Number Bases II
Mathematics S S S 1 First Term
WEEK 2
Number Bases II
Performance Objectives
Students should be able to;
1. Perform the mathematical operations
2. Convert decimal fractions to base 10 and one base to another.
- Apply number base system to computer programming
Addition, Subtraction, Multiplication and Division of number Bases
Content
Operation in other bases other than base ten are carried out like what is obtained in base ten. We can illustrate the procedure as shown in the example below.
Example 1: 167eight + 145eight
Solution
1 6 7eight+ 1 4 5 eight
7 + 5 = 12. This exceeds the value of the base. 12 contain a bundle of 8 and 4 units. That one bundle of 8 is carried to the next column as 1
1 + 6 + 4 = 11
11 is another single bundle of 8 and three, Hence we write 3 and carry the bundle to the next column as
1 6 7eight+ 1 4 5 eight 3 3 4 eight
Example 2: 501twelve – 3Btwelve
5 0 1twelve- 3 B twelve
Recall B in base twelve is eleven.
If 1 is ‘borrowed from 5 in the third column, getting to the next column on the right becomes a twelve. From it, we can take one to the next column to the right again. To get 12 + 1 = 13 from which we finally subtract B (i.e eleven)
5 0 1twelve- 3 B twelve 4 8 2 twelve
Notice that after borrowing 1 from the middle column, eleven was left. It is out of this eleven that 3 is subtracted to get 8 in the second column of the answer.
Example 3: Simply 154six × 5six
1 5 4sixx 5six
5 × 4 = 20 i.e 3 bundles of 6 plus 2 units. Write 2 add 3 to the product of 5 × 5 of second column to get 28. 28 = 4(sixes) plus 4. Take the 4 bundles to next column. 4 + 5 × 1 = 9 which is 13six. So 154six × 5six = 1342 six
1 5 4sixx 5six 1 3 4 2 six
Example 4: Simplify 134five × 24five
1 3 4fivex 2 4 five
4 × 4 = 16 i.e 3(fives) and 1 unit.
These 3 bundle of 5 is added to the product of (3 × 4) of the second column. 3 × 4 + 3 = 15.
15 = 3(fives) and zero
This new 3 bundles of 3 is to be added to the product 1 × 4 of the third column
1 × 4 + 3 = 7 which will be written as 12five similar thing is done with 134five times the distance 2, thus
1 3 4fivex 2 4 five 1 2 0 1 3 2 4 4 4 3 1 five
Division of Numbers Bases
Since in binary (base two) system, the digits we have are 0 and 1. Each digit of the quotient 110111 ÷ 101 must be either 1 or 0. Therefore, 110111two ÷ 101two is done as follows:
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1011 |
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101 |
110111 |
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101 |
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111 |
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101 |
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101 |
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101 |
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Once you start the division, the digits are brought down one after the other.
Example: 240six ÷ 20six
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12 |
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20 |
240 |
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20 |
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40 |
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40 |
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So 240six ÷ 20six = 12six
Conversion of Decimal Fractions in one Base to Base Ten
Sometimes we are faced with numbers which are not whole numbers. Hence it is very necessary to study also the conversion of fractional parts of numbers. The following examples can be used in our study of the conversion of fractional parts of other bases to decimal system.
Example 1: Convert 6.4 7 to denary number
Solution
6.47= 6 × 70 + 4 × 7−1
= 6 × 1 + 4 × 1/7
= 6 + 4/7
= (647
)10
Example 2: Convert 101.011two to base ten.
Solution
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101.011two =(1 × 22) + (0 × 21) + (1 × 20) + (0 × 2−1) + (1 × 2−2) + (1 × 2−3) = 1 × 4 + 0 × 2 + 1 × 1 + 0 × 1/21 + 1/22 + 1 × 1/23 = 4 + 0 + 1 + 0 + ¼ + 1/8 = 5 + ¼ + 1/8 =538 |
Conversion of Fractions in Base Ten to any other Base
Fraction in base ten can be converted to other bases using various methods.
Example 1:
Express 7/8 to decimals
Solution:
Change 7/8 to decimal fraction i.e 7/8 = 0.875 and multiply by 2.
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0.875 |
↓ |
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× |
2 |
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1750 |
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× |
2 |
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1500 |
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× |
2 |
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1000 |
As we multiply 2 × 0.875 we get 1.750. Keep the 1 and multiply 750 by 2, and get 1.500. Keep the 1 and multiply 500 by 2 and get 1.000. Stop when all is zero. The value of 7/8 = 0.111two or convert 7 and 8 to base two and then divide.
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2 |
7 |
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2 |
3 |
R |
1 |
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2 |
1 |
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1 |
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0 |
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1 |
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7ten=111two
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2 |
8 |
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2 |
4 |
R |
0 |
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2 |
2 |
R |
0 |
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2 |
1 |
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0 |
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0 |
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1 |
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8ten=1000two
∴1111000 = 0.111two
Example 2:
Express 191325ten to base five.
Solution:
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5 |
19 |
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5 |
3 |
R |
4 |
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0 |
R |
3 |
↑ |
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= 345
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5 |
13 |
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5 |
2 |
R |
3 |
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0 |
R |
2 |
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= 235
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5 |
25 |
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5 |
5 |
R |
0 |
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5 |
1 |
R |
0 |
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0 |
R |
1 |
↑ |
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=1005 = [34+23/100]5 = [34+0.23]5 = 34.235
Application to Computer Programming
Binary system is very important because of its use in most electronic devices.
Digital computers perform their two-way functions because only two digits 0 and 1, called bits are coded into them as a programming language. For instance, if a device is ON, it is represented by a 1 and if it is OFF, a 0 is represented.
Other situations with only two possibilities include: UP or DOWN, TRUE or FALSE, MAGNETIZE or DEMAGNETIZE