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SS1: MATHEMATICS - 1ST TERM

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  1. Number Base System I | Week 1
    6 Topics
    |
    2 Quizzes
  2. Number Base System II | Week 2
    3 Topics
  3. Number Base System III | Week 3
    2 Topics
    |
    1 Quiz
  4. Modular Arithmetic I | Week 4
    2 Topics
  5. Modular Arithmetic II | Week 5
    3 Topics
  6. Modular Arithmetic III | Week 6
    4 Topics
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    1 Quiz
  7. Indices I | Week 7
    3 Topics
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    1 Quiz
  8. Indices II | Week 8
    1 Topic
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    1 Quiz
  9. Logarithms I | Week 9
    3 Topics
  10. Logarithms II | Week 10
    4 Topics
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    1 Quiz

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Topic Content:

  • Addition & Subtraction in Modular Arithmetic
Screenshot 2023 09 14 at 06.42.14

The diagram above shows addition in modulo 6 in which an element on the extreme left is added to an element on the top row, in that order. You will observe that the first column on the extreme left and the top row contain the same set of integers.

Example 5.2.1:

\( \scriptsize 3 \: \bigoplus \: 2 \)

Solution

Look for the number 3 on row 4 and the number 2 on column 3, they intersect at number 5.

Screenshot 2023 09 14 at 06.44.12

i.e.  \( \scriptsize 3 \: \bigoplus \: 2 = 5\:(mod\:6) \)

In the same vein, subtraction can be considered as the inverse operation of addition. So, in mod 6, we can use the addition table to work out subtraction:

Example 5.2.2:

\( \scriptsize 2 \: \circleddash \: 5 \)

Solution

 

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Solution (i)

⇒ \( \scriptsize 54 \: \bigoplus \:  65 \: (mod\: 6) \)

Step 1

Add 54 and 65

i.e. 54 + 65 = 119 (mod 6)

Step 2

Take the answer of the addition to mod 6

= \( \frac{119}{6} \)

= 19 r 5

Step 3

119 = 19 × 6 + 5 (mod 6)

= 0 + 5 (mod 6)

∴ 119 (mod 6) = 5 (mod 6)

Solution (ii)

⇒ – 25 (mod 8)

=  (– 4 × 8) + 7 (mod 8)

= – 32 + 7 (mod 8)

= 0 + 7 (mod 8)

∴  – 25 (mod 8) = 7 mod 8

Solution (iii)

⇒ – 56 (mod 12)

= (– 5 × 12) + 4 (mod 12)

= – 60 + 4 (mod 12)

= 0 + 4 (mod 12)

∴   – 56 (mod 12) = 4 (mod 12)

Solution (iv)

⇒ 2 ⊖ 11 (mod 5)

2 – 11 (mod 5) = – 9 (mod 5)

=  – 5  × 2 + 1 (mod 5)

=  – 10 + 1 (mod 5)

= 0 + 1 (mod 5)

∴  2 – 11 (mod 5) = 1 (mod 5)

Solution (v)

⇒ 12 ⊖ 5 (mod 6)

12 – 5 (mod 6) = 7 mod 6

= 6 + 1 (mod 6)

= 0 + 1 (mod 6)

∴   12 – 5 (mod 6) = 1 (mod 6)

Solution (vi)

⇒ 5 ⊖ 20 (mod 11)

5 ⊖ 20 (mod 11) = – 15 (mod 11)

          =  –2 × 11 + 7 (mod 11)

          =  – 22 + 7 (mod 11)

           = 0 + 7 mod 11

∴     5 ⊖ 20 (mod 11) = 7 (mod 11)

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