1. COMPUTATION

FRACTIONS

Representing fractions Note that the value of a fraction is not changed if

We can represent a fraction using different forms.

1. As part of a whole, for example, the fraction

3 is really 3 parts of a whole divided into 8

8

equal parts. The top number, 3, is called the

numerator and the bottom number, 8, is

called the denominator.

both the numerator and denominator are multiplied or

divided by the same non-zero number.

2 = 2´4 = 8 8 = 8÷4 = 2

3 3´4 12 12 12 ÷ 4 3

When division is used, this process is called reducing

the fraction to its lowest terms.

2. As part of a set, Mixed numbers and improper fractions

A proper fraction is part of a whole. In this case, the

numerator is of a lesser value than the denominator.

3. As a position on a number line, An improper fraction is more than a whole. The

numerator, in this case, is of a greater value than the

denominator.

4. As a division of two whole numbers, Improper fractions can be written as a combination of

whole numbers and fractions. In this form, they are

called mixed numbers.

3 = 3÷ 4 2 = 2÷7 12 = 12 ÷5 If both the numerator and the denominator are equal,

4 7 5

the fraction is actually a whole or one. For example,

5. As a comparison between two quantities 2 , 3 and 5 are all equivalent to1.

2 3 5

(ratio). For example, if the ratio of girls to

boys in a class is 2:3 (for every 2 girls there

are 3 boys), we may say that the number of Example 1

!"ist !"h ethneunmubmebreorfobf ogyirslso. r

girls is the number Convert 3 5 to an improper fraction.

of boys 6

Equivalent fractions Solution

Two fractions are equivalent if they represent the 3 5 = 3+ 5

6 6

same quantity or value but come from different

= 18 + 5 = 23 3 wholes = 18 sixths

families (fraction families have the same 6 6 6

! '

denominator). For example, the fractions " and (! 5 23

6 6

are equivalent since both represent the same portion So 3 = as an improper fraction

of one whole, although they come from different

families.

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Example 2 Solution

Convert 19 to a mixed number. We must convert both fractions to a common

8

denominator. We chose 14 because it is the lowest

common denominator (LCM) of 7 and 2.

Solution Expressing both fractions in fourteenths, we have:

() = '' + '' + " 2 = 4 1 = 7

' ' 7 14 2 14

a)

So, () = 2 " as a mixed number 2 + 1 2 + 1

' ' 7 2 7 2

= 4 + 7 OR = 4+7

14 14 14

Addition and subtraction of fractions = 11 = 11

14 14

Fractions can only be added or subtracted if their b)

denominators are the same. When this is so, we

simply add or subtract the numerators. 4 - 1 4 - 1

5 4 5 4

Same denominators

= 16 - 5 OR = 16 - 5

20 20 20

a) 4 + 3 = 7 b) 7 − 5 = 2 = 1 = 11 = 11

9 9 9 12 12 12 6 20 20

c) 3 + 4 + 1 = 8 = 153 d ) 11 − 4 = 7 Example 4

5 5 5 5 15 15 15

a) 31 - 1 1 b) 4 1 - 2 1

3 7 4 3

Related denominators

If one denominator is a multiple of the other, then we Solution

must make change one so that both fractions have the

same denominator. a) b)

3 1 -1 1 4 1 - 2 1

3 7 4 3

a) 1 + 1 = 2 + 1 = 3 = 1 since 1 = 1´2 = 2 = 2 7-3 = 3 5 - 2 1

3 6 6 6 6 2 3 3´2 6 21 4 3

= 2 4 = 115 - 4

21 12

b) 11 - 5 = 5 - 5 = 15 - 5 = 10 = 5 since

4 12 4 12 12 12 12 6 = 11121

5 = 15

4 12

Multiplication of fractions

Different denominators Earlier in this chapter, we saw that any fraction of the

If the denominators are unrelated, we must change form - = ÷ . For example, ( = 1 ÷ 2. Hence,

both so that the denominators are the same. . !

(

multiplying by ! is the same as multiplying by 1 and

Example 3 dividing by 2.

a) 2 + 1 b) 4 - 1 3´ 1 = 3´1 = 3 = 1 1

7 2 5 4 2 2 2 2

3 ´ 1 = 3´1 = 3

7 2 7´2 14

2

Example 5

a) 35 b) 54 Mixed operations involving fractions

4´7 9 ´ 11

When we are required to perform computations on

Solution fractions, involving more than one operation, we

simplify operations within the brackets first.

a) 3 5 3´5 15

4 ´ 7 = 4´7 = 28

Example 8

b) 5 4 5´4 20 Calculate the exact value of æ 19 ÷ 6 ö - 1 2 .

9 11 9 ´11 99 èç 5 5 ÷ø 3

´ = =

Example 6 Solution

1 1 3 2 = æ 19 ÷ 6 ö - 1 2

8 3 èç 5 5 ø÷ 3

´

= æ 19 ´ 5 ö -1 2

èç 5 6 ø÷ 3

Solution

Mixed numbers must be expressed as improper = 19 -1 2 = 3 1 -1 2

fractions when computing. 6 3 6 3

1 1 × 3 2 = 9 × 11 = 99 = 4 3 = 4 1 = 2 7 -1 4 = 1 3

8 3 8 3 24 24 8 6 6 6

Division of fractions = 1 1

2

In order to perform the operation of division on Example 9

fractions, we must apply the inverse property, which

Calculate the exact value of 31 + æ 15 ´ 4 ö .

connects multiplication and division. 2 çè 7 5 ÷ø

15 ÷ 3 = 15 × 1 2 ÷ 5 = 2 × 1 = 2 Solution

3 3 3 5 15

3 1 + æ 15 ´ 4 ö = 3 1 + 12 = 3 1 +1 5

2 èç 7 5 ÷ø 2 7 2 7

Using the inverse property, we note that dividing by

any number is equivalent to multiplying by its = 3174 + 11140 = 5 3

multiplicative inverse. When our divisors are 14

fractions, the law still applies.

Example 10

30 ÷ 3 = 30 ´ 5 = 50 The multiplicative inverse 2 2 + 1 3

5 3 3 5

of 5 is 3 , Calculate the exact value of 2 .

3 5 3

75 ÷ 5 = 75 ´ 9 = 135 4

9 5 The multiplicative inverse

59 Solution

of 9 is 5 .

2 2 1 3 2 10 1195

3 + 5 15 +

Example 7 2 = 2

3 3

5 3 1 2 4 4

7 4 5 3

a) ÷ b) 3 ÷1 4

15

4 64 14

15 3

Solution = 2 = ÷

3

53 54 20 4

7÷4 7´3 21

a) = = = 64 ´ 3

15 14

31 2 16 5 16 3 48 23

b) 5 ÷1 3 = 5 ÷ 3 = 5 ´5 = 25 = 1 25 = 32

35

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DECIMALS Solution

Decimals are convenient forms of expressing 61 = 6 + 1 = 0.61 2 = 4 = 0.4

fractions without a numerator or denominator. They 100 10 100 5 10

are also known as base ten fractions, since they have

denominators that are multiples of ten, such as10, 3 = 375 = 0.375

100, 1000 etc. 8 1000

Expanded Notation In the examples above, the decimal equivalent is an

exact value of the fraction. We refer to these as

The place value of decimal numbers follows the same terminating decimals. Not all fractions can be

pattern as whole numbers, decreasing in powers of expressed as terminating decimals.

ten as one moves to the right of the decimal point.

æ 1 ö æ 1 ö æ 1 ö Example 13

çè 10 ø÷ èç 100 ÷ø ç 1000 ÷

1.024 = (1´1) + 0 ´ + 2 ´ + è 4 ´ ø Express as decimals: a) 2 b) 5

3 7

æ 1 ö æ 1 ö

34.56 = ( 3 ´10 ) + (4 ´1) + èç 5 ´ 10 ÷ø + çè 6 ´ 100 ÷ø Solution

Expressing decimals as common fractions a) It is not possible to express the denominator of

these fractions as powers of ten. So, we must use

another strategy.

Since decimals are base ten fractions, we can express Recall, 2 =2÷3. Hence, 2 = 0.66666... = 0.6!

them as common fractions first by writing them in 3 3

expanded notation. We then add the base ten

fractions to obtain a single fraction. A single dot, written over the 6 indicates that the

digit 6 is being repeated indefinitely (ad

Example 11 infinitum).

Express as a fraction in its lowest terms b) We can use the division meaning of fractions to

a) 0.58 b) 0.175 c) 4.08 obtain:

Solution 5 = 5÷7

7

5 8 58 29

a) 0.58 = 10 + 100 = 100 = 50 ••

= 0.714285 714285 714285... = 0.714285

b) 0.175 = 1 + 7 + 5 = 175 = 35 = 7 When more than two digits recur, the dots are

10 100 1000 1000 200 40 placed on the first and last digits in the ‘string’.

c) 4.08 = 4 + 0 + 8 = 4 8 = 4 2 In the above examples, the decimal equivalents are

10 100 100 25 not the exact values of the fractions, although they

are all close to the exact values. We refer to these as

recurring decimals.

Expressing common fractions as decimals Mixed operations involving decimals

Common fractions can have denominators of any In performing operations on decimals where more

value except zero. If their denominators are easily than one operation is involved, we simplify

expressed as a power of ten, then they are in a form operations within the brackets first.

that allows ready conversion to base ten fractions.

Roots and Powers

Example 12

Express as decimals: 61 , 2 , 3 . In performing mixed operations on numbers, we are

sometimes required to evaluate square roots, cube

100 5 8 roots and numbers with powers. These computations

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can be performed with a calculator but it is important Example 16

to interpret the meaning of roots and powers.

Calculate the exact value of

A number raised to the power 2 is the same as 79.38 + 12.96 .

6.3´ 4.2

multiplying the number by itself twice or squaring

the number. For example,

15! = 15 × 15 2.3! = 2.3 × 2.3 Solution

A number raised to the power 3 is the same as 79.38 + 12.96

multiplying the number by itself three times or 6.3´ 4.2

cubing the number. For example,

= 79.38 + 3.6

15" = 15 × 15 × 15 2.3" = 2.3 × 2.3 × 2.3 26.46

= 3 + 3.6

= 6.6

The square root of a number is that number when

multiplied by itself gives the number. For example,

4 × 4 = 16 Hence, √16 = 4 APPROXIMATIONS

1.2 × 1.2 = 1.44 Hence, √1.44 = 1.2

When working with very large and very small

The cube root of a number is that number when numbers, we may wish to round off the number to a

multiplied by itself three times gives the number. For given number of significant figures, decimal places,

example, or write the number using scientific notation

(standard form). Some general rules to follow in

4 × 4 × 4 = 64 Hence, >√64 = 4 performing approximations are:

1.2 × 1.2 × 1.2 = 1.728 Hence, >√1.728 = 1.2

1. Ensure that the place value of the digits in

Note that for cube roots we must insert a 3 at the left the new number remains unchanged.

of the root sign. If we were interested in the fourth 2. Decide which is the target digit – if we are

root we would insert a 4 in the same position. For approximating to the nearest ten, the target

digit is in the tens position.

square roots only, it is not necessary to insert the 2.

3. Round up when the digit on the immediate

Example 14 right of the target digit is 5, 6, 7, 8 or 9.

Calculate the exact value of Round down when it is 0, 1, 2, 3 or 4.

8.64 (3.27-1.12) + (1.2)2

In rounding 5 362 to the nearest hundred, the target

Solution digit is 3. The digit to the immediate right of 3 is 6

and is the deciding digit. Since 6 belongs to the set

8.64 (3.27-1.12) + (1.2)2 {5, 6, 7, 8, 9} we round up to 5 340.

= 8.64 (2.15) + 1.44

= 18.576 + 1.44 Decimal Places

= 20.016

When approximating a number to a given number of

Example 15 decimal places, we use the decimal point as the

Calculate the exact value of starting point and count to the right of the point.

2.34 + 1.08 . 15.6574, correct to three decimal places is 15.657

0.65 0.0756, correct to two decimal places is 0.08

1.2365, correct to one decimal place is 1.2

Solution

Calculate the exact value of

2.34 +1.08 = 3.42 = 5.26

0.65 0.65

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Example 17 Significant figures

Simplify

The first non-zero digit in a number, reading from

(4.14 ÷ 5.75) + (1.62)! left to right is the first significant figure. It is the digit

stating your answer correct to two decimal places. with the largest place value and so gives the best

indication of the size of the number.

Solution

(4.14 ÷ 5.75) + (1.62)! Large Numbers

= 0.72 + 2.6244

= 3.3444 6 843, correct to three significant figures is 6 840.

=3.34 (correct to 2 decimal places) (Round down)

6 843, correct to two significant figures is 6 800.

Standard forms (scientific notation) 6 843, correct to one significant figure is 7 000.

In writing a number in standard form or scientific Small Numbers

notation, we represent it using index notation,

A´10n, where n is an integer and 1 £ A < 10 0.01728, correct to three significant figures is

0.0173.

For example, to write the number 256 in standard 0.01728, correct to two significant figures is 0.017.

form, we place a decimal point so that its value lies 0.01728, correct to one significant figure is 0.02.

between 1 and 10. The value of A will now be 2.56.

However, this is not our original number and we Example 19

must multiply it by 100 or 102 to obtain its original Simplify

value. (2.56 + 0.65) + 0.451!, giving your answer

Hence, 256 = 2.56 ´ 102, which is in standard form. correct to 3 significant figures.

In the examples below, note that when A is larger Solution

than the original number, we must divide by powers (2.56 + 0.65) + 0.451!

of 10 to obtain the original value and so the power of = 3.21 + 0.203401

10 is negative in such cases. = 3.413401

= 3.41 (correct to 3 significant figures)

67 = 6.7 × 10¹

458 = 4.58 × 102

0.00568 = 5.68 × 10-3

0.0124 = 1.24 × 10-2

Example 18

Simplify

(12.8)! − (30 ÷ 0.375)

stating your answer in standard form.

Solution

(12.8)! − (30 ÷ 0.375)

= 163.84 − 80

= 83.84

= 8.38 × 10 (in standard form)

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