice that
these answers are
?integers
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When an expression involves division by a surd, we can write the expression with an integer denominator
which does not contain surds.
If the denominator contains a simple surd such as
p
a then we use the rule
p
a £
p
a = a:
For example:
6
p
3
can be written as
6
p
3
£
p
3
p
3
since we are really just multiplying
the original fraction by 1.
6
p
3
£
p
3
p
3
then simplifies to
6
p
3
3
or 2
p
3.
Express with integer denominator: a
7
p
3
b
10
p
5
c
10
2
p
2
a
7
p
3
=
7
p
3
£
p
3
p
3
=
7
p
3
3
b
10
p
5
=
10
p
5
£
p
5
p
5
= 10
5
p
5
= 2
p
5
c
10
2
p
2
=
10
2
p
2
£
p
2
p
2
=
10
p
2
4
=
5
p
2
2
Express
1
3 +
p
2
with integer denominator.
1
3 +
p
2
=
µ
1
3 +
p
2
¶ µ
3 ¡
p
2
3 ¡
p
2
¶
=
3 ¡
p
2
32 ¡ (
p
2)2
fusing (a + b)(a ¡ b) = a2
¡ b2
g
=
3 ¡
p
2
7
DIVISION BY SURDS [1.10]H
We are really
multiplying by one,
which does not change
the value of the
original expression.
If the denominator has the form a +
p
b then we can remove the surd from the denominator by multiplying
both the numerator and the denominator by its radical conjugate a ¡
p
b. This produces a rational
denominator, so the process is called rationalisation of the denominator.
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Write
1 ¡ 2
p
3
1 +
p
3
in simplest form.
1 ¡ 2
p
3
1 +
p
3
=
µ
1 ¡ 2
p
3
1 +
p
3
¶ µ
1 ¡
p
3
1 ¡
p
3
¶
=
1 ¡
p
3 ¡ 2
p
3 + 6
1 ¡ 3
=
7 ¡ 3
p
3
¡2
=
3
p
3 ¡ 7
2
EXERCISE 6H
1 Express with integer denominator:
a
1
p
2
b
2
p
2
c
4
p
2
d
10
p
2
e
p
7
p
2
f
1
p
3
g
3
p
3
h
4
p
3
i
18
p
3
j
p
11
p
3
k
1
p
5
l
3
p
5
m
p
3
p
5
n
15
p
5
o
125
p
5
p
p
10
p
2
q
1
2
p
3
r
2
p
2
p
3
s
15
2
p
5
t
1
(
p
2)3
2 Rationalise the denominator:
a
1
3 ¡
p
5
b
1
2 +
p
3
c
1
4 ¡
p
11
d
p
2
5 +
p
2
e
p
3
3 +
p
3
f
5
2 ¡ 3
p
2
g
¡
p
5
3 + 2
p
5
h
3 ¡
p
7
2 +
p
7
3 Write in the form a + b
p
2 where a, b 2 Q :
a
4
2 ¡
p
2
b
¡5
1 +
p
2
c
1 ¡
p
2
1 +
p
2
d
p
2 ¡ 2
3 ¡
p
2
e
1p
2
1 ¡ 1p
2
f
1 + 1p
2
1 ¡ 1p
2
g
1
1 ¡
p
2
3
h
p
2
2 + 1
1 ¡
p
2
4
Review set 6A #endboxedheading
1 Find the integer equal to: a 34
b 5 £ 23
2 Write as a product of primes in index form: a 36 b 242
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95
100
100

3 Simplify:
a (¡2)3
b ¡(¡1)8
c (¡1)3
£ (¡2)2
£ ¡33
4 Simplify, giving your answers in simplest rational form:
a 3¡3
b
¡4
3
¢¡2
c 30
¡ 31
5 Write 1
16 as a power of 2.
6 Simplify, using the exponent laws:
a 56
£ 5 b b7
¥ b2
c (x4
)3
7 Express in simplest form, without brackets:
a (4c3
)2
b (2a2
b)3
c
³ s
3t2
´4
a 9 b 34 900 c 0:0075
a 2:81 £ 106
b 2:81 £ 100
c 2:81 £ 10¡3
10 Simplify, giving your answer in standard form:
a (6 £ 103
) £ (7:1 £ 104
) b (2:4 £ 106
) ¥ (4 £ 102
)
11 The Earth orbits around the Sun at a speed of approximately 1:07 £ 105
km/h. How far does the
Earth move, relative to the Sun, in:
a 1 day b 1 week c 1 year?
Give your answers in standard form with decimal part correct to 2 decimal places. Assume that
1 year = 365 days.
12 a Simplify (3
p
2)2
: b Simplify ¡2
p
3 £ 4
p
3:
c Simplify 3
p
2 ¡
p
8: d Write
p
48 in simplest surd form.
a 2
p
3(4 ¡
p
3) b (3 ¡
p
7)2
c (2 ¡
p
3)(2 +
p
3)
d (3 + 2
p
5)(2 ¡
p
5) e (4 ¡
p
2)(3 + 2
p
2)
14 Rationalise the denominator:
a
8
p
2
b
15
p
3
c
p
3
p
2
d
5
6 ¡
p
3
15 Write
q
1
7 in the form k
p
7.
p
3 where a, b 2 Q :
a
p
3
2 + 1
1 ¡
p
3
2
b
2 +
p
3
2 ¡
p
3
¡
2
p
3
2 +
p
3
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Review set 6B #endboxedheading
1 Find the integer equal to: a 73
b 32
£ 52
2 Write as a product of primes in index form: a 42 b 144
3 Simplify:
a ¡(¡1)7
b ¡43
c (¡2)5
£ (¡3)2
4 Simplify, giving your answers in simplest rational form:
a 6¡2
b
¡
11
2
¢¡1
c
¡3
5
¢¡2
5 Simplify, using the exponent laws:
a 32
£ 36
b a5
¥ a5
c (y3
)5
6 Write as powers of 2, 3 or 5:
a 16
25 b 40
81 c 180 d 111
9
7 Express in simplest form, without brackets or negative indices:
a (5c)¡1
b 7k¡2
c (4d2
)¡3
a 263:57 b 0:000 511 c 863 400 000
a 2:78 £ 100
b 3:99 £ 107
c 2:081 £ 10¡3
10 Simplify, giving your answer in standard form:
a (8 £ 103
)2
b (3:6 £ 105
) ¥ (6 £ 10¡2
)
11 How many kilometres are there in 0:21 millimetres? Give your answer in standard form.
12 Simplify:
a 2
p
3 £ 3
p
5 b (2
p
5)3
c 5
p
2 ¡ 7
p
2
d ¡
p
2(2 ¡
p
2) e (
p
3)4
f
p
3 £
p
5 £
p
15
13 Write in simplest surd form: a
p
75 b
q
20
9
a (5 ¡
p
3)(5 +
p
3) b ¡(2 ¡
p
5)2
c 2
p
3(
p
3 ¡ 1) ¡ 2
p
3 d (2
p
2 ¡ 5)(1 ¡
p
2)
15 Express with integer denominator:
a
14
p
2
b
p
2
p
3
c
p
2
3 +
p
2
d
¡5
4 ¡
p
3
p
5 where a, b 2 Q :
a
1 +
p
5
2 ¡
p
5
b
3 ¡
p
5
3 +
p
5
¡
4
3 ¡
p
5
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Y:HAESEIGCSE01IG01_06145IGCSE01_06.CDR Wednesday, 5 November 2008 2:15:20 PM PETER

Challenge #endboxedheading
Discovery Continued square roots#endboxedheading
X =
s
2 +
r
2 +
q
2 +
p
2 +
p
2 + :::::: is an example of a
continued square root.
Some continued square roots have actual values which are integers.
1 Use your calculator to show that
p
2 ¼ 1:41421
p
2 +
p
2 ¼ 1:84776
q
2 +
p
2 +
p
2 ¼ 1:96157:
2 Find the values, correct to 6 decimal places, of:
a
r
2 +
q
2 +
p
2 +
p
2 b
s
2 +
r
2 +
q
2 +
p
2 +
p
2
3 Continue the process and hence predict the actual value of X.
4 Use algebra to find the exact value of X.
Hint: Find X2
in terms of X, and solve by inspection.
5 Can you find a continued square root whose actual value is 3?
1 Find
s
3 + 2
p
2
3 ¡ 2
p
2
giving your answer in the form a + b
p
2 where a, b 2 Q .
2 If x =
p
5 ¡
p
3, find x2
and x4
. Hence find the value of x4
¡ 16x2
.
Copy and complete: x =
p
5 ¡
p
3 is one of the solutions of the equation x4
¡ 16x2
= 0
3 a We know that in general,
p
a + b 6=
p
a +
p
b
Deduce that if
p
a + b =
p
a +
p
b then at least one of a or b is 0.
b What can be deduced about a and b if
p
a ¡ b =
p
a ¡
p
b?
4 a Find the value of
µ
1 +
p
5
2
¶n
¡
µ
1 ¡
p
5
2
¶n
for n = 1, 2, 3 and 4.
b What do you suspect about
µ
1 +
p
5
2
¶n
¡
µ
1 ¡
p
5
2
¶n
for all n 2 Z +
?
What to do:
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Y:HAESEIGCSE01IG01_06146IGCSE01_06.cdr Friday, 31 October 2008 9:51:20 AM PETER

Chapter 6 exponents and surds

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