FIRST TERM WEEK 1: NUMBER BASE SYSTEM Conversion can be done from other bases to base 10 For instance i. Denary or decimal number which is base 10 and the highest digit is 9 i.e. 0, 1, 2, 3, 4, 5, 6, 7, 8 and 9. ii. Binary system: base 2 which are 0 and 1. iii. Octal System: base 8 which has digits as 0, 1, 2, 3, 4, 5, 6 and 7 iv. Base 12 (duo decimal) 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A and B v. Base 16 (Hexadecimal System) the digit used are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A(10), B(11), C(12), D(13), E(14), and F(15) A. Conversion from other bases to base ten Example 1 a. Convert 5603eight to base 10 b. Convert 3B9twelve to base 10 Solution: a. Convert 5603eight to base 10 = 5 x 83 + 6 x 82 + 0 x 81 + 3 x 80 = 5x512 + 6 x 64 + 0 + 3 x 1 = 2560 + 384 + 0 + 3 Therefore, 5603eight = 2947ten b. Convert 3B9twelve to base 10 = 3 x 122 + B x 121 + 9 x 120 = 3 x 144 + 11 x 12 + 9 x 1 = 432 + 132 + 9 Therefore, 3B9twelve = 573ten Example 2 Convert 1011.101two to base 10 Solution: 1011.101 = 1x23 + 0 x 22 + 1 x 21 + 1x20 + 1x2-1 + 0x2-2 + 1x2-3 = 8 + 0 + 2 + 1 + ½ + 0 + 0.125 Therefore, 1011.101two = 11.625ten

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) B. Converting Numbers in Base ten to other bases Example Express 208ten in i. Binary ii. Hexadecimal Solution: i. 208ten in binary 2 208 Remainder 2 104 0 2 52 0 2 26 0 2 13 0 26 1 23 0 21 1 20 1 Therefore, 208ten = 11010000two ii. 208ten in hexadecimal 16 208 Remainder 16 13 0 0D Therefore, 208ten = D0sixteen WEEK 2: NUMBER BASE Addition and subtraction in other bases Addition and subtraction in other bases are carried out exactly the same way as base ten numbers. For example to add or subtract in base two: Hints 1+0 = 1 1–0=1 0+1 = 1 1–1=0 1+1=10 10 – 1 = 1 2

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Examples: Evaluate the following a. 56eight + 243eight b. 1001two – 111two Solution: Procedure: 1st Column 6 + 3 = 9( i.e. 1 eight and 1) a. 56 write 1 carry 1 forward. + 243 2nd Column: 5 + 4 + 1= 10( i.e. 1 eight and 321eight 2) , write 2 carry 1 forward 3rd Column: 2 + 1=3 (carry forward = 3) b. 1001 - 111 1st Column: 1 – 1 = 0 010 2nd Column: 0 – 1 is not possible, we borrow 1 from next column to give 2 i.e. 10. Then 10 – 1 = 1. 3rd Column: 1-1 = 0. Multiplication and Division in other bases Multiplication and Division in other bases are carried out exactly the same way as base ten numbers Hints: a. To multiply in base 2 i. 0 x 0 = 0 ii. 0x 1 = 0 iii. 1x 1=1 3

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) b. To multiply in base 8 i. 2 x 4 = 8 but 8 = 1eight and 0 Thus 2 x 4 = 10 ii. 3 x 4 = 12, But 8 = 1eight and 4 Thus 3 x 4 = 14 Example Calculate the following: a. 321five x 21five b. 110.11two x 11.1two Solution a. 321five x 21five 321 x 21 321 1142 12241five Hence 321five x 21five = 12241five b. 110.11two x 11.1two First, ignore the binary point and multiply 1011two by 111two as follows: 11011 x 111 11011 +11011 11011 10111101two 4

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Now insert the binary point (b.p) by counting the number of digits before binary point in the question 10111.101 Thus: 110.11two x 11.1two = 10111.101two APPLICATION TO COMPUTER PROGRAMMING Computers usually store information as a series of bits. A bit is the smallest unit information stored in a computer. A group of 8bits is called a byte, each byte corresponding to one character. Binary system contains two numbers 0 for OFF and 1 for ON. Therefore, a bit is either a 0 or a 1. The amount of data that can be stored on a disk is measured in kilobytes (i.e.One thousand bytes is 1024 kilobytes (KB) or 1048576 bytes .220 is 1 megabyte (MB). WEEK 3: MODULAR ARITHMETIC Concept of modular arithmetic: Modular arithmetic (sometimes called clock arithmetic) is a system of arithmetic in which numbers wrap around after they reach a certain value. It can also be regarded to as arithmetic that involves integers.The plural of modulus is moduli. Example 1: Simplify the following a. 76 (Mod 5) b. 340 (mod 7) Solution a. 76 ÷ 5 = 15 Remainder 1 5

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) OR 76 = 15 x 5 + 1 Therefore, 76 = 1 (mod 5) b. 340 (mod 7) 340 ÷ 7 = 48 Remainder 4 OR340 = 4 (mod 7) In general, two integers a and b are said to be contingent modulo m, written asa ≡ b (mod m). Operation in Modular Arithmetic Addition and Subtraction in modular arithmetic : Addition in modulo 5 +01234 001234 112340 223401 334012 440123 Also, Subtraction in modulo 5 6

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) - 01234 004 321 110432 221043 332104 443210 Example 2: (mod 6) a. Evaluate 36+20 Solution: 36 + 20 = 56 56 = 9 x 6 + 2 = 2 (mod 6) b. Work out the simplest positive form of –3(mod 8) Solution: –3 = -1 x 8 + 5 = 5 (mod 8) c. Evaluate the following: ii. i. 12 – 3 (mod 5) Solution: 11 – 24 (mod 7) i. 12 – 3 (mod 5) 12 – 3 = 9 7

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) ii. 9 = 5 + 4 = 4 (mod 5) Multiplication 11 – 24 (mod 7) 11 – 24 = -13 = -2 x 7 + 1 = 1 (mod 7) Multiplication can be interpreted as repeated addition. The table below shows multiplication in modulo 5 X01234 000000 101234 202413 303142 404321 Example 3: Calculate the following in (mod 5) a. 3 x 2 b. 28 x 12 Solution: a. 3 x 2 = 6 6 =1x5 +1 = 1 (mod 5) b. 28 x 12= 336 336 = 67 x 5 + 1 = 1 (mod 5) 8

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Division Since division is the inverse of multiplication, to solve division problem, for example : if 2 ÷ 4 =x (mod 5), then 2/4 = x 4x = 2, therefore x = 2 Multiplication inverse The multiplication inverse of a number x is a number y such that xy = 1 e.g. multiplication inverse of 7 is 1/7 because 7 x 1/7 = 1. We can apply the above to find the inverse of residue in modular arithmetic as shown in the following example. Example 1: Find the inverse of: a. 5(mod 7) b. 3 (mod 4) Solution a. 5(mod 7) The inverse of 5 in (mod 7) is the number that when multiplied by 5 gives 1 thus 5 x 3 = 15 = 1 (mod 7) b. 3 (mod 4) The inverse of 3 in (mod 4) is the number that when multiplied by 3 gives 1 Thus 3 x 3 = 9 therefore, = 1 (mod 4) This means the inverse of 3 in modulo 4 is 3 9

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Example 2: Simplify the following inmod 5 a. 3 ÷ 4 b. 3 ÷ 3 Solution: a. 3 ÷ 4 (mod 5) Let 3 ÷ 4 = x Then 4x = 3 ………* Multiply both sides by a number that will make the coefficient of x unityi.e 1 Try 4 x 1 =4 = 4 (mod 5) 4 x 2 = 8 = 3 (mod 5) 4 x 3 = 12 =2 (mod 5) 4 x 4 = 16 = 1 (mod 5) Multiply both sides of * by 4 i.e. the multiplicative inverse of 4 in modulo 5 (4 x 4)x = 3 x 4 1x = 12 = 2 x 5 + 2 x = 2 (mod 5) b. 3 ÷ 3 Let 3 ÷ 3 =x then 3x = 3 (mod 5) To make the coefficient of x 1 0

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) unity, multiply both sides by 2 (3 x 2)x = 3 x 2 1x = 6, therefore, x = 1 (mod 5) Solving Simple Equations in Modular Arithmetic. Example : Given that x + 5 = 2 (mod 7), find x Solution: x + 5 = 2 (mod 7) Look for a number in arithmetic (mod 7) that can be added to 5 tomake it 0 . Thus, add the additive inverse of 5 to both sides which is 2, to both sides. We have, x + 5 + 2 = 2 + 2 (mod 7) x + 0 = 4 (mod 7) x = 4 (mod 7) Check 4 + 5 = 9 = 2 (mod 7) Application of Modular Arithmetic Example: In one year, New Year Day is on Wednesday. On what day of the week would the New Year day be in the following year if it is: (a) Common Year? (b) A leap year? 1 1

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Solution: Since there are 7 days in a week, so we work in arithmetic (mod 7). We take Wednesday as 0 as shown in the number cycle below: Wed Tue 6 0 Thur 1 Mon 5 Sun 4 3 2 Fri Sat a. Common year = 365 days = 52 weeks +1 day = 52 x 7 + 1 =1 (mod 7) i.e. 1 day after Wednesday is Thursday Therefore, the day of the week is Thursday. b. Leap year, 366 days = 52 weeks + 2 days = 52 x 7 + 2 = 2 (mod 7) i.e. 2 days after Wednesday is Friday Therefore,the day of the week is Friday. WEEK 4: STANDARD FORM AND INDICES Objectives: By the end of the lesson students should be able to: 1 2

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) i. Write numbers in standard form ii. Relate indices to standard form iii. State the laws of indices and deduce their meaning iv. Solve problems using the laws of indices v. Solve problems on simple indices equation STANDARD FORM The standard form is generally expressed in term of a x 10n . Where a is a number between 1 and 10 and n is a positive or negative integer. Example 1: a. Express 7090000 in standard form b. Express 0.00007 in standard form Solution: a. 7090000 = 7.09 x 1000000 = 7.09 x 106 b. 0.00007 = 7.0 x ( Example 2: Change these standard forms to ordinary numbers a. 6.3 x 105 b. 4.08 x 10-5 Solution: a. 6.3 x 105 1 3

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) = 630000 b. 4.08 x 10-5 = 0.0000408 INDICES Laws of Indices An index tells us how many times the base number has been multiplied by itself. e.g. 35 = 3 x 3 x 3 x 3 x 3 In general , we have, am = axaxax…. (Where a = base and m = power). Law 1: amx an = am+n (addition law) i.e. multiplication involving same base results in addition of powers of the common base. Example i. a5 x a7 = a5+7 = a12 ii. 85 x 87 = 85+7 = 812 1 4

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Law 2 :am÷ an = a m/ an= a m-( Division law) i.e division involving same base results in subtraction of powers of the common base. When dividing in indices we are subtracting Example: i. a8 ÷ a5 = a8-5 =a3 ii. 109 ÷ 107 = 109-7 = 102 law 3 : a0 = 1( Zero index) Example: i. a8 – a8 =a8–8 =a0 = 1 (since a0 =1) law 4: a-m= 1÷ a m = 1/am (Negative index) Example 1 5

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) a. a-5= 1/a5 b. c-8= 1/c8 General Examples : Simplify the following : a. 6a-4 b. 3x3 y-4 z-2 c. 27x-1 x 271-x d. 48e-5 ÷ 12e-10 Solution: a. 6a-4 = 6 x 1/a4 = 6/a4 b. 3x3 y-4 z-2 = 3x3 x 1/y4 x 1/z2 = 3x3/y4z2 c. 27x-1 x 271-x = 33(x-1)x33(1-x) = 33x-3+3-3x = 30 = 1 d. 48e-5 ÷ 12e-10 = 48e-5/12e-10 x (e-5/ e-10 ) = 4e-5-(-10) = 4e-5+10 1 6

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) = 4e5 Law5 :Raising an index to another power In general ,(am)n = amn Example : Simplify the following : a. [(3g)2]-2 b. (-2x6)4 Solution : a. [(3g)2]-2 = [(34g)4] -! = 1/(34g4) = 1/81g4 b. (-2x6)4 = ( -2 4 x 24 ) = 16 x 24 OR (-2x6)4 = -2x6 x -2x6 x -2x6 x -2x6 = 16x6+6+6+6 = 16x24 1 7

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) WEEK 5: FRACTIONAL INDICES SQUARE ROOTS Law 6 (Root Law) a½ = a i.e. 251/2 = 25 Similarly in algebra a½ x a½ = a½+½ = a x a = a Cube roots and higher roots using the same method as above a1/3 x a1/3 x a1/3 = a1/3+1/3+1/3 = a1 = a But 3 a x 3 a x3 a = a1 Therefore, 3 a = a1/3 Also 4 a = a1/4 Note also that:, ma = a1/m When the numerator of the fractional index is not 1 for instance a. a2/3= a2 x 1/3 =(a2)1/3 =(a 1/3)2 =(3√a)2or3 a2 In general, 1 8

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) am/n = ( n√a )m a. 161/2 b. 25b4 Solution: Example: = ( 16 )1=( 4)1 =4 Evaluate: a. 161/2 b. 25b4 =25x b 4 = (52)1/2x b1/2x4= 5b2 Simple exponential equation rules Examples : 1. Solve these equations a. x1/3 = 5 b. 3x –1/2 = -27 c. 8x = 0.125 Solution: a. x1/3 = 5 x1/3 x 3 = 53 (Take the cube of both sides), x1 = 53 x = 125 b. 3x –1/2 = -27 x-1/2 = -9 ( dividing both sides by 3) (x-1/2 ) 2= (-9)2 (Squaring both sides ) x-1= 81 since a –m= 1/a m,then X= 81 c. 8x= 0.125 1 9

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 8 x = 8 -1 26 = 27 Example: If 92x + 1 = 81x-2/3x, find x. Solution 92x + 1 = 81x-2/3x 32(2x+1) = 34(x-2)-x 34x+2 = 34x-8-x 34x+2 = 33x-8 Equate the base 4x +2 = 3x – 8 4x – 3x = –8 – 2 x = –10 WEEK 5 LOGARITHM Behavioral objectives: By the end of the lesson students should be able to: i. Show logarithms as the inverse of indices ii. Define logarithm 2 0

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) iii. Relate indices to standard form iv. Find the logarithm and antilogarithms of numbers greater than one from four figure tables v. Use logarithm and antilogarithms tables in calculation vi. Apply logarithm to solve real life problems Relationship between indices and logarithms Recall in index notation 8 = 23 Similarly if 81 = 34 then Log381 = 4 In general If y = ax, then x = logay Remarks :, i. If 1 = 100 then Log101 = 0 ii. If 10 = 101 then log1010 = 1 Logarithm written to base 10 is called common logarithm and can be written as log1010 =log10 and,log10100 = log100. Definition : Logarithm simply means the inverse operation of indices. Using tables to find the logarithms of a number : 2 1

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) If a number is expressed as a power of 10, then the index is called the logarithm of the number e.g. 69 = 101.8388 Example Use logarithm table to find the log of the following a) 8.4 b) 840 Solution a) 8.4 Express 8.4 in form of ax10ni.e 8.4 x 100 in standard form. From thelogarithm table ,check 84 under 0. It corresponds to 0.9243 8.4 = 100.9243 :. Log 8.4 = 0.9243 b) 840 Express in form of a x 10n 840 = 8.4 x 102 in standard form From logarithm table, 840 = 102+0.9243 :. log 840 = 2.9243 Note : The logarithm of a number contain two parts i.e. the characteristics and the mantissa e.g. Log 840 = 2.9243( 2 is the characteristics and .9243 is the mantissa ). 2 2

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Reading of logarithm and antilogarithm (antilog) of a number means to find the number whose logarithm is given. Therefore antilogarithm is the reverse of logarithm. Illustration: log101 = 0, then antilog of 0 = 100 = 1 Also log1010 = 1, then antilog of 1 = 101 = 10 etc Examples : Use the antilogarithm tables to find the following : a) The antilog of 0.6894 b) The number whose logarithm is 4.56 Solution : a. Antilog 0.6894 The mantissa .6894 in the antilog table = 4887 + 4 = 4891. The characteristic is 0. This means there must be 1 digit before the decimal point. :. antilog of 0.6894 = 4.891 b. Antilog 4.56 The mantissa .56 in the antilog table = 3631. The characteristic is 4, then 4+1 = 5. This means there must be 5 digits before the decimal point. :. antilog of 4.56 = 36310. WEEK 7: Use of logarithm table in solving problems. Multiplication : To find the product of number ; 2 3

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) i. Find the logarithm of each number ii. Add the logarithm together then iii. Find the antilogarithm of the result Example: a). Use logarithm tables to evaluate 9.456 x 867.4 Solution : Number Log 9.456 0.9757 867.4 +2.9382 3.9139 8202  Antilog of 3.9139 Division : b). Use logarithm table to evaluate Solution : Number Log 786.8 2.8959 12.54 -1.0983 1.7876 Antilog of 1.7976 = 62.75 Power : Examples : 2 4

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Evaluate the following using logarithm tables : (a) (89.67)4 b) (7.0698)3 Solution: a) (89.67)4 Number Log 89.67 1.9526 89.674 1.9526 x4 64620000 7.8104 Roots: Examples : Evaluate the following using logarithm tables : a) 79.84 b) 4 894.5 Solution a) 79.84 Number Log 79.48 1.9002 79.48¼ 1.9002 ÷ 2 0.9501 Antilog of 0.9501 = 8.915 b) 4 894.5 Number Log 894.5 2.9516 894.5¼ 2.9516 ÷ 4 0.7379 Antilog of 0.7379 = 5.469 2 5

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Example 3: Use algorithm table to evaluate 8.35523 X.5 893.4 4 7.245X25.342 Solution 8.35523 X.5 893.4 8.35523 X.893.4 4 7.245X25.342 = 7.245X24.35 X2 8.35523 X.893.4 = 7.245X24.35 Number Log 8.3523 0.9218 x 3 = 2.7654 (893.4)1/5 2.9511 ÷ 5 = 0.5902 Numerator 3.3556 3.3556 7.245 0.8600 X 25.34 +1.4038 - 2.2638 (7.245 x 25.34)1/2 2.2638 ÷ 2 = 1.1319 1.1319 2.2237 Antilog of 2.2237 = 167.4 Exercise Use logarithm and antilogarithm tables to evaluate the following: 1. 4.9272 x 45.1 2. 18.63 X8.09 5 85000 WEEK 8: SETS Behavioral Objective : By the end of this lesson, students should be able to define and identify various types of sets. 2 6

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Definition 1 : A set can be defined as a group or a collection of well defined objects or numbers. E.g. A = {Set of fruits} i.e. A = {Mango, Guava etc} . Set is usually denoted by a capital letters i.e. A, B, C, D etc. Definition2: Elements of a set The elements of a set are the objects or numbers that belong to the set. Elements of a set are also called members of a set. Example: If M = {1, 3, 5, 7}, then n(M) = 4 and 3 is a member of Mis written as 3 Є M A set can be described by: i. Listing all its elements or members within the brackets { } e.g. F = { mango, grape, orange, guava} ii. In words e.g. M = {Multiple of 3 from 3 to 18} C = {Consonants} iii. Algebraically E.g. B = {x: -10 < x ≤ 3, x is an integer} this means B is the set of numbers x such that x is a whole number greater than -10 but less than or equal to 3. A set specified in this way is known as a set builder notation. Hence, B = {9,-8,…..3}. Types of Sets 1. Finite Set: A finite set is a set in which all its members can be listed. E.g. If A = {factors of 6}, then A = {1,2,3,6} 2. Infinite Set: An infinite set is a set in which all its members cannot be listed E.g. If B = {Odd numbers greater than 5}, then B = {7, 9, 11, 13, 15…} 3. Empty Set (Or Null Set) A Set without any members or elements, is called a null set or an empty set. It is usually represented by { } or φ E.g. M = {Months with two letters} :. M = φor M={ } 4. Subsets: 2 7

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Set A is a subset of set B if and only if every member of A is a member of B. It is denoted by C. e.g. B c S reads as B is a subset of S S ɔB reads asS is a superscript of B Consider the set below S = {2, 4, 6, 8, 12, 14, 16} B = {2, 6, 10, 12, 14} BCS SƆB i.e. S B 2 46 26 10 8 12 12 10 14 16 14 S BCS 4 8 16 2 6 10 12 14 B 5. Power Set and the Number of Subsets The power set of a set is a list of all its possible subset. It is denoted asP(A) If A = {1, 2, 3, ...} the power set of A is 24 Thus, if A = {1, 3}, the power of set A = 22 = 4 and the list of all possible subsets are ; {1}, {3}, {1,3}, {} Example A set contains 7 numbers; find the number of subsets that can be obtained from it Solutions: 2 8

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Number of possible subsets N = 2n Where n = 7, then N = 27 = 128 The number of possible subsets of the sets is 128. 6. Universal set It is denoted by ƹ or Ʋ The Universal set for any given problem is a set that contains all the members, which can be used for that problem E.g. if ƹ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} A = {2, 3, 5, 6, 7} A is a subset of ƹ i.e. Aϲ Ʋ 7. Set Builder Notation A set specified by using an algebraic expression such as A = {x : x ≤ 8}. Also ifA = {x : x2 – 4x = 5}, then it means , A is the set of x such that x2 – 4x = 5. We can solve this expression by factorization: x2 – 4x = 5 x2 – 4x – 5 = 0 (x + 1) (x – 5) = 0 x+1 =0 0r x -5) =0 X = -1 or x = 5 :. X=-1 or 5 Hence x = { -1, 5 } 8. The number of elements in a set is the total number of elements in a particular or given set , say A, and it is denoted by n(A). It reads numbers of elements in set A or cardinality of set A. Consider the set P = {2, 3, 5, 7, 11} . :. n(P) =5 .Also, ifM = {Datsun, Toyota, Mazda, Nisan} . :. n(P)=4 9. Equivalent sets Two sets are said to be equivalent if they have equal number of elements Let P = {a, b, c} and 2 9

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Q = {mango, apple, guava} Since each set contains three elements, P is equivalent to Q i.e. P ≡ Q since n(P) = n(Q) Example Determine whether the following are equal or not  C = {4, 6, 2, 8} E = {2, 4, 6, 2, 8, 4}  M = {x : x is factors of 12} N = {1, 3, 2, 12, 6, 4} Solution a) C = {4, 6, 2, 8} and E = {2, 4, 6, 2, 8, 4}. In E = {2, 4, 6, 2, 8, 4} 2 and 4 are repeated, remember that any element that is repeated is counted once only . Now, N (C ) =4 and n(E) = 4 . Hence, n ( C) = n(E) C ≡ E. b) M = {x : x is factors of 12} : . M = {1, 2, 3, 4, 6, 12} and N = {1, 3, 2, 12, 6, 4} Both sets have the same number of elementsi.e n(M)=n(N)=6. Since n(M) =n(N) M ≡ N. 3 0

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 3 1

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) WEEK 9 SET OPERATIONS :Union and Intersection Of Sets . 1. Union of sets (U) The union of sets A and B is the set of all members that belong to A or B or to both A and B.It is denoted by U. The Union of sets A and B can be written as A Illustrations : If A = {a, b, c, d} and B = {b, d, e, f}then, A U B = {a, b, c, d, e, f} 2. Intersection of Sets The intersection of sets A and B is the set of members that are common to both A and B. The Intersection of A and B is usually written as A∩ B when ‘∩’ is the symbol used to denote intersection. Illustrations : If A = {a, b, c, d} and B = {a, c, e}, then the elements common to A and B are “a and c” . Hence, A ∩ B = {a, c} If A = {2, 5, 8, 11, 14, 17, 20} B = {2, 3, 5, 7, 9, 11} C = {5, 11, 17} and D = {1, 2, 3, 4} Using Venn diagram find (a) A U C (b) A ∩ B (c) C U D (d) Present the answers on Venn Diagram Solution: A = {2, 5, 8, 11, 14, 17, 20} B = {2, 3, 5, 7, 9, 11} C = {5, 11, 17} and D = {1, 2, 3, 4} (a) A U C = {2, 5, 8, 11, 14, 17, 20}.

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Here , A and C have same elements in common. Hence, the Venn diagram becomes (b) A ∩ B= {2, 5, 11} A ∩B A8 2 B 14 5 3 11 7 8ƹ 17 20 ( c ) C U D= {1, 2, 3, 4, 5, 11, 17} Hence, the two sets C and D are disjoint because they have no elements in common. C D 5 11 12 ƹ 17 34 3. Difference of Sets AUC 8 A 2 ƹ 5C 17 11 14 330 230

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) The difference of two sets of B and D denoted by B – D (i.e. B minus D or B difference D) is the set of members which only belongs to B but do not belong to D e.g Let B = { 2, 3, 5, 7} and D ={1, 2, 4, 8 } :. B – D ={ 3, 5, 7} 4. Disjoint Sets Two sets A and B are said to be disjoint, if they have no common members i.e. if no member of A is in B and vice versa. In other words, two sets A and B are said to be disjoint if and only if A ∩ B = φ For example (a) If A = {4, 7, 11, 13, 16} and B = {2, 3, 5, 8} then A and B are disjoint because they have no members in common. : . A ∩ B =φ This is shown in the Venn diagram below : A B A∩B = φ 4 7 10 23 13 16 58 ƹ (b) If Q = {Odd numbers} and M = {multiples of 4} then, Q = { 1, 3, 5, 7, 9 …} M = {4, 8, 12, 16, 20, 24 …} Notice that multiple of 4 are even numbers, since sets Q and M cannot have common elements. : .Q∩M =φ. 5. Complement of Sets The members in the universal set ƹ that are not in set A are called the complement of set A and is denoted by A’ or Ac . 34

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Illustration: If ƹ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {6, 7, 8, 9, 10} Then A’ = {1, 2, 3, 4, 5} The Venn diagram of the above relationship is shown below. The shaded portion represents A’ 3 67 4 2 8 9 10 ƹ 1 A A’ 5 ‘ Example If ƹ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} , A = {2, 3, 5, 7} ,B = {2, 4, 5, 7} and D = {1, 2, 5, 7} , find (a) A’ (b) (A ∩ B)’ (c) (A U D)’ Solution (a) The set of members in ƹ which are not in A is given by A’ : . A’ = {1, 4, 6, 8, 9, 10} (b) (A ∩ B)’ A ∩ B = {2, 5, 7} : . (A∩B)’ = {1, 3, 4, 6, 8, 9, 10} (c) (A U D)’ A U D = {1, 2, 3, 5, 7} : . (A U D)’ = {4, 6, 8, 9, 10} WEEK 10: VENN DIAGRAM AND APPLICATION UP TO 3 SET PROBLEMS Definition : 3 5

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Venn diagram is a diagrammatical way of solving problems on set. It consists of circles and rectangular box. Example1 : In a class of 30 students 25 like football and 15 like wrestling, every pupil like at least one of these sports. How many students like both sports? Solution : Let F represent football, and W represent wrestling :.n (F) = 25 n(W) =15 n (ƹ ) = 30, Let the number of students who like both sports be x. :. n(F∩W) = x Also, number of students who like Football only = 25 – x i.e. n(F∩W)’ = 25 – x and number of students who like wrestling only = 15 – x : . n(F∩W) = 15 – x n(ƹ) = 30 Then the total number of students in the class is 30 = (25 – x) + x + (15 – x) 30 = 25 – x + x + 15 – x 30 = 40 – x x = 40-30 X = 10 Therefore, 10 students like both sports. Solving problems involving three sets Example2 : F W 25 - x x 15 - x ƹ 36

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 2000 people were asked which make of cars would they like to have. 300 saidPeugeot(P) only, 500 said Toyota (T) only, 450 said Datsun (D) only , 200 said Peugeot and Toyota, 180 said Peugeot and Datsun and 250 said Toyota and Datsun.If 420 said none of these cars, (a). draw a Venn diagram to illustrate this information and hence find the number of people who like the three types of cars. (b). How many people like at least two of these cars? (c). How many people like both Toyota and Datsun? Solution : (a). Let the number who like the three types of cars be xn(P∩T∩D) = x Those who like Peugeot only = 300n(P∩T’∩D’) = 300 Toyota onlyn(P’∩T∩D’) = 500 Datsun onlyn(P’∩T’∩D) = 450 Peugeot and Toyotan(P∩T∩D’) = 200 – x Toyota and Datsunn(P’∩T∩D) = 250 – x Peugeot and Datsunn(P∩T’∩D) = 180 – x Those who like none of these carsn(P’∩T’∩D’) or n(P∩T∩D)’ = 420 n (ƹ) = 2000 Now, these information are shown in the Venn diagram below; 3 7

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) P 200-x T 300 500 x ƹ 180-x 250-x 450 420 D Since n (ƹ ) = 2000, we have 300 + 500 + 450 + 200 – x + 180 –x + 250 – x + x + 420 = 2000 2300 – 2x = 2000 Collecting like terms, we have -2x = 2000 – 2300 -2x = -300 Dividing both sides by -300, we have x= x = 150 Hence, the number of people who like the three types of cars is 150 b) Those who like at least two of the cars = 200 – x + x + 250 – x + 180 – x = 630 – 2x Substituting for x = 150 from ( a ) above, we have = 630 – (2 x 150) 38

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) = 630 – 300 = 330 : . those who likes at least two types of car is 330 b) From the Venn diagram, those who like both Toyota and Datsun = 250 – x = 250 – 150 = 100 CLASSWORK In a certain class, 22 students take one or more of Chemistry (C), Economics (E) and Government (G). 12 take Economics (E), 8 take Government (G) and 7 take Chemistry (C). Nobody takes Economics and Chemistry and 4 students take Economics and Government. a (i). Using set notation and letter indicated above write down the two statements in the last sentences (ii). Draw a Venn diagram to illustrate the information b. How many students take: (i). Both Chemistry and Government (ii). Government only 3 9

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) SECOND TERM 1. SIMPLE LINEAR EQUATION Behavioral Objectives By the end of the lesson student should be able to: i. Identify and solve linear equations ii. Solve linear equations with brackets and fractions iii. Substitute values into a given formula iv. Change the subject of a given formula Illustrations : An equation is simply a statement showing that two algebraic expressions are equal in value. A. Equation with brackets Examples : 1. Solve the equation 3(4c – 7) – 4(4c – 1) = 0 Solution 3(4c – 7) – 4(4c – 1) = 0 Remove the brackets, we have 12c – 21 – 16c + 4 = 0 –4c – 17 = 0 Add 17 to both sides, we have –4c = 17 Divide both sides by –4,we have c= c = – 4¼ B. Equations with Fractions Examples 1. Find the value of x in the equation 40

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) To clear the fraction , multiply through by the l.c.m of 5 and 4 i.e 20 x20 x20 – 2x20 2(6x -1) x 4 = 3(3x+2) x 5 – 40 8(6x –1) = 15(3x + 2) – 40 Clear brackets, we have 48x – 8 = 45x + 30 – 40 Collect like terms, we have 48x – 45x = 30 – 40 + 8 3x = – 2 Dividing both sides by 3, we have x =- C. Change of subject of formulae (Transposition) To transpose a formula, it means to rearrange it so that a different letter becomes the subject Examples 1. Make x the subject of the formula a = b(1–x) Solution: a = b(1–x) By opening the bracket, we have a = b - bx To make x the subject , we have a - b = –bx -bx = a - b Divide 42

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 2x3 p =( 2x3 x y2) + 2x3 x 2x3 p = 2x3 x y2 + 1 To make x the subject, we have 2x3 p – 2x3 x y2 = 1 Factorizing, we have x3(2p – 2y2) = 1 Dividing both sides by 2p – 2y2 , we have x3 = Taking cube root of both sides, we have 1 x=3 2 2(py ) 1 x= 3 2(py2) D. Substitution in formulae Example Calculate the value of x in the equation mn q = if q = 4, m = –5, n = 11 and y = –8.5 yx Substituting q = 4, m = –5, n = 11 and y = –8.5 into the given equation, we have 511 4= 44

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 8.5x Eliminate the fraction by cross multiplication 4(–8.5 – x) = –5 + 11 Remove the bracket, we have –34 –4x = 6 Collect like terms, we have –4x = 6 + 34 –4x = 40 Divide both sides by -4, we have x= : . x = –10 CLASSWORK Solve the following equations 1. 4a – (3–a) = 17 2. 8n – (5n + 13) = 7 3. x x 1 23 2 4. 2d 7 d 5 0 63 5. Make x the subject of the following equations a. ax + bx = c b. a(b–x) = cx c. a = 2b3x 3b2x 6. If y = 2x2 – 5x – 3, find the value of y when a. x = –1 (b)x = 0 (c)x = 1 2. VARIATIONS Behavioral Objectives 4 5

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) By the end of the lesson, students should be able to solve problems involving direct, inverse, joint and partial variations. Definition : Variation is the relationship that exits between two or more quantities in which a change in one quantity leads to change in the others. Variations can be classified into direct, inverse, joint and partial variations. A. Direct Variation Y varies directly as x is written as y α x, yα x also means y is directly proportional to x, meaning y=kx Where k is a constant of proportionality or simply a constant. Example1: If y α h and h = 16 when y =9 , Find (a) The relationship between y and h (b) The value of y when h = 36 (c) The value of h when y = 27 Solution (a) If y α h Then y = k h ( where k is a constant ) Substitute h = 16and y = 9 into the relation, we have 9 = k 16 9 = 4k k =  The required relation is y =  h (b) When h = 36, then the above relation is y= 36 y = ±2.25 x 6 y = ±13.5 46

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) (c) When y = 27, then the relation becomes 27 = h Cross multiply by 4 to clear the fraction, we have 27 x 4 = 9 h Divide both sides by 9, we have 27 x 4/ 9 = h 12 = h Square both sides, we have ( h )2 = 122 : . h = 144 B. Inverse Variations y varies inversely as x is written as y α :. y = (where k is constant ) Example : Ifp is inversely proportional to 3 q , Find a. the constant of proportionality when p=8 and q = 27 b. the equation connecting p and q c. the value of q when p = 3 Solution : a. p α 1 then 3q k p = where k is constant 3 q When p = 8, q = 27 then k 8=3 4 7

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 27 k 8= 3 Cross multiply, we have k = 8 x 3 :. k = 24 b. The required equation of the variation is 24 p= 3q c. To find the value of q when P = 3,recall that the equation connecting d. p and q is 24 p= 3q When p=3 then, we have 24 3= 3q Cube both sides, we have 33= 241/33 q )3 33 = 243 q 48

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) Cross multiply, we have 33 q = 243 Divide both sides by 3 3 ,we have q = 243/33 q= q=8x8x8q = 83 : .q = 512 C. Joint Variation In joint variation, three or more variables are involved for example, if A varies directly as F and inversely as M then A or A = , Where K is a constant. Therefore, joint variation is a combination of direct and inverse variation. Note that (a) If p α q, then q α p (b) If p α 1/q then q α 1/p (c) If p α q2 then q α p1/2etc Example P varies directly as Q and inversely as the square of R and Q = 40 When R = 25 and P = 2/5 (a) Find the formula connecting the variables (b) Find R when Q = 8 and P = 2 Solution P α Q and P α 1/R 2  P α Q/R2 : . P = KQ/R2 Where K is a constant (a). To get the formula connecting the variables : P = KQ/R2 When Q = 40, R = 25and P = 2/5 , we have 4 9

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 2/5 = 40K/252 Cross multiply to clear the fraction, we have 5 x 40K = 2 x 252 200K = 2x 625 Divide both sides by 200, we have K = 2 x 625/200 K : . The required formula isP = 25Q/4R2 (b) To find the value of R when Q = 8 and P = 2 using the formula P = 25Q/4R2 , we have 2 = 25 x 8/4R2 2 = 25x2/R2 Cross multiply byR2, we have 2R2 = 25 x 2 Divide both sides by 2, we have R2 = 25 x 2/2 R2 = 25 Taking square root of both sides, we have R = 25 R=5 Partial Variation When a variation is expressed as a sum of two or more parts it is called partial variation. Illustration: If y varies directly as x and partly varies inversely as P2, then we write 50

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) yα x and y α 1/p2 or y = ax + ( where a and b are constants of proportionality). Example: X is partly constant and partlyvaries as y. When y=5, x = 7 and when y = 8, x = 10. Find the relationship between y and x. Hence, find the value of x when y = 15. Solution : xα k + y x = K + c y ( where k and c are constants) To find the constants in the formula, we substitute Y = 5 and x = 7, we have 7 = k + 5 c ………..( i ) and also y =8 when x = 10. We have 10 = k + 8 c…………( ii) To eliminate one of the constants , we subtract equation (i) from equation (ii) i.e _ 10 = k + 8 c…………( ii) 7 = k + 5 c ………..( i ) 3 = 0 + 3c 3 = 3c Divide both sides by 3, we have 5 1

AJUMOSE LECTURE NOTES (MATHEMATICS SS1) 1= c :. c = 1 Put c = 1 into either (i) or ( ii), we have 10 = k + 8 c…………( ii) 10 = k + 8x1 10 = k + 8 Collect like terms, we have 10-8 = k 2 =k :. k = 2 The relationship between y and x in the formulax = K + c y is x = 2 + 1 y or x = 2 + y To find the value of x when y = 15, we use the formula x = 2 + y and we have X = 2 + 15 :. X = 17 52