The decimal numeral system is the system humans use for counting and commerce. Counting in groups of ten makes sense to us as we have ten fingers (digits). Decimal is therefore a base-10 system, using the numerals (0,1,2,3,4,5,6,7,8,9).
Each of these Hindu-Arabic numerals has a face value. By this I mean that each symbol has an intrinsic, built-in number value, from zero to nine. The symbol ‘1’ represents the concept of one, the symbol ‘2’ represents the concept of two, etc.
The decimal number that follows 9 is 10, and similarly 99 is followed by 100.
Notice that we are now using groups of numerals, written from left to right, to represent numbers.
The position of these numerals determines their worth beyond their face-value. In positional numeral systems like decimal this positional worth is called a place-value.
The place-value is the numerals face-value multiplied by the numerals place-value-weight.
To expand the number, we work through the columns from left to right, multiplying the face-value of each numeral with its place-value-weight. We then add all those products together.
Now we work our way left to right (just like we did when we expanded our decimal number) multiplying the face-values of our binary numerals with their place-value-weights. Finally, we sum all those products.
101101 [binary] = (32 x 1) + (16 x 0) + (8 x 1) + (4 x 1) + (2 x 0) + (1 x 1) [decimal]
Note:If our binary number contained more numerals we would have to add extra columns to the left-side of our chart. Can you figure out the place-value-weights of the next three columns to the left?
Reverse Binary Conversion
We can also use the place value weights from the Place Value Table to convert from decimal to binary.
Let’s try the decimal number 54.
We start with the left most column and ask if we can use each weight to build the number 54. Each time we say yes, we’ll subtract that weight from our number.
Question
Answer
Reason
Do we need a 32?
yes!
That leaves us with 54-32 = 22
Do we need a 16?
yes!
That leaves us with 22-16 = 6
Do we need an 8?
no!
We only have 6 remaining.
Do we need a 4?
yes!
That leaves us with 6-4 = 2
Do we need a 2?
yes!
That leaves us with 2-2 = 0
Do we need a 1?
no!
We have no more left.
So that’s (yes,yes,no,yes,yes,no) or if we replace yes by 1 and no by 0: 110110
Note:If we wanted to convert a decimal number larger that 63 we would need to add another column to the left of our table, with a place-weight-value of 64. The larger the number, the more columns required.
Octal (Base-8)
Octal is the base-8 numeral system
Eight digits are used: 0 through 7
Like hexadecimal, octal is a way of representing binary data (but using 3 bits per character instead of 4)
Octal is used in certain computing applications
One application is the Unix chmod command which is used to assign file permissions
Now we work our way left to right multiplying the face-values of our octal numerals with their place-value-weights. Finally, we sum all those products.
2376 [octal] = (512 x 2) + (64 x 3) + (8 x 7) + (1 x 6) [decimal]
We can also use the place value weights from the Place Value Table to convert from decimal to octal.
Let’s try the decimal number 2547.
We start with the left most column. For each column we ask how many of that weight can evenly fit into our number. We then subtract the remainder from our number.
Question
Answer
Reason
How many 32768s fit in 2547?
none
That still leaves us with 2547
How many 4096s fit in 2547?
none
That still leaves us with 2547
How many 512s fit in 2547?
4
That leaves us with 2547 - (512 x 4) = 499
How many 64s fit in 499?
7
That leaves us with 499 - (64 x 7) = 51
How many 8s fit in 51
6
That leaves us with 51 - (8 x 6) = 3
How many 1s fit in 3
3
That leaves us with 3 - (1 x 3) = 0
So that’s 4763 [octal]
Hexadecimal (Base-16)
Hexadecimal (also known as hex) is a base-16 numeral system
It has 16 symbols: 0 to 9 and A to F
Each hex digit represents a nibble, which is 4 binary digits (bits) - half a byte
It is used to easily represent binary (base-2) values in a more-human readable form
Now we work our way left to right multiplying the face-values of our hexadecimal numerals with their place-value-weights. Finally, we sum all those products.
3D2A [hex] = (4096 x 3) + (256 x D) + (16 x 2) + (1 x A) [decimal]
3D2A [hex] = (4096 x 3) + (256 x 13) + (16 x 2) + (1 x 10) [decimal]
We can also use the place value weights from the Place Value Table to convert from decimal to hexadecimal.
Let’s try the decimal number 6082.
We start with the left most column. For each column we ask how many of that weight can evenly fit into our number. We then subtract the remainder from our number.
Question
Answer
Reason
How many 1048576s fit in 6082?
none
That still leaves us with 6082
How many 65536s fit in 6082?
none
That still leaves us with 6082
How many 4096s fit in 6082?
1
That leaves us with 6082 - (4096 x 1) = 1986
How many 256s fit in 1986?
7
That leaves us with 1986 - (256 x 7) = 194
How many 16s fit in 194
12 (C)
That leaves us with 194 - (16 x 12) = 2
How many 1s fit in 2
2
That leaves us with 2 - (1 x 2) = 0
So that’s 17C2 [hexadecimal]
Conversion Among Bases
It’s simple to convert Octal or Hexadecimal to Binary (and vis-versa). Each Octal numeral represent groups of three binary bits. While each Hexadecimal numeral represents groups of four binary bits.
The Octal number 137 converted to binary:
First convert each numeral to a three bit binary number: 1 = (001) 3 = (011) 7 = 111
Then string those numbers together: 137 [octal] = 001011111 = 1011111 [binary]
The Hex numer E30A converted to binary:
First convert each numeral to a four bit binary number: E = (1110) 3 = (0011) 0 = (0000) A = (1010)
Then string those numbers together: E30A [hex] = 1110001100001010 [binary]
Let’s work in reverse and convert the binary numeral 110011110101 into both octal and hexadecimal.
Octal:
Group the bits into groups of three from right to left: (110)(011)(110)(101)
Convert each group of bits into an octal numeral: (6)(3)(6)(5)
110011110101 [binary] = 6365 [octal]
Hex:
Group the bits into groups of four from right to left: (1100)(1111)(0101)
Convert each group of bits into a hex numeral: (C)(F)(5)
110011110101 [binary] = CF5 [hexadecimal]
Note:If the left-most group does contain enough bits you can pad it with zeros. For example:
Think back to the previous two modules (the history of computing and how computer’s work) and then enjoy this video of the Digi-Comp II, a billiard ball computer.
