Question 7.1: Represent the decimal number (183)10 in binary and in hex.
Represent the decimal number (183)_{10} in binary and in hex.
Initially we are not sure how many bits the number (183)_{10} requires in binary representation. To sort of figure out the number of bits required, let us list the powers of two:
Note from Table 7.3 that the number (128)_{10} can be represented in binary simply by writing:
(1000_{-}0000)_2=(128)_{10}
Similarly 128 + 64 = (192)_{10} is written as:
(1100_{-}0000)_2=(192)_{10}
Since we want to write the number (183)_{10} in binary representation we know that 8 bits will suffice. To convert from decimal to binary, algorithmically we proceed as follows:
We divide (183)_{10} by 2 to give an integer quotient of 91 and a remainder of ½. This process is repeated until the integer quotient becomes zero. We record all the operations as shown below:
| Integer quotient | Remainder | Bit position | Weight | |
| 183/2 = 91 | + | ½ | b0 = 1(LSB) | 1 |
| 91/2 = 45 | + | ½ | b1 = 1 | 2 |
| 45/2 = 22 | + | ½ | b2 = 1 | 4 |
| 22/2 = 11 | + | 0 | b3 = 0 | 8 |
| 11/2 = 5 | + | ½ | b4 = 1 | 16 |
| 5/2 = 2 | + | ½ | b5 = 1 | 32 |
| 2/2 = 1 | + | 0 | b6 = 0 | 64 |
| 1/2 = 0 | + | ½ | b7 = 1(MSB) | 128 |
From above we conclude that (183)_{10} = (1011_{-}0111)_2.
| Table 7.3 Some powers of two and bit position in a binary number |
|
| Powers of 2 | Binary Bit Position |
| 2^0=1 | b_0 |
| 2^1=2 | b_1 |
| 2^2=4 | b_2 |
| 2^3=8 | b_3 |
| 2^4=16 | b_4 |
| 2^5=32 | b_5 |
| 2^6=64 | b_6 |
| 2^7=128 | b_7 |
Now to convert (183)_{10} to hex we simply translate each group of four bits into their hex equivalent starting with the LSB position according to Table 7.2,
| Table 7.2 List of 16 uniquely defined hex digits, their binary and decimal equivalents | ||
| Hexadecimal (Hex, Base 16) | Binary (Base 2) | Decimal (Base 10) |
| 0 | 0000 | 0 |
| 1 | 0001 | 1 |
| 2 | 0010 | 2 |
| 3 | 0011 | 3 |
| 4 | 0100 | 4 |
| 5 | 0101 | 5 |
| 6 | 0110 | 6 |
| 7 | 0111 | 7 |
| 8 | 1000 | 8 |
| 9 | 1001 | 9 |
| A | 1010 | 10 |
| B | 1011 | 11 |
| C | 1100 | 12 |
| D | 1101 | 13 |
| E | 1110 | 14 |
| F | 1111 | 15 |
thus:
(183)_{10}=(1011_0111)_2=(B7)_{16}
In Chapter 8 we will address some interesting ways of representing positive and negative binary numbers. This will be useful to design digital arithmetic circuits.