**Number bases**

Most people count in tens. For instance, the place value of the digits in number 4956 in this system

Thousands hundreds tens units

↓ ↓ ↓ ↓

4 9 5 6

The place values are in powers of ten. It is called the **base ten **system.

4956 = **4** X 1000 + **9** X 100 +** 5 **X 10 + **6 **X 1

= **4** X10^{3} + **9** X 10^{2} + **5 **X 10^{1} + **6 **X 1^{0 }

Some people traditionally count in 5s, other in 20s. Using the method of the base ten system, a **base five **system is in powers of five. So, for example, in base five, 342 would be

342_{five }= **3** X 25 + **4** X 5 + **2** X 1

= **3** X 5^{2} + **4** X 5^{1}+ **2** X 5^{0}

Notice that 342_{five }is short for 342 in base five.

Converting base ten numbers to other bases

To convert from base ten to another base, express the given number in powers of the new base.

**Example**

Convert 37_{ten } a. to base eight, b. to base five.

- Since 37 < 64, there are no sixty-fours in 37. To find the number of eights in 37, divide by 8.

37 ÷ 8 = 4, remainder 5

37 = 4 eights + 5 units

37_{ten }= 45_{eight }

Check: 45_{eight }= 4 X 8 + 5 X 1

= 32 + 5

= 37

Since 37 > 25, there must be a twenty-five in 37.

37 ÷ 25 = 1, remainder 12

37 = 1 twenty-five + 12 units.

Consider the 12 units. Since 12 > 5, there must be some fives in 12.

12 ÷ 5 = 2, remainder 2

12 = 2 fives + 2 units

∴ 37 = 1 twenty-five + 2 fives + 2 units

= 1 x 5^{2} + 2 X 5^{1} + 2 X 1

37_{ten }= 122_{five}

Check: 122_{five }= 1 X 25 + 2 X 5 + 2 X 1

= 25 + 10 + 2

= 37

The method in part** b** of Example 2 can be shortened as follows:

5 | 37

5 | 7 + 2 (i.e 7 X 5 + 2 X 1)

5 | 1 + 2 ( i.e. 1 X 5^{2 }+ 2 X 5^{1})

5 | 0 + 1 ( i.e. 0 X 5^{3 }+ 1 X 5 ^{2})

↑ ↑

Continued division by 5 gives remainders. Reading the remainders upwards gives 37_{ten }= 122_{five}

(see the arrows above).

To change from base ten to another base:

- Divide the base ten number by the new base number.
- Continue dividing until 0 (zero) is reached, writing down the remainder each time.
- Start at the last remainder and read upwards to get the answer.

**Convert from other bases to base ten**

**Convert an Octal into a Decimal**

This is a little bit of the same as before except our base is now 8 instead of 16. Try it with the number 723_{8}_{ }also written 0723 in computer programming notation. Go ahead and convert it. I’ll outline the steps below again.

723_{8} = (7 X 8^{2}) + (2 X 8^{1}) + (3 X 8^{0})

= 448 + 16 + 3

= 467_{10}

723 octal is 467 decimal. See, that’s not so tough. Now, the astute reader may have noticed another way this can be accomplished using the same information from the original formula. Take the previous octal number as an example here. Since you know what the base and exponent will be for all digits’ place, you can go ahead and write that out: 82 = 64, 81 = 8, 80 = 1. Write your number beneath and multiply out, then add the products.

**Operations with binary numbers**

To add, subtract and multiply binary number, use the method that you use with base ten numbers. However, you must remember that you are working with powers of two, not powers of ten. The following identities are very useful:

Addition

0 + 0 = 0 1 + 0 = 1

0 + 1 = 2 1 + 1 = 10

Multiplication

0 X 0 = 0 1 X 0 = 0

0 X 1 = 0 1 X 1 = 1

The example below shows how to use these identities when operating with binary numbers. Follow the notes carefully.

**QUESTIONS**

Lets see how much you’ve learnt, attach the following answers to the comment below

Calculate the following in base two:

1. 1 001

+ 1 011

—————

—————

2. 1 110

– 101

—————–

—————–

3. 110

x 101

—————–

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