MATHEMATICS SS1 FIRST TERM LESSON PLAN SCHEME OF WORK
MATHEMATICS SS1 FIRST TERM LESSON PLAN SCHEME OF WORK
INDICES
CONTENT

Revision of Standard Form

Laws Of Indices and Application of Indices

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 × 10^{0}
(b) 53.7 = 5.37 × 10^{1}
(c) 537 = 5.37 × 100
= 5.37 × 10 × 10
= 5.37 × 10^{2}
(d) 35.65 = 3.565 × 10
= 3.565 × 10^{1}
(e) 7500 = 7.5 × 1000
= 7.5 × 10^{3}
(f) 1403420 = 1.403420 × 1000000
= 1.403420 × 10^{6}
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.

(a) 2037000 (b) 00469 (c) 0.000513 (d) 146200
Write each of the following numbers in full

(a) 6 × 10^{4 }(b) 5.142 × 10^{7 }(c) 0.7883 × 10^{– 5 }(d) 2.095 × 10^{9}
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:

(278)−23

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:

4−3x=164

(x+7)3=27
Solution:

4−3x=1434−3x=4−3−3x=−3∴x=1

(x+7)3=27(x+7)3=33x+7=3x=3−7x=−4
Practice Exercises

Write 3.654×10−10 in full

Simplify 32×1048×107

Simplify 196m3n74m5n−−−−−−√

Simplify 2−n×82n+1×162n43n

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

Solve the equation 3x−4=2342
ASSIGNMENT

Given that: 2×41−x=8−x, find x

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

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

Solve for xif 4x×82x−1=1161−x

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

If 8x2=238×434, find x WAEC SSCE
Glossary
Number, Index, Power, Roots, Base etc.
Logarithms I
LOGARITHMS
CONTENT

Deducing logarithm from indices and standard form

Definition of Logarithms

Definition of Antilogarithms

The graph of y = 10^{x}

Reading logarithm and Antilogarithm tables
Deducing Logarithm From Indices and Standard Form
There is a close link between indices and logarithms
100 = 10^{2}. This can be written in logarithmic notation as log_{10}100 = 2.
Similarly 8 = 2^{3} and it can be written as log_{2}8 = 3.
In general, N = b^{x} in logarithmic notation is Log_{b}N = 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.
Rewrite using logarithmic notation (i) 1000 = 10^{3} (ii) 0.01 = 10^{−2} (iii) 2^{4} = 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 × 10^{1}. 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 × 10^{n} where A is a number such that 1 ≤ A < 10 and n is an integer.
27.5 = 2.75 × 10^{1}, 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 4figure table. The 4figure table is at the back of your New General Mathematics textbook.
Below is a row from the 4figure 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 × 10^{2}
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 × 10^{0}
= 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 = 10^{x} 
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:

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

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:

Log 436.2

Log 25.38

Log 3.258

10^{0.0148}

Find the number whose log is 2.6021
ASSIGNMENT
Use the Logarithm table to evaluate each of the following:

10^{1.1844}

10^{3.0631}

10^{5.1047}
Use antilogarithm tables to find the numbers whose common logarithms are:

0.0254

1.4662

6.0129
Glossary:

Character/Integer

Mantissa/Decimal fraction

Logarithm,

Antilogarithm, etc
Logarithms II
LOGARITHMS
CONTENT:

Use Of Logarithm Table And Antilogarithm Table In Calculations Involving

Multiplication

Division

Powers

Roots

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;

When numbers are multiplied, we add their logarithms

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 10^{1} 
1.9667 

2.914 
2.914 x 10^{0} 
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 10^{1} 
1.542 

5.427 
5.427 x 10^{0} 
0.7346 
add 

2.2766 
Antilog of log 2.2766 = 189.1
Class Activity:
Evaluate the following:

6.26 × 23.83

409.1 × 3.932

8.31 × 22.45 × 19.64

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 10^{2} 
2.5529 

87.23 
8.723 x 10^{1} 
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 10^{1} 
1.8765 

2.581 
2.581 x 10^{0} 
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 10^{1})^{3} 
1.7304 

X 3 
Multiply log by 3 

5.1912 
Antilog of 5.1912 = 155300
Example 2: 64.59^{2}
Solution
Number 
Standard form 
Log 
Operation 

(64.59)^{2} 
(6.459 x 10^{1})^{2} 
1.8102 
Multiply Log by 2 

X 2 

3.6204 
Antilog of 3.6204 = 4173
Class Activity:
Evaluate the following:

5.6324

35.81−−−−√

19.183

(67.95.23)3

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 10^{1})^{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 10^{2}) 
2.3385 
subtraction 

3.12 
(3.12 x 10^{0}) 
0.4942 

1.8443 ÷3 
division 

0.6148 
Antilog of 0.6148 = 4.119
Example 3: Evaluate 63.75^{2} – 21.39^{2}
We can use difference of two squares
i.e A^{2} – B^{2} = (A + B)(A – B)
63.75^{2} – 21.39^{2} = (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:

(39.65^{2 }– 7.43^{2})^{1/2}

84.35^{2}

36.95^{2}

64.74^{2} – 55.26^{2}

94.68^{2} – 43.25^{2}

25.14^{2} – 7.52^{2}
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:

On Wednesday 8^{th}August 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 10^{8} 
8.418 

261,800,000 
Hence the investor’s profit was ₦262 million

₦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.

17.83×24.692.56×32.8

43.67×45.86−−−−−−−−−−−√

(987.3)13

218×37.295.43−−−−−−√3

94311.64×7.189
ASSIGNMENT
Use tables to find the values of the following:

(85.329.82)2

17.42×4.42858,000√3

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

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.

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:

Character/Integer

Mantissa/Decimal fraction

Logarithm

Antilogarithm etc
Sets I
SETS
CONTENTS:

Definition of sets

Set notations

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:

Listing method

Rule method or word description

Set builders notation.

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.

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

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

Define Set

C = {x : 3x – 4 = 1, x є ƶ}

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

Q = {The set of alphabets}

R = {x : x ≥ 5, x is an odd number}
Set – Builders Notations (contd.)
Example 1:
List the elements of the following sets

A = {x : 2 < x ≤ 7, x є ƶ}.

B = {x : x > 4, x є ƶ}

C = {x : −3 ≤ x ≤ 18, x є ƶ}.

D = {x : 5x −3 = 2x + 12, x є Z}.

E = {x : 3x −2 = x + 3, x є I}

F = {x : 6x −5 ≥ 8x + 7, x є ƶ}

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

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

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.

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)

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).

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}

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

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}

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.

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 setbuilders notation can be obtained from :
x^{2} – (sum of roots)x + product of roots = 0
x^{2 }–(–1)x + (1 x –2) = 0
x^{2} + x – 2 = 0
P = {x : x^{2} + x – 2 = 0, x є ƶ}
(vi) Q = {x : x is a vowel}
Class Activity

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}

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 2^{n }where n is the number of elements in the set
Class Activity

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}

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

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

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}

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 }

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}

Find all the possible subsets of each of the following sets
(i) A = {1,2}
(ii) B = { 7,9}
(iii) C = {2,4,6}

Find the power set of each of the following set
(a) A = {0,5}
(b) B = {7,8,9}
ASSIGNMENT

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}

Calculate the number of possible subsets of each of the following set
(a) {a,b}
(b) {x,y,z}
(c) {m,n,o,p}

Write down the possible subsets of each of the following set
(a) {a,b}
(b) {x,y,z}
(c) {m,n,o,p}

Given that A = {x;x is an even number
and B = {2,3,4,6}
Show that B is not a subset of A

If A ⊂ { }, then what is A?
Glossary:

finite

infinite

subset

superset

cardinality

powerset etc.