Lesson Notes / Scheme of work

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MATHEMATICS SS1 FIRST TERM LESSON PLAN SCHEME OF WORK

INDICES

CONTENT

1. Revision of Standard Form

2. Laws Of Indices and Application of Indices

3. Simple Indicial Equations

Revision of Standard Form

Example 1: Express the following in standard form.

(a) 5.37 (b)  53.7 (c) 537 (d) 35.65  (e) 7500  (f) 1403420

Solution:

(a) 5.37 = 5.37 × 1

= 5.37 × 10⁰

(b) 53.7 = 5.37 × 10¹

(c) 537 = 5.37 × 100

= 5.37 × 10 × 10

= 5.37 × 10²

(d) 35.65 = 3.565 × 10

=  3.565 × 10¹

(e) 7500 = 7.5 × 1000

= 7.5 × 10³

(f) 1403420 = 1.403420 × 1000000

=  1.403420 × 10⁶

Example 2:  Express the following in standard form.

(a) 0.037 (b)  0.00065   (c)  0.0058   (d)  0.61

Method 1:

(a) 0.037 = 3.7 × 0.01

= 3.7 × 10⁻²

(b) 0.00065 = 6.5 × 0.0001

= 6.5 × 10⁻⁴

(c) 0.0058 = 5.8 × 0.001

= 5.8 × 10⁻³

(d) 0.61 = 6.1 × 0.1

= 6.1 × 10⁻¹

Method 2:

(a) 0.037=0.0371=3.7100=3.7102=3.7×10−2

(b) 0.00065=0.000651=6.510000=6.5104=6.5×10−4

(c) 0.0058=0.00581=5.81000=5.8103=5.8×10−3

(d) 0.061=0.611=6.110=6.1×10−1

Class Activity:

Express the following in standard form.

1. (a) 2037000 (b) 00469 (c) 0.000513 (d) 146200

Write each of the following numbers in full

2. (a) 6 × 10⁴ (b) 5.142 × 10⁷  (c) 0.7883 × 10^{– 5} (d)  2.095 × 10⁹

Laws of Indices and Application of Indices

The following are the laws governing the mathematical operations involving index numbers. These laws are true for all values of m, n and x ≠ 0.

(1) ax×ay=ax+y

(2) ax÷ay=ax−y

(3) a−x=1ax

(4) a0=1

(5) (ax)y=axy

(6) a1n=a−−√n

(7) axy=ax−−√y

Example 1: Simplify (181)−14

Solution:

(811)14(34)14344=3

Example 2: Simplify 12523

Solution:

(125−−−√3)2(53−−√3)252=25

Class Activity:

Simplify the following:

1. (278)−23

2. 3a2b×(2ab)−3

Simple Indicial Equations

Indicial equations are equations which have the variable or unknown quantity as an index or exponent.

Examples: Solve for x in the following:

1. 4−3x=164

2. (x+7)3=27

Solution:

1. 4−3x=1434−3x=4−3−3x=−3∴x=1

2. (x+7)3=27(x+7)3=33x+7=3x=3−7x=−4

Practice Exercises

1. Write 3.654×10−10 in full

2. Simplify 32×1048×107

3. Simplify 196m3n74m5n−−−−−−√

4. Simplify 2−n×82n+1×162n43n

5. Solve the equation, 9x=13(27x)

6. Solve the equation 3x−4=2342

ASSIGNMENT

1. Given that: 2×41−x=8−x, find x

2. Solve for the value of n in the equation, 9n×13n−1=27n+3

3. Without using calculator, work out 0.0125×0.005760.0015×0.32 leaving your answer in standard form.

4. Solve for xif 4x×82x−1=1161−x

5. Given that m8×n3m4×n5=mx×ny, calculate the value of 2x−y

6. If 8x2=238×434, find x WAEC SSCE

Glossary

Number, Index, Power, Roots, Base etc.

Logarithms I

LOGARITHMS

CONTENT

1. Deducing logarithm from indices and standard form

2. Definition of Logarithms

3. Definition of Antilogarithms

4. The graph of y = 10^x

5. Reading logarithm and Antilogarithm tables

Deducing Logarithm From Indices and Standard Form

There is a close link between indices and logarithms

100 = 10². This can be written in logarithmic notation as log₁₀100 = 2.

Similarly 8 = 2³ and it can be written as log₂8 = 3.

In general, N = b^x in logarithmic notation is Log_bN = x.

We say the logarithms of N in base b is x. When the base is ten, the logarithms is known as common logarithms.

The logarithms of a number N in base b is the power to which b must be raised to get N.

Re-write using logarithmic notation (i) 1000 = 10³ (ii) 0.01 = 10⁻² (iii) 2⁴ = 16 (iv) 1/8 = 2⁻³

Change the following to index form

(i) Log416=2

(ii) Log3(127)=−3

The logarithm of a number has two parts and integer (whole number) then the decimal point. The integral part is called the characteristics and the decimal part is called mantissa.

To find the logarithms of 27.5 from the table, express the number in the standard form as 27.5 = 2.75 × 10¹. The power of ten in this standard form is the characteristics of Log 27.5. The decimal part is called mantissa.

Remember a number is in the standard form if written as A × 10ⁿ where A is a number such that 1 ≤ A < 10 and n is an integer.

27.5 = 2.75 × 10¹, 27.5 when written in the standard form, the power of ten is 1. Hence the characteristic of Log 27.5 is 1. The mantissa can be read from a 4-figure table. The 4-figure table is at the back of your New General Mathematics textbook.

Below is a row from the 4-figure table

Differences

	X	0	1	2	3	4	5	6	7	8	9	1	2	3	4	5	6	7	8	9
	2	431	433	434	436	437	439	440	442	444	445	1	3	4	6	7	9	1	1	1
	7	4	0	6	2	8	3	9	5	0	6							1	2	4

To check for Log27.5, look for the first two digits i.e. 27 in the first column.

Now look across that row of 27 and stop at the column with 5 at the top. This gives the figure 4393.

Hence Log27.5 = 1.4393

To find Log275.2,   2.752 × 10²

The power of 10 is the standard form of the number is 2. Thus, the characteristic is 2. Log275.2 = 2. ‘Something’

For the mantissa, find the figure along the row of 27 with middle column under 5 as before (4393). Now find the number in the differences column headed. This number is 3. Add 3 to 4393 to get 4396. Thus Log275.2 = 2.4396.

Example 1: Use table to find (i) Log 37.1 (ii) Log 64.71 (iii) Log 7.238

If the logarithm of a number is given, one can determine the number for the antilogarithm table.

Example 2: Find the antilogarithms of the following (a) 0.5670 (b)2.9504

Solution:

(a) The first two digits after the decimal point i.e .56 is sought for in the extreme left column of the antilogarithm table then look across towards right till you, get to the column with heading 9. (Read 56 under 9) there you will see 3707. Since the integral part of 0.5690 is 0, it means if the antilog (3707) is written in the standard form the power of 10 is zero.

i.e antilog of 0.5690 = 3.707 × 10⁰

= 3.707

(b) The antilog of 2.9504 is found by checking the decimal part .9504 in the table.

(c) Along the row beginning with 0.95 look forward right and pick the number in the column with 0 at the top. This gives the figures 8913 proceed further to the difference column with heading 4 to get 8. This 8 is added to 8913 to get 8921. The integral part of the initial number (2.9504) is 2. This shows that there are three digits before the decimal point in the antilog of 2.9504 so antilog 2.9504 = 892.1.

The graph of y = 10^x can be used to find antilogarithm (and logarithm). Below is the table of values.

		X	0	0.1	0.2	0.3	0.4	0.5	0.6	0.7	0.8	0.9	1
		Y = 10x	1	1.3	1.6	2	2.5	3.2	4	5	6.3	7.9	10

For example the broken line shows that the antilog 0.5 is approximately 3.2 or inversely that Log 3.2 ≅0.5

Class Activity:

1. Find the logarithms of (i) 32.7 (ii) 61.02 (iii) 3.247

2. Use antilog tables to find the numbers whose logarithms are (i) 1.82  (ii) 2.0813 (iii) 0.2108

Practice Exercises

Find the results of the following:

1. Log 436.2

2. Log 25.38

3. Log 3.258

4. 10^0.0148

5. Find the number whose log is 2.6021

ASSIGNMENT

Use the Logarithm table to evaluate each of the following:

1. 10^1.1844

2. 10^3.0631

3. 10^5.1047

Use antilogarithm tables to find the numbers whose common logarithms are:

1. 0.0254

2. 1.4662

3. 6.0129

Glossary:

1. Character/Integer

2. Mantissa/Decimal fraction

3. Logarithm,

4. Antilogarithm, etc                   

Logarithms II

LOGARITHMS

CONTENT:

1. Use Of Logarithm Table And Antilogarithm Table In Calculations Involving

• Multiplication

• Division

• Powers

• Roots

2. Application of logarithm in capital market and other real life problems

Logarithm and antilogarithm tables are used to perform some arithmetic basic operations namely: multiplication and division. Also, we use logarithm in calculations involving powers and roots.

The basic principles of calculation using logarithm depends strictly on the laws on indices. Recall that,

(a) LogMN=LogM+LogN

(b) LogMN=LogM−LogN

Hence, we conclude that in logarithm;

1. When numbers are multiplied, we add their logarithms

2. When two numbers are dividing, we subtract their logarithms.

Multiplication of Numbers using Logarithm Tables

Example 1: Evaluate 92.63 x 2.914

Solution

	Number	Standard form	Log	Operation
	92.63	9.263 x 101	1.9667
	2.914	2.914 x 100	0.4645	add
			2.4312

Antilog of 2.4312 = 269.9

Example 2: Evaluate 34.83 x 5.427

Solution

	Number	Standard form	Log	Operation
	34.83	3.483 x 101	1.542
	5.427	5.427 x 100	0.7346	add
			2.2766

Antilog of log 2.2766 = 189.1

Class Activity:

Evaluate the following:

1. 6.26 × 23.83

2. 409.1 × 3.932

3. 8.31 × 22.45 × 19.64

4. 431.2 × 21.35

Division of Numbers Using Logarithm

Example 1: Evaluate 357.2 ÷ 87.23

Solution

	Number	Standard form	Log	Operation
	357.2	3.572 x 102	2.5529
	87.23	8.723 x 101	1.9406	subtraction
			0.6123

Antilog of 0.6123 = 4.096

Example 2: Use a logarithm table to evaluate 75.26 ÷ 2.581

Solution

	Number	Standard form	Log	Operation
	75.26	7.526 x 101	1.8765
	2.581	2.581 x 100	0.4118	subtraction
			1.4647

Antilog of 1.4647 = 29.16

Class Activity:

Use table to evaluate the following:

(1) 53.81 ÷ 16.25 (2) 632.4 ÷ 34.25 (3) 63.75 ÷ 8.946 (4) 875.2 ÷ 35.81

Powers Using Logarithm

Study these examples.

At times, calculations can involve powers and roots. From the laws of logarithm, we have

(a) LogMn=nLogM

(b) LogM1n=1nLogM=LogM−−√n

(c) LogMxn=xnLogM=xlogMn

Example 1: Evaluate the following (53.75)³

Solution

	Number	Standard form	Log	Operation
	(53.75)3	(5.375 x 101)3	1.7304
			X      3	Multiply log by 3
			5.1912

Antilog of 5.1912 = 155300

Example 2: 64.59²

Solution

	Number	Standard form	Log	Operation
	(64.59)2	(6.459 x 101)2	1.8102	Multiply Log by 2
			X      2
			3.6204

Antilog of 3.6204 = 4173

Class Activity:

Evaluate the following:

1. 5.6324

2. 35.81−−−−√

3. 19.183

4. (67.95.23)3

5. 679.5−−−−√×92.6

Roots Using Logarithms

Example 1: Use tables of Logarithms and antilog to calculate 27.41−−−−√5

Solution

	Number	Standard form	Log	Operation
	27.41−−−−√5	(2.741 x 101)1/5	1.4380 ÷ 5	Divide Log by 5
			0.2876

Antilog of 0.2876 = 1.939

Example 2: Use table to find 2183.12−−−√3

Solution

Hint: Workout 218 ÷ 3.12 before taking the cube root.

	Number	Standard form	Log	Operation
	218	(2.18 x 102)	2.3385	subtraction
	3.12	(3.12 x 100)	0.4942
			1.8443 ÷3	division
			0.6148

Antilog of 0.6148 = 4.119

Example 3: Evaluate 63.75² – 21.39²

We can use difference of two squares

i.e A² – B² = (A + B)(A – B)

63.75² – 21.39² =  (63.75 + 21.39)(63.75 – 21.39)

= (85.14)(42.36)

= 85.14 × 42.36

	Number	Log	Operation
	85.14	1.9301
	42.36	1.6270	addition
		3.5571

Antilog of 3.5571 = 3607

 

Class Activity:

Evaluate the following:

1. (39.65² – 7.43²)^1/2

2. 84.35²

3. 36.95²

4. 64.74² – 55.26²

5. 94.68² – 43.25²

6. 25.14² – 7.52²

Application of Logarithm in the Capital Market and Other Real Life Problems

To start a big business or an industry, a large amount of money is needed. It is beyond the capacity of one or two persons to arrange such a huge amount. However, some persons partner together to form a company. They then, draft a proposal, issue a prospectus (in the name of the company), explaining the plan of the project and invite the public to invest money on this project. They then pool up the form from the public, by selling them shares from the company.

Examples:

1. On Wednesday 8^thAugust 2008 an investor bought 6,274,383 shares on the floor of a stock market exchange at ₦92.85 per share. Four years thereafter, he sold them at ₦134.76 per share. Calculate his profit, correct to three significant figures.

Solution:

On 8th August, 2008:

1 share = ₦92.85

Then let, 6,274,383 shares = ₦x

This gives, x = 6,274,383 × 92.85

Four years thereafter:

1 share = ₦134.76

Then let, 6,274,383 = ₦y

This gives, y = 6,274,383 × 134.76

Thus, his profit four years thereafter is (y – x) naira,

That is, 6,274,383 × 134.76 − 6274 383 × 92.85 = 6,247,383 (41.91)

≅ 6,247,000 × 41.91

	Number	Log
	6247000	6.7957
	41.91	1.6223
	2.618 x 108	8.418
	261,800,000

Hence the investor’s profit was ₦262 million

2. ₦67,200 are invested in ₦100 shares which are quoted at ₦120. Find the income if 12% dividend is declare in the shares.

Solution:

Sum invested = ₦67,200

And M.V of each share = ₦120

Therefore No. of shares bought = ₦67,200 ÷ #120

	Number	Log
	67200	4.8274
	120	2.0792
	560.02	2.7482

Given: Dividend (income) on 1 share = 12% of N.V = 12% of ₦100 = ₦12

Therefore, total income from the shares = ₦560 × ₦12 = ₦6,720

PRACTICE EXERCISES:

Find the values of the following using logarithm tables.

1. 17.83×24.692.56×32.8

2. 43.67×45.86−−−−−−−−−−−√

3. (987.3)13

4. 218×37.295.43−−−−−−√3

5. 94311.64×7.189

ASSIGNMENT

Use tables to find the values of the following:

1. (85.329.82)2

2. 17.42×4.42858,000√3

3. (38.32×2.9648.637×6.285)2−−−−−−−−−−−√3

4. In one day a total of 478,900 shares were traded on the floor of a stock exchange. If the value of each share was ₦23.50, use tables to calculate, to three significant figures, the value of the traded shares.

5. An investor buys 2,650 shares at ₦1.44 each. Use logarithm tables to calculate his profit to the nearest naira if he sells them 3 years later at ₦1.565 each.

Glossary:

1. Character/Integer

2. Mantissa/Decimal fraction

3. Logarithm

4. Antilogarithm etc

Sets I

SETS

CONTENTS:

1. Definition of sets

2. Set notations

3. Types of sets

DEFINITION OF SETS

A set is a general name for any group or collection of distinct elements. The elements of a set may be objects, names, points, lines, numbers or idea

The elements must have unique characteristics (specification) that can help to distinguish them from any other element outside the group or set. Hence, a set is a collection of well defined objects e.g.

(i) a set of mathematics text books

(ii) a set of cutleries

(iii) a set of drawing materials etc.

Sometimes there may be no obvious connection between the members of a set. Example: {chair, 3, car, orange, book, boy, stone}.

Each item in a given set are normally referred to as member or element of the set.

SET NOTATION

This is a way of representing a set using any of the following:

1. Listing method

2. Rule method or word description

3. Set builders notation.

1. Listing Method: A set is usually denoted by capital letters and the elements in it can be defined either by making a list of its members. Eg A = {2, 3, 5, 7}, B = {a, b, c, d, e, f, g, h, i} etc.

Note that the elements of a set are normally separated by commas and enclosed in curly brackets or braces.

2. Rule Method: The elements in a set  can be defined also by describing the rule or property that connects its members. Eg  C = {even number between 7 and 15. D= {set of numbers divisible by 5 between 1 and 52.}, B = {x : x is the factors of 24}etc

3. Set–Builders Notations

A set can also be specified using the set – builder notation. Set – builder notation is an algebraic way of representing sets using  a mixture of word, letters , numbers and inequality symbols e.g. B = {x : 6 ≤ x < 11, x є ƶ} or B = {x/6 ≤ x < 11, x є I}. The expression above is interpreted as “B is a set of values x such that 6 is less than or equal to x and x is less than 11, where x is an integer (z)”

— The stroke (/) or colon (:) can be used interchangeably to mean “such that”

— The letter Z or  I if used represents integer or whole numbers. Hence, the elements of the set B = {x : 6 ≤ x < 11, x є ƶ} are B = {6, 7, 8, 9,10}.

NB:  

— The values of x starts at 6 because 6 ≤ x

— The values ends at 10 because x < 11 and 10 is the first integer less than 11.

The set builder’s notation could be an equation, which has to be solved to obtain the elements of the set. It could also be an inequality, which also has to be solved to get the range of values that forms the set.

Class Activity

1. Define Set

2. C = {x : 3x – 4 = 1, x є ƶ}

3. P = {x : x is the prime factor of the LCM of 15 and 24}

4. Q = {The set of alphabets}

5. R = {x : x ≥ 5, x is an odd number}

Set – Builders Notations (contd.)

Example 1:

List the elements of the following sets

1. A = {x : 2 < x ≤ 7, x є ƶ}.

2. B = {x : x > 4, x є ƶ}

3. C = {x : −3 ≤ x ≤ 18, x є ƶ}.

4. D = {x : 5x −3 = 2x  + 12, x є Z}.

5. E = {x : 3x −2 = x  + 3, x є I}

6. F = {x : 6x −5 ≥ 8x + 7, x є ƶ}

7. P = {x : 15 ≤ x < 25, x are numbers divisible by 3}

8. Q = {x : x is a factor of 18, }

Solution:

1. A = {3, 4, 5, 6, 7}

Note that:

— The values of x start at 3, because 2 < x

— The values of x  ends at 7 because x ≤ 7 i.e. because of the equality   sign.

2. B = {5, 6, 7, 8, 9, .. .}

Note that:

The values of x starts from 5 because 5 is the first number greater than 4 (i.e. we are told that x is greater than 4)

3. C = {–3, –2, –1, 0, 1, . .  , 15, 16, 17, 18}

Note that:

— The values of x starts from −3 because −3 ≤ x, and ends at 18 because x ≤ 18 (there is equality sign at both ends).

4. To be able to list the elements of this set, the equation defined has to be solved i.e.

5x – 3 = 2x + 12

5x – 2x = 12 + 3

3x  =  15

x = ¹⁵/₃

∴ x = 5

∴ D = {5}

5. We also need to solve the equation to get the set values

3x − 2 = x + 3

3x – x  = 3 + 2

2x = 5

∴ x =  ⁵/₂

Since ⁵/₂ is not an integer (whole number) therefore the set will contain no element.

∴ є = { } or Ø

6. Solving the inequality to get the range of values for the set, we have

6x – 5 ≥ 8x + 7

6x – 8x ≥ 7 + 5

–2x ≥ 12

x ≤ ¹²/_–2

∴ x  ≤ –6

∴ F = {…, –8, –7, –6}

7. P = {15, 18, 21, 24}

Note that:

The values of x start at 15 because it is the first number divisible by 3 and falls within the range defined.

8. Q = {1, 2, 3, 6, 9, 18}

Example 3:

Rewrite the following using set builder notation

(i)      A = {8, 9, 10, 11, 12, 13, 14}

(ii)     B = {3, 4, 5, 6 . . . }

(iii)    C = {. . . 21, 22, 23, 24}

(iv)    D = {7, 9, 11, 13, 15, 17 . . .}

(v)     P = {1, -2}

(vi)    Q = {a, e, i, o, u}

Solution:

(i)      A = {x : 7 < x < 15, x є ƶ} OR

A = {x : 8 ≤ x < 15, x є ƶ} OR

A = {x : 7 < x ≤ 14, x є ƶ} OR

A = {x : 8 ≤ x ≤ 14, x є ƶ}

(ii)     B = {x : x > 2, x є ƶ} OR

B = {x : x ≥ 3, x є ƶ}

(iii)    C = {x : x < 25, x є ƶ} OR

C = {x : x ≤ 24, x є ƶ}

(iv)    D = {x : x > 8 or x ≥ 7, x is odd,  x є ƶ}

(v)     P = {1,2} suggests the solutions of a quadratic equation. Therefore, the equation or set-builders notation can be obtained from :

x² – (sum of roots)x + product of roots  = 0

x² –(–1)x + (1 x –2) = 0

x² + x  – 2 = 0

P = {x : x² + x  – 2 = 0, x є ƶ}

(vi)  Q = {x : x is a vowel}

Class Activity

1. List the elements in the following Sets

(a)  A = {x : -2 ≤ x < 4, x є ƶ}

(b)  B = {x : 9 < x < 24, x є N}

(c)  C = {x : 7 < x ≤ 20, x is a prime number, x є I}

(d)  D = {x / 2x – 1 = 10, x є Z}

(e)   P = {x : x  are the prime factor of the LCM of 60 and 42}

2. Rewrite the following using Set – builder notations.

(a)  Q = {. . . 2, 3, 4, 5}

(b)  A = {2, 5}

(c)  B = {2, 4, 6, 8, 10, 12 . . .}

(d)  A = {−2, –1, 0, 1, 2, 3, 4, 5, 6}

(e)  C = {1, 3, –2}

TYPES OF SETS

Finite Sets: Refers to any set, in which it is possible to count all the elements that make up the set. These types of sets have end. E.g.

A = {1, 2, 3,  . . , 8, 9, 10}

B = {18, 19, 20, 21, 22}

C = {Prime number between 1 and 15} etc.

Infinite Sets: Refers to any set, in which it is impossible to count all the elements that make up the set. In other words, members or elements of  these types of set have no end. These types of set, when listed are usually terminated with three dots or three dots before the starting values showing that the values continue in the order listed. E.g.

A = {1, 2, 3, 4,  . . }

B = {…,–4,–3,–2,–1,0,1,2,3,…}

C = {Real numbers} etc.

Empty or null Set: A set is said to be  empty  if it contains no element. Eg {the set of whole number that lies between 1 and 2}, {the set of goats that can read and write}, etc. Empty sets are usually represented using ø or { }.

It should be noted that {0} is NOT an empty set because it contains the element 0, Another name for empty set is null set.

Cardinality of a Set/Number of Elements in a Set

Given a set A = {–2, –1, 0, 1, 2, 3, 4, 6} the number of elements in the set A denoted by n(A) is 9;  i.e. n (A) = 7

If B = {2,3, 5} then n (B) = 3

If Q = {0} then n (Q) = 1

Other examples are as follows:

Example 4:

Find the number of elements in the set:

P =  {x : 3x –5 < x + 1 < 2x + 3, x є ƶ }

Solution:

	3x – 5 < x + 1	and	x + 1 < 2x + 3
	3x – x < 1 + 5	and	x – 2x < 3 – 1
	2x < 6		– x < 2
	x < 6/2		x > -2
	x < 3		-2 < x
		2 < x < 3

The integers that form the solution set are

P =  {–1, 0, 1, 2}

∴ n {P} = 4

Example 5:

Find the number of elements in the set

A = {x : 7 < x < 11, x is a prime number}

Solution:

The set A = { } or Ø since 8, 9, 10 are no prime numbers.

∴ n(A) = 0

Example 6:

Find the number of elements in the following sets:

(i) B = {x : x ≤ 7, x є ƶ}

(ii) C = {x : 3 < x ≤ 8, x is a number divisible by 2}.

Solution:

(i) B =  {. . . 3, 4, 5, 6, 7}.

The values of the set B has no end, hence it is an infinite set i.e. n(B) = ∞

(ii) C =  {4, 6, 8}.

∴ n(C) = 3

The Universal Set

This is the Set that contains all the elements that are used in a given problem. Universal Sets vary from problem to problem. It is usually denoted using the symbols ξ or μ.

Note that when the Universal Set of a given problem is defined, all values outside the universal set cannot be considered i.e. they are invalid.

Equivalent Sets

Two Sets are said to be equivalent if the Sets have equal number of elements. E.g.

If A = {2, 3, 4, 5} and B = {a, b, c, d} then the Sets A ≡ B (A is equivalent to B) since n(A) = n(B).

Equality of Sets

Equal sets are special cases of equivalent sets. They have exactly equal elements on both sets. The order of writing the elements is not important. We use equality sign “=” to indicate equal sets

Given two sets P = {3,4,5} and Q = {4,3,5}. The elements of the two sets are equal. Therefore P = Q

Subset and Superset

Suppose that A = {1,2,3} and B = {1,2,3,4,5}. Notice that every element of set A is also an element of set B. We say that A is a subset of B written as A ⊂ B. While B is a superset of A is written as B ⊃ A.

Power Sets

Given a set A then the power set of A, denoted by P(A) is the set of all possible subsets of A

The number of possible subsets in P(A) is given by 2ⁿ where n is the number of elements in the set

Class Activity

1. List the elements in the following Sets

(a)  A = {x : -2 ≤ x < 4, x є ƶ}

(b)  B = {x : 9 < x < 24, x є N}

(c)  C = {x : 7 < x ≤ 20, x is a prime number, x є I}

(d)  D = {x / 2x – 1 = 10, x є Z}

(e)  P = {x : x  are the prime factor of the LCM of 60 and 42}

2. Find the number of elements in the  sets in question (1) above

3. If

(a)   A=  {3,5,7,8,9,10,}, Then n(A) =

(b)   B=  {1, 3, 1, 2, 1, 7}, Then n(B) =

(c)   Q=   {a, d, g, a, c, f, h, c,} , Then n(Q) =

(d)   P=   {4,5,6,7,…,12,13}, Then n(P) =

(e)   D ={ days of the week} , then n(D)=

PRACTICE EXERCISES

1. State if the following are finite, infinite or null set

(i)  Q = {x : x ≥ 7, x Є Z}

(ii)  P = {x : -4 ≤ x < 16, x Є I}

(iii)  A = {x : 2x – 7 = 2, x Є Z}

(iv)  B = { sets of goats that can fly}

(v)   D ={sets of students with four legs}

2. List the following sets in relation to the universal set. Given that the universal set ξ = {x: 1< x< 15, x ε z}

(i)  A= {x; -3≤ x≤7, x ε Z}

(ii)  B= {x: 5<x<6, xεz},

(iii)  C={x: X≥ 5, xε z }

3. List the elements of the following universal sets.

(i) The set of all positive integers

(ii) The set  of all integers

(iii) ξ = { x : 1 <  x < 30, x are multiples of 3}

(iv) ξ = { x : 7 ≤ x < 25, x are odd numbers}

(v)  ξ = { x : x ≥ 10, xεz}

4. Find all the possible subsets of each of the following sets

(i) A = {1,2}

(ii) B = { 7,9}

(iii) C = {2,4,6}

5. Find the power set of each of the following set

(a) A = {0,5}

(b) B = {7,8,9}

ASSIGNMENT

1. Identify each of the following set as null, finite or infinite set

(a) A = {x : 0<x<10,x is an integer}

(b) B = {x: x>2, x is an integer}

2. Calculate the number of possible subsets of each of the following set

(a) {a,b}

(b) {x,y,z}

(c) {m,n,o,p}

3. Write down the possible subsets of each of the following set

(a) {a,b}

(b) {x,y,z}

(c) {m,n,o,p}

4. Given that A = {x;x is an even number

and B = {2,3,4,6}

Show that B is not a subset of A

5. If A ⊂ { }, then what is A?

Glossary:

1. finite

2. infinite

3. subset

4. superset

5. cardinality

6. power-set etc.

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