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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
    |
    1 Quiz
  7. Indices I | Week 7
    3 Topics
    |
    1 Quiz
  8. Indices II | Week 8
    1 Topic
    |
    1 Quiz
  9. Logarithms I | Week 9
    3 Topics
  10. Logarithms II | Week 10
    4 Topics
    |
    1 Quiz
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Topic Content:

  • Subtraction in Other Bases

Reminder of the rules of binary subtraction:

0 – 0 = 0
1 – 0 = 1 
1 – 1 = 0
10 – 1 = 1 
11 – 1 = 10 

When you borrow a number during subtraction, it becomes the number of the base you are subtracting from.

For example, in Base 2 if you borrow a number the number you borrow becomes 2, Likewise in base 5 if you borrow a number it becomes 5.

Now, let’s take some examples.

Examples 1.5.1:

Evaluate the following:

1. 11000two – 111two
2. 20204six – 5125six
3. 72438 – 45368
4. 1A3516 – F7E16

1. 11000two – 111two

Solution

Screenshot 2023 09 13 at 16.52.34

Explanation

1st Column (from right to left): 0 is smaller than 1 so we will have to borrow from the 2nd column. Since it is in base 2 any number borrowed is 2. So borrowing two from the second will give (2 + 0) – 1 = 1

2nd Column: This reduces 0 to -1 in the second column. -1 is smaller than 1 so we have to borrow again from the third column. We now have:

= 2 + (- 1) – 1
= 1 -1
= 0

3rd Column: This also reduces 0 to – 1 in the 3rd column. This too is smaller than 1 so like the second column what we have is 2 – 1 – 1 = 0

4th Column: This reduces 1 to 0 in the fourth column. We then have 0 – 0 = 0

5th Column: This column requires no borrowing since 1 is greater than 0. We simply have 1 – 0 = 1

This yields 10001two

2. 20204six – 5125six

Solution

Screenshot 2023 09 13 at 16.50.32

Explanation

1st Column (from right to left): 4 is smaller than 5 so we will have to borrow from the 2nd column. Since it is in base 6 any number borrowed is 6. So borrowing six from the second will give:
(6 + 4) – 5 = 5

2nd Column: This reduces 0 to -1 in the second column. -1 is smaller than 2 so we have to borrow again from the third column. We now have: 6 – 1 – 2 = 3

3rd Column: This also reduces 2 to 1 in the 3rd column. We can subtract 1 from 1 so we now have 1 – 1 = 0

4th Column: We have 0 – 5. We are going to have to borrow from the 5th column i.e 6 – 5 = 1

5th Column: This reduces 2 in the 5th column to 1. We now have 1 – 0 = 1

This yields 110356

3. 72438 – 45368

Solution | Octal Subtraction

Screenshot 2023 09 13 at 16.59.34

Explanation

1st Column (from right to left): 3 is smaller than 6 so we will have to borrow from the 2nd column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the 2nd column will give (8 + 3) – 6 = 5

2nd Column: This reduces 4 to 3 in the second column. We can subtract 3 from 3 so we now have 3 – 3 = 0

3rd Column: 2 is smaller than 5 so we will have to borrow from the 4th column. Since it is in base 8 any number borrowed is 8. So borrowing 8 from the 4th column will give (8 + 2) – 5 = 5

4th Column: This reduces 7 in the 4th column to 6. 6 is greater than 4 so we can subtract normally. We now have 6 – 4 = 2

This yields 25058

4. 1A3516 – F7E16

Solution | Hex Subtraction

Screenshot 2023 09 13 at 17.08.00

Explanation

Hex subtraction can be computed much the same way as hex addition; by performing the operation while converting between hex and decimal values. The most significant difference between hex and decimal subtraction involves borrowing. When borrowing in hex, the “1” that is borrowed represents 16

Remember A = 10, B = 11, C= 12, D = 13, E = 14, F = 15

1st Column (from right to left): 5 is smaller than E (14) so we will have to borrow from the 2nd column. Since it is in base 16 any number borrowed is 16. So borrowing 16 from the second will give (16 + 5) – 14 = 21 – 14 = 7

2nd Column: This reduces 3 to 2 in the second column. 2 is smaller than 7 so we have to borrow again from the third column. We now have (16 + 2) – 7 = 11. B = 11

3rd Column: This also reduces A (10) to 9 in the 3rd column. 9 is smaller than F (15) we are going to have to borrow from the 4th column. We have (16 + 9) – 15 = 10. A = 10

4th Column: This reduces 1 to 0 in the 4th column. We then have 0 – 0 = 0

This yields AB716

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Giovanni Ohaeri
9 months ago

Becoming simpler by the lesson

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