CCNA Number System

Introduction

A number system defines a set of values used to represent quantity. You may regard each digit as a box that can hold a number. In the binary system, there can be only two choices for this number -- either a "0" or a "1". In the octal system, there can be eight possibilities:

"0", "1", "2", "3", "4", "5", "6", "7".

In the decimal system, there are ten different numbers that can enter the digit box:

"0", "1", "2", "3", "4", "5", "6", "7", "8", "9".

In the hexadecimal system, we allow 16 numbers:

"0", "1", "2", "3", "4", "5", "6", "7", "8", "9", "A", "B", "C", "D", "E", and "F".

As demonstrated by the following table, there is a direct correspondence between the binary system and the Hexa system, with four binary digits translate directly into one hexadecimal digit. In computer usage, hexadecimal notation is especially common because it easily replaces the binary notation, which is too long and human mistakes in transcribing the binary numbers are too easily made.

Base Conversion Table

     BIN    HEX   DEC
  ----------------
  0000    0     0
  0001    1     1
  0010    2     2
  0011    3     3
  0100    4     4
  0101    5     5
  0110    6     6
  0111    7     7
  ----------------
  1000    8     8
  1001    9     9
  1010    A    10
  1011    B    11
  1100    C    12
  1101    D    13
  1110    E    14
  1111    F    15

Convert From Any Base To Decimal

Let's think more carefully what a decimal number means. For example, 1234 means that there are four boxes (digits); and there are 4 one's in the right-most box (least significant digit), 3 ten's in the next box, 2 hundred's in the next box, and finally 1 thousand's in the left-most box (most significant digit). The total is 1234:

Original Number: 1 2 3 4 | | | | How Many Tokens: 1 2 3 4 Digit/Token Value: 1000 100 10 1 Value: 1000 + 200 + 30 + 4 = 1234 or simply, 1*1000 + 2*100 + 3*10 + 4*1 = 1234

Thus, each digit has a value: 10^⁰=1 for the least significant digit, increasing to 10^¹=10, 10^²=100, 10^³=1000, and so forth.

Likewise, the least significant digit in a hexadecimal number has a value of 16^⁰=1 for the least significant digit, increasing to 16^¹=16 for the next digit, 16^²=256 for the next, 16^³=4096 for the next, and so forth. Thus, 1234 means that there are four boxes (digits); and there are 4 one's in the right-most box (least significant digit), 3 sixteen's in the next box, 2 256's in the next, and 1 4096's in the left-most box (most significant digit). The total is:

1*4096 + 2*256 + 3*16 + 4*1 = 4660

Example. Convert the hexadecimal number 4B3 to decimal notation.

Solution:

     Original Number  :   4    B    3
                       |    |    |
  How Many Tokens  :   4   11    3
  Digit/Token Value: 256   16    1
  Value:            1024 +176  + 3 = 1203

Another way is to think of a cash register with different slots, each holding bills of a different denomination

Convert From Decimal to Any Base

Again, let's think about what you do to obtain each digit. As an example, let's start with a decimal number 1234 and convert it to decimal notation. To extract the last digit, you move the decimal point left by one digit, which means that you divide the given number by its base 10.

1234/10 = 123 + 4/10

The remainder of 4 is the last digit. To extract the next last digit, you again move the decimal point left by one digit and see what drops out.

123/10 = 12 + 3/10

The remainder of 3 is the next last digit. You repeat this process until there is nothing left. Then you stop. In summary, you do the following:

Quotient Remainder ----------------------------- 1234/10 = 123 4 --------+ 123/10 = 12 3 ------+ | 12/10 = 1 2 ----+ | | 1/10 = 0 1 --+ | | | (Stop when the quotient is 0) | | | | 1 2 3 4 (Base 10)

Now, let's try a nontrivial example. Let's express a decimal number 1341 in binary notation. Note that the desired base is 2, so we repeatedly divide the given decimal number by 2.

Quotient Remainder ----------------------------- 1341/2 = 670 1 ----------------------+ 670/2 = 335 0 --------------------+ | 335/2 = 167 1 ------------------+ | | 167/2 = 83 1 ----------------+ | | | 83/2 = 41 1 --------------+ | | | | 41/2 = 20 1 ------------+ | | | | | 20/2 = 10 0 ----------+ | | | | | | 10/2 = 5 0 --------+ | | | | | | | 5/2 = 2 1 ------+ | | | | | | | | 2/2 = 1 0 ----+ | | | | | | | | | 1/2 = 0 1 --+ | | | | | | | | | | (Stop when the | | | | | | | | | | | quotient is 0) 1 0 1 0 0 1 1 1 1 0 1 (BIN; Base 2)

Let's express the same decimal number 1341 in hexadecimal notation.

Quotient Remainder ----------------------------- 1341/16 = 83 13 ------+ 83/16 = 5 3 ----+ | 5/16 = 0 5 --+ | | (Stop when the quotient is 0) | | | 5 3 D (HEX; Base 16)

Example. Convert the decimal number 3315 to hexadecimal notation.

Solution:

Quotient Remainder ----------------------------- 3315/16 = 207 3 ------+ 207/16 = 12 15 ----+ | 12/16 = 0 12 --+ | | (Stop when the quotient is 0) | | | C F 3 (HEX; Base 16) Thus, 3315 (DEC) --> CF3 (HEX)

Note that from the Base Conversion Table, you can easily get the binary notation from the hexadecimal number by grouping four binary digits per hexadecimal digit, or from or the octal number by grouping three binary digits per octal digit, and vice versa.

HEX 5 3 D BIN 0101 0011 1101 OCT 2 4 7 5 BIN 010 100 111 101

Note: In 640-801 exam there is one question to convert a number from one base to others.