Lesson Note on Mathematics

Modular Arithmetics

week 3-5

Subject: Mathematics

Theme: Modular Arithmetics

Topic: congruence and Residues of Modular

Sub Topic: Operation with Modula Arithmetic (Residues)

Date: dd/mm/yyyy

Class: S.S.S 1

Duration: 30 Minutes

No of Learners: 30

Learning Objectives:

By the end of the lesson learners should be able to:
  1. Defined Modular arithmetic.

    Is defined as the arithmatic of remainder where an integer is divided by a fixed non-zero integer. Moduli is the plural of modulus.
    In general, in modular arithmetic, you do arithmetic with the numbers from 1 to n as normal, and if your answer goes outside that range you add or subtract multiples of n to get back into the range.
    The number n is called the modulus, and the calculations are said to be done modulo n or mod n.
    Instead of the usual equality sign (=), the symbol ≡ is used in modular arithmetic.
    E.g
    b ≡ r(mod k)
    2 ≡ 7(mod 5)
    6 ≡ 13(mod 7)
  2. Defined the congruent and residue of modular.

    Two integers are congruent to each other concerning a module if their difference is a multiple of the modulus.
    Or
    If two integers b and r have the property that b-r is divided by k, then b and r are called congruent modulo k while r is called the residue written as
    b ≡ r(mod k)
    NOTE; 0, 1, 2, 3, 4, --- k-1 is the set of residues mod k.
    i.e, in mod 6, the largest residue is 5
    in mod 10, the largest residue is 9
    in mod y, and the largest residue is y-1
    Examples

    1. 2 ≡ 7(mod 5)
      2-7 = -5
      -5÷5 = -1
      Hence; 2 ≡ 7(mod 5) is a congruence of mod 5

    2. 6 ≡ 13(mod 7)
      6-13 = -7
      -7÷7 = -1
      Hence; 6 ≡ 13(mod 7) is a congruence of mod 7

    3. 25 ≡ 11(mod 7)
      25-11 = 14
      14÷7 = 2
      Hence; 25 ≡ 11(mod 7) is a congruence of mod 7

    4. 13 ≡ 1(mod 12)
      13-1 = 12
      12÷12 = 1
      Hence; 13 ≡ 1(mod 12) is a congruence of mod 12

    5. 15 ≡ 0(mod 15)
      15-0 = 15
      15÷15 = 1
      Hence; 15 ≡ 0(mod 15) is a congruence of mod 15

    6. 5 ≡ -5(mod 5)
      5-(-5) = 10
      10÷5 = 2
      Hence; 5 ≡ -5(mod 1) is a congruence of mod 5

    7. -3 ≡ 11(mod 4)
      -3-11 = -8
      -8÷4 = -2
      Hence; -3 ≡ 11(mod 4) is a congruence of mod 4

    8. -3 ≡ 7(mod 4)
      -3-7 = -10
      -10÷4 = 2.5
      Hence; -3 ≡ 7(mod 4) is NOT a congruence of mod 4

  3. explain and compute a residue that is equivalent to a given residue.

    When two residue leaves the same remainder upon division by the modulus, the such residue is said to be equivalent residue.
    i.e Whenever two residues x and y are found to be equivalent relative to a fixed modulus k, we write; that if

    x ≡ y(mod k) then,
    y ≡ x(mod k)
    where nk ≡ 0
    where n is an integer


    Example
    1. Show that 5 and 2 are equivalent in mod 3.
      Solution
      5 ≡ 1x3+2(mod 3) ----- the residue is = 2
      2 ≡ 0x3+2(mod 3) ----- the residue is =2
      Hence 5 and 2 are equivalent in mod 3.

    2. Find the positive integer of the least absolute value that is equivalent to 33 in mod 7.
      Solution
      33 ≡ 4x7+5(mod 7)
      5 ≡ 0x7+5(mod 7)
      Therefore 33 ≡ 5 (mod 7)
      Hence, the integer is 5

    3. Find the smallest integer than 0 that is equivalent to
      1. 34 in mod 5
      2. 6 in mod 9
      3. 130 in mod 12

      Solution
      1. 34 ≡ 6x5+4(mod 5)
        4 ≡ 0x5+4(mod 5)
        Therefore 34 ≡ 4 (mod 5)

      2. 6 => 6+9 = 15 (you can add multiple modular to an integer if the integer is less than the modular)
        Therefore, 15 ≡ 1x9+6(mod 9)
        6 ≡ 0x9+6(mod 9)
        Therefore 15 ≡ 6(mod 9)

      3. 130 ≡ 10x12+10(mod 12)
        10 ≡ 0x12+10(mod 12)
        Therefore 130 ≡ 10(mod 12)
  4. carry out basic operations (addition, subtraction, multiplication and division) of residue


    1. ADDITION IN MODULAR

      Mod 7 Addition Table

      0 1 2 3 4 5 6
      0 0 1 2 3 4 5 6
      1 1 2 3 4 5 6 0
      2 2 3 4 5 6 0 1
      3 3 4 5 6 0 1 2
      4 4 5 6 0 1 2 3
      The symbolshow addition in modular arithmetic. This shows that it is different from ordinary arithmetic.
      In the table, multiples of 4 are ignored and remainders are written.
      1. Evaluate 5 6 in mod 8
        Solution
        5 6 = 11
        Therefore, 11 ≡ 1x8+3(mod 8)
        11 ≡ 3(mod 8)
        Therefore 5 6 ≡ 3(mod 8)

      2. Simplify 1. 12 7 and 2. 13 9 given that all the residues are in mod 5.
        Solution
        1. 12 7 in mod 5
          12 7 = 19
          19 ≡ 3x5+4(mod 5)
          19 ≡ 4(mod 5)
          Therefore 12 7 ≡ 4(mod 5)

          ALTERNATIVELY
          12 7 in mod 5
          12 in mod 5 is 12 ≡ 2x5x2(mod 5)
          7 in mod 5 is 7 ≡ 1x5+2(mod 5)
          Therefore
          12 ≡ 2(mod 5)
          7 ≡ 2(mod 5)
          Hence, 12 7 in mod 5
          12 7(mod 5) ≡ 2 2(mod 5)
          12 7(mod 5) ≡ 4(mod 5)


        2. 13 9(mod 5)
          Solution
          13 9 = 22
          22 ≡ 4x5+2(mod 5)
          22 ≡ 2(mod 5)
          Therefore, 13 9(mod 5) ≡ 2(mod 5)
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    3. SUBTRACTION IN MODULAR

      Mod 6 Subtraction Table

      0 1 2 3 4 5
      0 0 5 4 3 2 1
      1 1 0 5 4 3 2
      2 2 1 0 5 4 3
      3 3 2 1 0 5 4
      4 4 3 2 1 0 5
      5 5 4 3 2 1 0
      The symbol show subtraction in modular arithmetic. This shows that it is different from ordinary arithmetic.
      In the table, multiples of 4 are ignored and remainders are written.
      NOTE: 2 1≠ 1 2
      1. Evaluate 8 3 (mod 6)
        SOLUTION
        8 3 (mod 6) ≡ 5(mod 6)

        ALTERNATIVELY
        8 3 (mod 6)
        8(mod 6) ≡ 1x6+2(mod 6)
        8(mod 6) ≡ 2(mod 6)
        3(mod 6) => 3 6(mod 6)
        9(mod 6) ≡ 1x6+3(mod 6)
        :. 3(mod 6) ≡ 3(mod 6)
        Hence, 8 3 (mod 6) ≡ 2(mod 6) 3 (mod 6)
        8 3 (mod 6) ≡ 2 3 (mod 6)
        8 3 (mod 6) ≡ -1(mod 6)
        -1(mod 6) => -1 6(mod 6)
        -1(mod 6) => 5(mod 6)
        Hence, 8 3 (mod 6) ≡ 5(mod 6)

      2. Simplify 23 2in mod 7
        SOLUTION
        23 2 (mod 7) => 21(mod 7)
        21(mod 7) ≡ 3x7+0(mod 7)
        21(mod 7) ≡ 0(mod 7)
        :. 23 2 (mod 7) ≡ 0(mod 7)

        ALTERNATIVELY
        23 2 (mod 7)
        23(mod 7) ≡ 3x7+2(mod 7)
        23(mod 7) ≡ 2(mod 7)
        2(mod 7) ≡ 2(mod 7)
        :. 23 2 (mod 7) ≡ 2(mod 7) 2 (mod 7)
        23 2 (mod 7) ≡ 2 2 (mod 7)
        23 2 (mod 7) ≡ 0(mod 7)

      3. Find the the inverse, concerning addition, of the following residues.
        1. -5(mod 5)
        2. 5(mod 10)
        3. -7(mod 8)
        4. 3(mod 13)

        SOLUTION
        1. -5(mod 5)
          the inverse of -5 is 5
          5(mod 5) ≡ 0(mod 5)

        2. 5(mod 10)
          the inverse of 5 is -5
          -5 10(mod 10) ≡ 5(mod 10)
          :. 5(mod 10) ≡ 5(mod 10)

        3. -7(mod 8)
          the inverse of -7 is 7
          7(mod 8) ≡ 7 (mod 8)

        4. 3(mod 13)
          the inverse of 3 is -3
          -3(mod 13) ≡ 3 13(mod 13)
          -3(mod 13) ≡ 10(mod 13)
    4. MULTIPLICATION IN MODULAR

      Mod 7 Multiplication Table

      0 1 2 3 4 5 6
      0 0 0 0 0 0 0 0
      1 0 1 2 3 4 5 6
      2 0 2 4 6 1 3 5
      3 0 3 6 2 5 1 4
      4 0 4 1 5 2 6 3
      5 0 5 3 1 6 4 2
      6 0 6 5 4 3 2 1
      The symbol show multiplication in modular arithmetic. This shows that it is different from ordinary arithmetic.
      In the table, multiples of 4 are ignored and remainders are written.

        Evaluate
      1. 5 6in mod 10
        SOLUTION
        5 6in mod 10
        30 in mod 10
        30 ≡ 5x6+0 (mod 10)
        30 ≡ 0 (mod 10)
        :. 5 6≡ 0 (mod 10)

      2. 5 6in mod 14
        SOLUTION
        30 in mod 14
        30 ≡ 2x14+2 (mod 14)
        30 ≡ 2 (mod 14)
        :. 5 6≡ 2 (mod 14)

      3. 7 1 9 in mod 3
        SOLUTION
        7 1 9 in mod 3
        133 in mod 3
        133 ≡ 44x3+1 (mod 3)
        133 ≡ 1 (mod 3)
        :. 7 1 9 ≡ 1 (mod 3)

        ALTERNATIVELY
        7 1 9 in mod 3
        7 ≡ 2x3+1 (mod 3)
        19 ≡ 6x3+1 (mod 3)
        :. 7x19 ≡ 1x1 (mod 3)
        7x19 ≡ 1 (mod 3)
    5. DIVISION IN MODULAR

      The modular division is trickier. We start with the idea that division is supposed o be the opposite of multiplication. The concept of multiplication inverse of the denomination is used in modular division.
      m n ≡ m n -1

      Mod 7 Division Table

      0 1 2 3 4 5 6
      0 NA 0 0 0 0 0 0
      1 NA 1 4 5 2 3 6
      2 NA 2 1 3 4 6 5
      3 NA 3 5 1 6 2 4
      4 NA 4 2 6 1 5 3
      5 NA 5 6 4 3 1 2
      6 NA 6 3 2 5 4 1
      The symbol show division in modular arithmetic. This shows that it is different from ordinary arithmetic.
      Division in modular arithmetic may give more than one value or even no value at all.

      If we can find two residues say m and n in mod k such that;
      m n≡ 1 (mod k)
      Then, n ≡ m-1 (RECAL: from indices) if n = m-1
      then n = 1/m
      and n x m = 1

      HENCE, n ≡ m-1
      We refer to n as the multiplicative the inverse of m in mod k. Therefore, if m is the multiplicative the inverse of n then n is also the multiplicative the inverse of m in the same modulus.
      :. n m ≡ n m-1
      In the mod 7 multiplication table above, the multiplication the inverse of the residues is obtained from the table.

      • 1-1 ≡ 1
      • 2-1 ≡ 4 hence 4-1 ≡ 2
      • 3-1 ≡ 5 hence 5-1 ≡ 3
      • 4-1 ≡ 2 hence 2-1 ≡ 4
      • 5-1 ≡ 3 hence 3-1 ≡ 5
      • 6-1 ≡ 6
      • 0 has no the inverse

        Examples:
      1. Evaluate 4 3 in mod 7
        SOLUTION
        4 3 ≡ 4 3 -1
        But 3-1 ≡ 5 (mod 7) "from the mod 7 multiplication table"
        :. 4 3 -1 ≡ 4 5 ≡ 20
        :. 4 3 (mod 7) ≡ 2x7+6 (mod 7)
        4 3 (mod 7) ≡ 6 (mod 7)

      2. Evaluate 4 3 in mod 5
        SOLUTION

        4 3 ≡ 4 3 -1
        But 3-1 ≡ 2 (mod 5) "from the mod 5 multiplication table"
        :. 4 3 -1 ≡ 4 2 ≡ 8
        :. 4 3 (mod 5) ≡ 1x5+3 (mod 7)
        4 3 (mod 5) ≡ 3 (mod 5)

      3. Evaluate 9 5 in mod 7
        SOLUTION

        9 5 ≡ 9 5 -1
        But 5-1 ≡ 3 (mod 7) "from the mod 7 multiplication table"
        :. 9 5 -1 ≡ 9 3 ≡ 27
        :. 9 5 (mod 7) ≡ 3x7+6 (mod 7)
        9 5 (mod 7) ≡ 6 (mod 7)

      4. Evaluate 6 2 in mod 8
        SOLUTION

        6 2 ≡ 6 2 -1
        But 2-1 in mod 8 does not exist "from the mod 8 multiplication table"
        Hence,
        6 2 (mod 8) has no solution.
    6. solve modular arithmetics with more than one operation.

      EXAMPLES
      1. Simplify 7 5 in mod 4
        SOLUTION
        7 5 (mod 4)
        Applying BODMAS
        2 5 7 (mod 4)
        7 7 (mod 4)
        ≡ 0 (mod 4)

      2. Simplify 2 5 4 (mod 7)
        SOLUTION
        3 2 5 4 (mod 7) ≡ (3 2) (5 4)
        3 2 5 4 (mod 7) ≡ 6 (5 4-1)
        But 4-1≡ 2 (mod 7)
        :. 3 2 5 4 (mod 7) ≡ 6 (5 2)
        3 2 5 4 (mod 7) ≡ 6 10
        3 2 5 4 (mod 7) ≡ -4
        3 2 5 4 (mod 7) ≡ -4 7 (mod 7)
        :. 3 2 5 4 (mod 7) ≡ 3 (mod 7)

      3. Simplify (3 7 1) 5 (mod 9)
        SOLUTION
        (3 7 1) 5 (mod 9)
        (3 7 1) 5-1 (mod 9)
        (21 1) 5-1 (mod 9)
        But 5-1 ≡ 2 (mod 9) that is; from the mod 9 multiplication table

        :. (21 1) 5-1 (mod 9) ≡ 22 2 (mod 9)
        44 (mod 9) ≡ 4 9 8 (mod 9)
        :. (3 7 1) 5 (mod 9) ≡ 8 (mod 9)

Rationale:

Modular arithmetic, also known as clock arithmetic, every time you tell the time, you are dealing with the modulus 12. There exist several events, conditions or quantities that usually vary in a sequence such that, whenever a certain fixed amount is attained, the sequence is repeated. Most of these events or conditions are related to time, such as
  • 1 rotation of the hour hand of the clock or records 12 hrs (half day).
  • 1 rotation of the minute hand of the clock or watch records 60 mins or 1 hr.
  • 1 rotation of the second hand of the clock or watch records 60 secs or 1 minute.
  • 1 rotation of 7 days records 1 week
  • 1 rotation of 24 hours record 1 day
  • Except for the leap year, 1 rotation of 365 days records 1 year
  • 1 rotation of 12 months record 1 year

Prerequisite/ Previous knowledge:

The learners have learnt about the time.

Learning Materials:

Clock, calender, textbooks.

Reference Materials:

  1. Man Mathematics for Senior Secondary Schools 1
  2. New General Mathematics Textbook for West Africa
  3. OnlineResources


Lesson Development:

STAGE
TEACHER'S ACTIVITY
LEARNER'S ACTIVITY
LEARNING POINTS
INTRODUCTION
full class session (10 mins)
Asking some questions:
  1. How many seconds make one minute?
  2. How many minutes makes one hour?
  3. How many hours make one day?
  4. How many days makes one week?
  5. How many weeks make one month?
  6. How many weeks make one year?
  7. How many months make one year?
  8. How many days makes one year?
Learners' expected response.
  1. 60 seconds makes one minute?
  2. 60 minutes makes one hour?
  3. 24 hours makes one day?
  4. 7 days makes one week?
  5. 4 weeks makes one month?
  6. 52 weeks makes one year?
  7. 12 calendar months or 13 lunar months make one year?
  8. 365 days makes one year, 366 days makes one leap year?
Learner’s entry points.
Competences: Critical thinking, Communication.

We provide educational resources/materials, curriculum guide, syllabus, scheme of work, lesson note & plan, waec, jamb, O-level & advance level GCE lessons/tutorial classes, on various topics, subjects, career, disciplines & department etc. for all the Class of Learners

STEP 1
5 mins.
Development
Ask students to form three groups. Form three groups and share responsibility. Inculcating leadership skills, competitive spirit, cooperation, teamwork and a sense of responsibility among learners.
Teacher to provide learners with materials (Wall clock, calendar) and instructions and allow them to identify the materials


The teacher thereafter asks learners.
  1. 3 hours after 10.00 clock is what?
  2. 17 hours after 9.00 clock is what?
  3. 9 hours before 4.00 clock is what?



The teacher explains the concept of modular to students and thereafter guild the students to define modular.
  1. 3 hours after 10.00 clock is 1.00 clock because 3 + 10 = 13, and 13 = 1 x 12 + 1 (in 12 rotation hand of the clock)
  2. 17 hours after 9.00 clock is 2.00 clock because 17 + 9 = 26, and 26 = 2 x 12 + 2 (in 12 rotation hand of the clock)
  3. 9 hours before 4.00 clock is 7.00 clock because 4 - 9 = -5, and -5 + 12 = 7 (in 12 rotation hand of the clock)
Students take materials needed for this activity and identify the materials as a Wall clock, and calender.




Hence respond to the teacher's questions
  1. 3 hours after 10.00 clock is 1.00 clock
  2. 17 hours after 9.00 clock is 2.00 clock
  3. 9 hours before 4.00 clock is 7.00 clock







The students define modular;
Modular is an arithmetic of remainder where an integer is divided by a fixed non-zero integer.
Developing the concept of modular arithmetic
Step 2
3 mins
The teacher asks learners to calculate with the aid of the calendar.
If today is Friday, what day of the week will it be in 19 days?

The teacher explains the day it will be using modular if today is Friday.
If 0 day ≡ Sunday
then Friday ≡ 5
The day of the week 19 days after is 5 19 ≡ 24 (mod 7) 'i.e days of the week is a rotation of 7'
24 ≡ 3 7 3
:. 24 ≡ 3 (mod 7)
:. The next day is 3 days after Sunday
Hence; The expected day is Wednesday.
Learners expected respond

HINTS:
Sun (0), Mon (1), Tue (2), Wed (3), Thurs (4), Fri (5), Sat (6)

Students counted 19 days from Friday and stopped on the 24th of Wednesday.
Hence; The expected day is Wednesday.
STEP 3
congruent and residue of modular (5 mins)
The teacher explains the congruent and residue of modular, thus;
If two integers b and r have the property that b-r is divided by k, then b and r are called congruent modulo k while r is called the residue written as
b ≡ r(mod k)
NOTE; 0, 1, 2, 3, 4, --- k-1 is the set of residues mod k.
i.e, in mod 6, the largest residue is 5
in mod 10, the largest residue is 9
in mod y, and the largest residue is y-1

Example

13 ≡ 1(mod 12)
13-1 = 12
12÷12 = 1
Hence; 13 ≡ 1(mod 12) is a congruence of mod 12

-3 ≡ 7(mod 4)
-3-7 = -10
-10÷4 = 2.5
Hence; -3 ≡ 7(mod 4) is NOT a congruence of mod 4

The teacher asks students questions.
Which of the following statement is true
  1. 4 ≡ 3 (mod 7)
  2. 5 ≡ -5 (mod 15)
  3. 2 ≡ 4 (mod 4)
  4. 6 ≡ 3 (mod 3)
  5. 32 ≡ 2 (mod 5)
Students pay attention and ask questions.

Learners expected respond
  1. 4 ≡ 3 (mod 7)
    4 - 3 = 1
    1 ÷ 7 = 0.1428
    Hence; 4 ≡ 3 (mod 7) is NOT a congruence of mod 7

  2. 5 ≡ -5 (mod 15)
    5 - (-5) = 10
    10 ÷ 15 = 0.6666
    Hence; 5 ≡ -5 (mod 15) is NOT a congruence of mod 15

  3. 2 ≡ 4 (mod 4)
    2 - 4 = -2
    -2 ÷ 4 = -0.5
    Hence; 2 ≡ 4 (mod 4) is NOT a congruence of mod 4

  4. 6 ≡ 3 (mod 3)
    6 - 3 = 3
    3 ÷ 3 = 1
    Hence; 6 ≡ 3 (mod 3) is a congruence of mod 3

  5. 32 ≡ 2 (mod 5)
    32 -2 = 30
    30 ÷ 5 = 6
    Hence; 32 ≡ 2 (mod 5) is a congruence of mod 5
Communication.
congruent and residue of modular.
STEP 4
Equivalent residue of a modulus
5 mins
Ask students,
  1. what do you understand by equivalent?
  2. prompt students to say what equivalent residue is.



  3. The teacher explains equivalent residues to students, thus;
    Whenever two residue x and y are found to be equivalent relative to a fixed modulus k, we write; that if

    x ≡ y(mod k) then,
    y ≡ x(mod k)
    where nk ≡ 0
    where n is an integer

Students expected to respond.
  1. The equivalent is a set of data having the same value/solution set
  2. When two residue leaves the same remainder upon division by the modulus, the such residue is said to be equivalent residue.


Example
  1. Show that 5 and 2 are equivalent in mod 3.
    Solution
    5 ≡ 1x3+2(mod 3) ----- the residue is = 2
    2 ≡ 0x3+2(mod 3) ----- the residue is =2
    Hence 5 and 2 are equivalent in mod 3.


  2. Find the positive integer of the least absolute value that is equivalent to 33 in mod 7.
    Solution
    33 ≡ 4x7+5(mod 7)
    5 ≡ 0x7+5(mod 7)
    Therefore 33 ≡ 5 (mod 7)
    Hence, the integer is 5

Collaboration among group members, manipulation of results and observing the equivalent residue.
Step 5
Basic operation (addition, subtraction, multiplication and division) of residue
10 mins
The teacher guides learners on how to carry out the addition of residue.
  1. ADDITION IN MODULAR

    Mod 7 Addition Table

    0 1 2 3 4 5 6
    0 0 1 2 3 4 5 6
    1 1 2 3 4 5 6 0
    2 2 3 4 5 6 0 1
    3 3 4 5 6 0 1 2
    4 4 5 6 0 1 2 3
Learners pay attention to the teacher and follow the teacher's explanation.
    ADDITION IN MODULAR
  1. Evaluate5 4 in mod 7
    Solution
    5 4 = 9
    Therefore, 9 ≡ 1x7+2(mod 7)
    9 ≡ 2(mod 7)
    Therefore 5 4 ≡ 2(mod 7)

  2. Evaluate54 65 in mod 7
    Solution
    54 65 = 119
    Therefore, 119 ≡ 17x7+0(mod 7)
    119 ≡ 0(mod 7)
    Therefore 54 65 ≡ 0(mod 7)
Addition, Subtraction, Multiplication and Division of Residue.
The teacher guides learners on how to carry out subtraction of residue.
  1. SUBTRACTION IN MODULAR

    Mod 6 Subtraction Table

    0 1 2 3 4 5
    0 0 5 4 3 2 1
    1 1 0 5 4 3 2
    2 2 1 0 5 4 3
    3 3 2 1 0 5 4
    4 4 3 2 1 0 5
    5 5 4 3 2 1 0
Learners pay attention to the teacher and follow the teacher's explanation.
    SUBTRACTION IN MODULAR
  1. Evaluate 4 3 (mod 6)
    SOLUTION
    4 3 (mod 6) ≡ 1(mod 6)

  2. Evaluate 8 3 (mod 6)
    SOLUTION
    8 3 (mod 6) ≡ 5(mod 6)

  3. Evaluate 28 13 (mod 6)
    SOLUTION
    28 13 = 15
    Therefore, 15 ≡ 2x6+3(mod 6)
    15 ≡ 3(mod 6)
    28 13 ≡ 3(mod 6)
The teacher guides learners on how to carry out the multiplication of residue.
  1. MULTIPLICATION IN MODULAR

    Mod 8 Multiplication Table

    0 1 2 3 4 5 6 7
    0 0 0 0 0 0 0 0 0
    1 0 1 2 3 4 5 6 7
    2 0 2 4 6 0 2 4 6
    3 0 3 6 1 4 7 2 5
    4 0 4 0 4 0 4 0 4
    5 0 5 2 7 4 1 2 3
    6 0 6 4 2 0 2 4 2
    7 0 7 6 5 4 3 2 7
Learners pay attention to the teacher and follow the teacher's explanation.
    MULTIPLICATION IN MODULAR
  1. Evaluate 4 3 (mod 8)
    SOLUTION
    4 3 = 12
    Therefore, 12 ≡ 1x8+4(mod 8)
    12 ≡ 4(mod 8)
    4 3 ≡ 4(mod 8)

  2. Evaluate 27 31 (mod 8)
    SOLUTION
    27 31 = 837
    Therefore, 837 ≡ 104x8+5(mod 8)
    837 ≡ 5(mod 8)
    27 31 ≡ 5(mod 8)
The teacher guides learners on how to carry out the division of residue.
  1. DIVISION IN MODULAR

    Mod 7 Multiplication Table

    0 1 2 3 4 5 6
    0 0 0 0 0 0 0 0
    1 0 1 2 3 4 5 6
    2 0 2 4 6 1 3 5
    3 0 3 6 2 5 1 4
    4 0 4 1 5 2 6 3
    5 0 5 3 1 6 4 2
    6 0 6 5 4 3 2 1

    If we can find two residues say m and n in mod k such that;
    m n≡ 1 (mod k)
    Then, n ≡ m-1 (RECAL: from indices) if n = m-1
    then n = 1/m
    and n x m = 1

    HENCE, n ≡ m-1
    :. n m ≡ n m-1
    We refer to n as the multiplicative the inverse of m in mod k. Therefore, if m is the multiplicative the inverse of n then n is also the multiplicative the inverse of m in the same modulus.
    • 1-1 ≡ 1
    • 2-1 ≡ 4
    • 4-1 ≡ 2
    • 3-1 ≡ 5
    • 5-1 ≡ 3
    • 4-1 ≡ 2
    • 2-1 ≡ 4
    • 5-1 ≡ 3
    • 3-1 ≡ 5
    • 6-1 ≡ 6
    • 0 has no the inverse

    Mod 7 Division Table

    0 1 2 3 4 5 6
    0 NA 0 0 0 0 0 0
    1 NA 1 4 5 2 3 6
    2 NA 2 1 3 4 6 5
    3 NA 3 5 1 6 2 4
    4 NA 4 2 6 1 5 3
    5 NA 5 6 4 3 1 2
    6 NA 6 3 2 5 4 1
Learners pay attention to the teacher and follow the teacher's explanation.
    DIVISION IN MODULAR
  1. Evaluate 4 3 in mod 7
    SOLUTION
    4 3 ≡ 4 3 -1
    But 3-1 ≡ 5 (mod 7) "from the mod 7 multiplication table"
    :. 4 3 -1 ≡ 4 5 ≡ 20
    :. 4 3 (mod 7) ≡ 2x7+6 (mod 7)
    4 3 ≡ 6 (mod 7)

  2. Evaluate 6 6 in mod 7
    SOLUTION
    6 6 ≡ 6 6 -1
    But 6-1 ≡ 6 (mod 7) "from the mod 7 multiplication table"
    :. 6 6 -1 ≡ 6 6 ≡ 36
    :. 6 6 (mod 7) ≡ 5x7+1 (mod 7)
    6 6 ≡ 1 (mod 7)

  3. Evaluate 6 5 in mod 7
    SOLUTION
    6 5 ≡ 6 5 -1
    But 5-1 ≡ 3 (mod 7) "from the mod 7 multiplication table"
    :. 6 5 -1 ≡ 6 3 ≡ 18
    :. 6 5 (mod 7) ≡ 2x7+4 (mod 7)
    6 5 ≡ 4 (mod 7)

  4. Evaluate 9 5 in mod 7
    SOLUTION
    9 5 ≡ 9 5 -1
    But 5-1 ≡ 3 (mod 7) "from the mod 7 multiplication table"
    :. 9 5 -1 ≡ 9 3 ≡ 27
    :. 9 5 (mod 7) ≡ 3x7+6 (mod 7)
    9 5 ≡ 6 (mod 7)

  5. Evaluate 29 2 in mod 7
    SOLUTION
    29 2 ≡ 29 2 -1
    But 2-1 ≡ 4 (mod 7) "from the mod 7 multiplication table"
    :. 29 2 -1 ≡ 29 4 ≡ 116
    :. 29 2 (mod 7) ≡ 16x7+4 (mod 7)
    29 2 ≡ 4 (mod 7)

  6. Evaluate 4 35 in mod 7
    SOLUTION
    4 35 ≡ 35 4
    35 4 ≡ 35 4 -1
    But 4-1 ≡ 2 (mod 7) "from the mod 7 multiplication table"
    :. 35 4 -1 ≡ 35 2 ≡ 70
    :. 35 2 (mod 7) ≡ 10x7+0 (mod 7)
    35 4 ≡ 0 (mod 7)
Step 6
Modular arithmetics with more than one operation
The teacher explains and guides learners in the application of BODMAS to solve modular arithmetic with more than one arithmetics operation.
  1. Simplify 7 5 in mod 4
    SOLUTION
    7 5 (mod 4)
    Applying BODMAS
    2 5 7 (mod 4)
    7 7 (mod 4)
    ≡ 0 (mod 4)

  2. Simplify 2 5 4 (mod 7)
    SOLUTION
    3 2 5 4 (mod 7) ≡ (3 2) (5 4)
    3 2 5 4 (mod 7) ≡ 6 (5 4-1)
    But 4-1≡ 2 (mod 7)
    :. 3 2 5 4 (mod 7) ≡ 6 (5 2)
    3 2 5 4 (mod 7) ≡ 6 10
    3 2 5 4 (mod 7) ≡ -4
    3 2 5 4 (mod 7) ≡ -4 7 (mod 7)
    :. 3 2 5 4 (mod 7) ≡ 3 (mod 7)

  3. Simplify (3 7 1) 5 (mod 9)
    SOLUTION
    (3 7 1) 5 (mod 9)
    (3 7 1) 5-1 (mod 9)
    (21 1) 5-1 (mod 9)
    But 5-1 ≡ 2 (mod 9) that is; from the mod 9 multiplication table

    :. (21 1) 5-1 (mod 9) ≡ 22 2 (mod 9)
    44 (mod 9) ≡ 4 9 8 (mod 9)
    :. (3 7 1) 5 (mod 9) ≡ 8 (mod 9)
solving modular arithmetics with more than one operation
CONCLUSION
2 mins
Teachers wrap up the learners' observations,
summarise the learned knowledge, and give more clarification on the learned content.
Participate actively in summarizing the content.
Making short notes.
Listening skills, Writing skills.
Assessment/ Assignment:
5 mins
The teachers give learners a take-home and engage students to work individually.
  1. Calculate the the inverse (concerning multiplication) of the following
    1. 3 in mod 11
    2. 2 in mod 5
    3. 7 in mod 9
  2. Simplify the following
    1. 3 2-1 mod 5
    2. 2-1 13 mod 17
  3. Simplify the following
    1. 5 4 3 in mod 11
    2. 13 (8 3) in mod 12
Do exercises. Competences: Creativity.

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Facts about Teachers

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● ● ● Teachers are like candles, they burn themselves to light others.

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● ● ● Every great mind was once taught by some brilliant teachers.

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● ● ● The good teacher explains. The superior teacher demonstrates. The great teacher inspires.

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