Mastering Binary Numbers

Decimal Representation of Numbers.

The most common numbering system we use is the decimal number system. in this system, numbers are expressed as digits from 0 to 9. Each number can be interpreted according to a system of place values, we proceed from right to left. The digit in the extreme right represents the number of 1s, the next digit the number of 10s, the next digit the number of 100s, and so forth. For example, the number 2375 stands for:
51s=1 * 5=5
710s=10 * 7=70
3100s=100 * 3=300
21000s=1000 * 2=2000
------
2375

If we carefully investigate this numbering representation, it can be seen in another way:

2 * 103+ 3 * 102+ 7 * 101+ 5 * 100= 2375

Note that we have arranged the digits in their order of decreasing powers of 10.

Binary Representation of Numbers.

In the binary number system, numbers are represented by either 0 or 1 digits. Just as the decimal number system is based on powers of 10, the binary number system is based on powers of two. Let us take a binary number like: 10001111 starting from the right most digit which is corresponds to the number of 1s the next digit to the number of 2s, the next digit to number of 4s and the next to 8s and so forth. The binary number 10001111 can be written as:

1 * 27+ 0 * 26+ 0 * 25+ 0 * 24+ 1 * 23+ 1 * 22+ 1 * 21+ 1 * 20=
128+ 0+ 0+ 0+ 8+ 4+ 2+ 1= 143

As a result the binary number 10001111 corresponds to the decimal number 143. The binary system of numbers are extremely important in the operations of computers as memory can be presented as on or off to correspond to binary digit of 0 or 1 this one binary digit is also called a bit. As we shall see, 8 bit and 16 bit binary numbers play a special role in internal working of computers. The number of 8 bit also called byte represents the number 0 through 255. Similarly the number of 16 bit binary number (2 byte) represents the decimal numbers 0 through 65535.

Converting From Decimal to Binary.

As we have seen, every binary number has a decimal equivalent, however, the reverse is also true, every decimal number has a binary equivalent. For example, consider the decimal number 41, in order to find the equivalent binary number we have to apply the following steps. In each step we divides the number by 2 and write down the remainder.

41=20 * 2+ 1Right most bit
20=10 * 2+ 0
10=5 * 2+ 0
5=2 * 2+ 1
2=1 * 2+ 0
1=0 * 2+ 1Left most bit

And the resultant bit presentation of 41 decimal is 101001.
This process can be translated into a simple computer program:

Visual Basic Program

       Function DecimalToBinary(ByVal Dec As Integer) As String
'Dec is the number to be converted
Dim Res as String
Do While Dec > 0
  Res = Cstr(Dec MOD 2) & Res
  Dec = Dec \ 2
Loop
return Res

       End Function

Hexadecimal Representation of Numbers.

The hexadecimal number system is closely connected with the binary number system and is much easier to work with in many applications, one hexadecimal digit is corresponding to 4 binary digits, and so 8 bit binary numbers can be represented by two hexadecimal digits. There are 16 possible hexadecimal digits:

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

The digits 0-9 have their usual decimal values and A, B, C, D, E, F have the respective values

A = 10B = 11C = 12D = 13E = 14F = 15

Let us take a hexadecimal number say A2BF and imply the translation to decimal number system, as we done before we receive:

10 * 163+ 2 * 162+ 11 * 161+ 15 * 160=
40960+ 512+ 176+ 15= 41663

Octal Representation of Numbers.

The octal numbering system is based on 8 numeric digits, namely from 0 to 7, and as seen before it represents powers of 8. For example the octal number 723 represent the decimal value of 467.
7 * 82+ 2 * 81+ 3 * 80
448+ 16+ 3= 467

In the following table we expressed the hexadecimal numbers by their corresponding binary, octal and decimal values:

HexadecimalBinaryOctalDecimal
0000000
1000111
2001022
3001133
4010044
5010155
6011066
7011177
HexadecimalBinaryOctalDecimal
81000108
91001119
A10101210
B10111311
C11001412
D11011513
E11101614
F11111715

Note that each octal digit contains three binary digits. Therefore converting a binary to octal can be performed by breaking the binary string up into consecutive groups of three starting from right, and then convert each group into an octal digit.
For example take the 16 bit binary number 1110011010100111 and convert it to octal number.
	                Binary number                   1110011010100111
	                Divide into groups of three     1 110 011 010 100 111
	                Convert each group into octal   1  6   3   2   4   7
	                Final octal number              163247
	            
Now converting from octal number to binary is the exact same thing except in reverse. For example take the octal number 27035 and convert it to binary.
                       Octal number                     2  7   0   3   5
                       Convert each digit into binary  10 111 000 011 101    
                       Final binary number             10111000011101
	            
Converting Numbers From Decimal to Any Base.

Now we can generalize the conversion process to include conversions from decimal to any base. For example convert the decimal number 6478 to base 24, The conversion process includes the following steps:

Step 1: Find the remainder of division by the new base (this is the Mod operation).
6478 Mod 24 = 22The corresponding character to 22 is M.
Step 2: Perform integer division of the original decimal value by the new base, the result is.
6478 \ 24 = 269
Step 3: Return to step 1 and Perform Mod operation on the division result, that gives.
269 Mod 24 = 5The corresponding character to 5 is 5.
This value will be added to the left of the previous character M to form 5M.
Step 4: Return to step 2 and Perform integer division on the previous Mod result.
269 \ 24 = 11
Step 5: Return to step 1 and Perform Mod operation of the last integer division.
11 Mod 24 = 11The coresponding character to 11 is B and it is
added to the left to yield the final conversion value of B5M


Visual Basic Program to Convert a Decimal Number to Any Other Base


Converting Numbers From Any Base to Decimal.

We can represent the conversion process mathematicaly as followes:

As seen in the above equation, the equivalent decimal number is the sum of the characters value in decimal, starting from right, multiplyed by the base value. If for example we take the number 6MA in base 24 the conversion to decimal will be:

Visual Basic Program to Convert a Number in Any Base to Decimal Number



Binary Right and Left Shift.

In some applications it is often required to move all the bits of a certain binary number to the left or to the right. Such operations are called a left shift or a right shift. For example consider this 8 bit (byte) binary number: 10110111, which is equivalent to 183 in decimal representation. If we apply a left shift we obtain the binary number 01101110 = 110 decimal note that a zero was added at the right most digit and the original left most bit has been pushed off the end. Similarly, a right shift applied to the binary number 10110111 yields 01011011 = 91 decimal then a second right shift can be apply to yield 00101101 = 45.
The shift operation can be summarized as in the next tables.

Left ShiftBinary Number1st shift2nd shift3rd shift4th shift 5th shift6th shift7th shift8th shift
Binary1011011101101110110111001011100001110000 11100000110000001000000000000000
Decimal183110220184112 2241921280


Right ShiftBinary Number1st shift2nd shift3rd shift4th shift 5th shift6th shift7th shift8th shift
Binary1011011101011011001011010001011000001011 00000101000000100000000100000000
Decimal18391452211 5210

A computational left shift can be done by multiplying the decimal value by 2, this can lead to the left most bit to be pushed to the 9th place. In order to eliminate the 9th bit we perform the MOD operation (division remainder) on the decimal number by the number 256, this operation will return only the 8 right most bits.



Left shift=2 * DecimalMOD256

A right shift on the other hand can be done by integer division of the decimal value by 2, that is

Right shift=Decimal\2
In the above examples we assumed that right and left shifts were operated on 8 bit (byte) binary number, but shift operation can be performed also on 16 bit (2 byte or a word) or 32 bit or on any number of bit that we wish, in this case the same rules of shifting applies, for example the 16 bit binary number 11100110 10111011 can be left shifted to obtain 11001101 01110110.

11100110 10111011 = 59067 Decimal
Left shift = 59067 * 2 = 118134 Decimal
118134 MOD 65536 = 52598 Decimal = 11001101 01110110
The purpose of the division by MOD 65536 is to eliminate the bit pushed to the 17th place by the shift operation. The next table summarizes the MOD operator values to obtain N number of bits:

BitsDecimal value
416
8256
1665536
2416777216
324294967296
N2 ^ N

Logical Operations on Bits.
If A and B are two binary numbers with similar length, then we can compare A and B bit by bit for a given bit position and perform a logical operation like AND, OR, XOR and NOT. The use of the logical operations are extremely important as it allows us to control switch states, by setting a certain bit on or off.

Logical AND Operation.

If two corresponding bits of A and B have a one, then the result is a one, if either A or B has a zero, the result is a zero. For example, consider the following 16 bit binary numbers:
A=1011 1011 1000 0011= 48003 decimal
B=1110 1010 1111 1010= 60154 decimal
A AND B=1010 1010 1000 0010= 43650 decimal
Switching control
Suppose we have an 8 bit number like 11110011, this array of bits can simulate 8 switches that each of them can be in on or off position. Lets think what happens if we need to turn bit 6 (from right or switch number 6) to off. This process is done by the AND operation, in order to find the second number needed for the AND operation, we will choose the binary number 11011111. All bits are 1 except the one we want to turn off.
A=11110011= 243 decimal
B=11011111= 223 decimal
A AND B=11010011= 211 decimal

Note that all the bits remained untouched, except bit number 6 that has been turned off. The same result would be obtained by the operation on the equivalent decimal numbers 243 AND 223 = 211
By applying the AND operation we can only turn switches off in order to turn a bit on we need to apply the OR operation as described in the next paragraph. We can turn off multiple switches in a single operation, this will be done by choosing zeroes in the second number in the places we want to turn them off, like 10001101 in this case we will turn off 4 bits (all the 0 values).

Logical OR Operation.

The OR operation performed on two identical length binary numbers is obtained by comparing A and B bit by bit. For a given bit position, if either A or B have a one, then the corresponding bit of A OR B is one. If both A and B have a zero, the corresponding bit of A OR B is a zero. For example, consider the two 8 bit binary numbers
A=01010011= 83 decimal
B=11010101= 213 decimal
A OR B=11010111= 215 decimal

Switching control
By performing the OR operation we can turn on a single switch, This is done as follows, Lets take the switch array 11010111 and try to turn on switch number 6 (from right). The second binary number is choose to be 00100000, all bits are zero except the one we want to turn on.
A=11010111= 215 decimal
B=00100000= 32 decimal
A OR B=11110111= 247 decimal

Note that switch number 6 is now turned on and all other switches remained untouched.

Logical XOR Operation.

The operation A XOR B is called the exclusive OR, it is performed on two identical length binary numbers by comparing A and B bit by bit. For a given bit position, if A and B have different bits values, then the corresponding bit of A and B is a one. If A and B have the same bits value, the corresponding bit A XOR B is a zero. For example consider the 8 bit binary number A = 11001001 and B = 10001101 then,
A=11001001= 201 decimal
B=10001101= 141 decimal
A XOR B=01000100= 68 decimal

If A and B are integers given in decimal form, we may also apply the operation XOR the same way like
201 XOR 141 = 68.
An interesting use of The XOR operation is in the field of
encoding text expressions.

Logical NOT Operation.

The NOT operation reverses all the bits of a given number. For example consider the binary number 00111001. We may compute this number by changing every 0 to 1 and every 1 to 0, see the next lines.
A=00111001= 57 decimal
NOT A=11000110= 198 decimal

The NOT operation can be performed on decimal numbers as well. For example NOT 57 = 198.
It is strictly important to define previously the length of the binary number, as different lengths has different NOT values. The above calculations were based on 8 bit length, if the same values were calculated on 16 bit length, the results would be as follows,

A=00000000 00111001= 57 decimal
NOT A=11111111 11000110= 65478 decimal

Logical operations on A and B
A B AND OR XOR NOR XNOR NAND NOT A NOT B
0 0 0 0 0 1 1 1 1 1
0 1 0 1 1 0 0 1 1 0
1 0 0 1 1 0 0 1 0 1
1 1 1 1 0 0 1 0 0 0

Encoding text.

One interesting use of the XOR operation is for encoding purposes, this encoding is based on the fact that by performing the XOR operation twice on a certain character, we receive back the original character. The value for the encoding can be selected to be any value as our imagination goes.
For example examine the letter "s", this letter has the equivalent Ascii value of 115 or in binary presentation A = 01110011. For the encoding code we choose the value 66 or in binary presentation B = 01000010, now we apply the XOR operation to receive.
A=01110011= 115 decimal
B=01000010= 66 decimal
C = A XOR B=00110001= 49 decimal

So we encoded the letter "s" by the Ascii code of 49 which is the digit "1". Now applying the encoding value once again to restore the original letter yields.
C=00110001= 49 decimal
B=01000010= 66 decimal
A = C XOR B=01110011= 115 decimal

As we expected the original letter has been restored. The only rule we have to remember is to apply the same encoding value twice. By applying this technique on each letter we can encode the entire document.
The above example uses the same encoding value to all the characters. To further improve the encoding process we can apply a trigonometric function or a varying encoding values. For example we can define an array of arbitrary values like 3, 44, 77, 295, 3393, 858, 44, 338 and so on, then apply this encoding values, each of them on a progressing character, the only thing to remember is to apply the same encoding values to restore the original letter.


Last updated 7.22.08