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JSS1: MATHEMATICS - 1ST TERM

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  1. Whole Numbers I | Week 1
    3 Topics
    |
    1 Quiz
  2. Whole Numbers II | Week 2
    1 Topic
    |
    1 Quiz
  3. Counting in Base Two | Week 3
    4 Topics
    |
    1 Quiz
  4. Arithmetic Operations | Week 4
    3 Topics
    |
    1 Quiz
  5. Lowest Common Multiple (LCM) | Week 5
    2 Topics
    |
    1 Quiz
  6. Highest Common Factor | Week 6
    1 Topic
    |
    1 Quiz
  7. Fraction | Week 7
    7 Topics
    |
    1 Quiz
  8. Basic Operations with Fractions I | Week 8
    3 Topics
    |
    1 Quiz
  9. Basic Operations with Fractions II | Week 9
    1 Topic
    |
    1 Quiz
  10. Directed Numbers | Week 10
    3 Topics
    |
    1 Quiz
  11. Estimation and Approximation I | Week 11
    3 Topics
    |
    1 Quiz
  12. Estimation and Approximation II | Week 12
    6 Topics
    |
    1 Quiz
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Lesson 3, Topic 3
In Progress

Converting Numbers in Other Bases to Base Ten

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Topic Content:

  • Converting Numbers in Other Bases to Base Ten
  • Worked Examples

A given number can be converted from any base to base 10 using the Expansion Method.

Once you are able to work out the place values, you can easily convert other base numbers to binary.

Let’s take a look at the following examples:

Worked Example 3.3.1:

Write the following numbers in expanded form. 

1. 356eight
2. 10321four 

Solution 

1st Step: To use the expansion method for conversion we first find the place numbers.

1. 356eight  

Screenshot 2023 08 18 at 13.17.09

Students often find it difficult to locate the place values. The easiest way is to start from the last digit to the first digit. The last Digit always starts with a 0, the digit that follows a 1, and so on. (0, 1 2, 3, etc)

Screenshot 2023 08 18 at 13.35.35

2nd Step: Multiply each digit by its base number raised to the power of its place value

Screenshot 2023 08 23 at 07.58.04

356eight  = 3 × 82 + 5 × 81 + 6 × 80 

Base eight is also called the octal system 

2. 10321four

Screenshot 2023 08 18 at 13.30.03
Screenshot 2023 08 18 at 13.34.32

Multiply each digit by its base number raised to the power of its place value

10321four = 1 × 44 + 0 × 43 + 3 × 42 + 2 × 41 + 1 × 40 

To convert any number base to base ten, simply express it as the power of the given base. As discussed earlier the place value is important here.

Worked Example 3.3.2:

Express the following binary numbers as numbers in base ten. 

a. 110100two
b. 534six
c. 4307eight

a. 110100two

Solution 

Using the expanding method 

Place
Value
543210
Binary
Number
110100

110100ten = 1 × 25 + 1 × 24 + 0 × 23 + 1 × 22 + 0 × 21 + 0 × 20

= (1 × 25) + (1 × 24) + (0 × 23) + (1 × 22) + (0 × 21) + (0 × 20)

= (1 × 32) + (1 × 16) + 0 + (1 × 4) + 0 + 0

= 32 + 16 + 0 + 4 + 0 + 0 

= 32 + 16 + 4 

= 52ten 

⇒ 110100two = 52ten 

b. 534six

Solution

Using the expansion method: 

Place
Value
210
Binary
Number
534

534six = 5 × 62 + 3 × 61 + 4 × 60

= (5 × 36) + (3 × 6) + (4 × 1) 

= 180 + 18 + 4 

= 202ten 

⇒ 534six  = 202ten

Note
62 = 6 × 6 = 36
61 = 6
60 = 1

c. 4307eight

Solution

Place
Value
3210
Binary
Number
4307

Using the expanding method: (same approach as Examples above) 

4307eight = 4 × 83 + 3 × 82 + 0 × 81 + 7 × 80

 = (4 × 512) + (3 × 64) + (0 × 8) + (7 × 1) 

 = 2048 + 192 + 0 + 7 

= 2247ten 

⇒ 4307eight = 2247ten 

Worked Example 3.3.3:

Find the value of (111two)2 in base two 

Solution

1st Step: The first thing to do is to convert the number to base ten.
2nd Step: After converting to base ten we will then square the value.
3rd Step: Finally we will convert the squared value back to base two.

1st Step:

Place
Value
210
Binary
Number
111

Using the expansion method;

111two = (1 × 22) + (1 × 21) + (1 × 20

= 1 × 4 + 1 × 2 + 1 × 1 

= 4 + 2 + 1 

= 7ten 

2nd Step:

Recall that (111two)2 = (7ten)2 = 7 × 7 = 49 

= 49ten  

3rd Step:

Then convert 49ten to base two (binary) using the method of repeated division

Screenshot 2023 08 23 at 07.48.23

⇒ 49ten = 110001two