I won't go into much detail about what exactly they are. But here they are.

Decimal: Can use any number between 0 and 9.

Hexadecimal: Can use any number between 0 and 9. Any number higher is represented by a letter. These letters are A to F. Where A = 10 and F = 15.

Octal: Can use any number between 0 to 7.

Decimal: Can use 0 or 1.

One thing i will mention which is important is these number systems use a base.

Denoted as (N)^what ever base

Decimal is base 10. (N)^10

Hexadecimal is base 16. (N)^16

Octal is base 8. (N)^8

Binary is base 2. (N)^2

The base is represented by how many numbers are in the number systems set.

So as for converting between the numbers.

EX Decimal to Binary

Lets have our decimal number as 1411 as today is the 14th and the year is 2011.

So we write (1411)^10 ---> (N)^2 to show our decimal number and what base we are converting to.

Heres where the conversion starts. Take whatever the base is of the system you are converting two. Then divide the number you are converting by that base. Keeping the remainder each time and add it on top of your divided number. We are actually using the operator Modulo instead.

1411 / 2 = 705+1

705/2 = 352+1

352/ 2 = 176 + 0

176/2 = 88 + 0

88/2 = 44 + 0

44/2 = 22 + 0

22/2 = 11 + 0

11/2 = 5 + 1

5 / 2 = 2 + 1

2 / 2 = 1 + 0

1 / 2 =

*+ 1*

**0**When we get to zero as in bold above we stop. Note that this is also similar to Euclid's algorithm. Now this is the kool part. Starting from the bottom of that long series of modulo. Take the remainders of each modulo and write those numbers next to each other as shown below.

(10110000011) this is the binary conversion. To test to see if we did this right. Starting from the right hand side of our binary number. We go up the powers of 2 on each number.

2^ 0 = 1 ; 2^ 1 = 2 ; 2 ^ 2 = 4; 2^ 3 = 8 ; 2 ^ 4 = 16 ; 2 ^ 5 = 32 ; 2 ^ 6 = 64 ; 2^ 7 = 128 ;

2 ^ 8 = 256 ; 2 ^ 9 = 512; 2 ^ 10 =1024 ; 2^ 11 = 2048 ;

So to break our binary up into individual bits.starting from the right hand side. If it is a one you give it the power of two determined.

1 + 2 + 128 + 256 + 1024 = 1411.

EX conversion from octal to binary.

This starts to get a little different. Octal has a base of 8. Which means that for each octal number binary requires 3 binary numbers.

Lets pick good old 63.

(63)^ 8 ---> (N)^2. Same idea but now we do each number individually. So starting at 6 then 3.

6/2 = 3 + 0

3/2 = 1 + 1

1=2 = 0 + 1 if you notice there is our 3 binary numbers required to represent one octal number.

3/2 = 1 + 1

1/2 = 0 + 1

Now in this case if you start with zero. But you know you have to have 3 numbers keep going till you have three numbers.

3/2 = 1 + 1

1/2 = 0 + 1

0 / 2 = 0 + 0

So taking our 6 binary numbers. we get (110) and (011)

Testing this out to see if we did it correctly.

(110) = 2+4 = 6

(011) = 1+ 2 = 3

Note that testing the two together will give you 51 and not 63. By together I mean 110011.

Hexadecimal to Octal

(FACE5) ^ 16 ---> (N)^8

For these conversion we either need to change hexadecimal to binary or decimal. If you change to binary for each hexadecimal requires 4 bits( 4 binary numbers, such as how octal requires 3 bits).

We will convert to binary.

(FACE5)^16 ---> (N)^ 2

F = 15 / 2 = 7 + 1

7 / 2 = 3 + 1

3 / 2 = 1 + 1

1 / 2 = 0 + 1

A = 10 / 2 = 5 + 0

5 / 2 = 2 + 1

2/2 = 1 + 0

1/2 = 0 + 1

C = 12 / 2 = 6 + 0

6 = 2 = 3 + 0

3 / 2 = 1 + 1

1/2 = 0 + 1

E = 14 / 2 = 7 + 0

7 / 2 = 3 + 1

3 / 2 = 1 + 1

1/2 = 0 + 1

5/2 = 2 + 1

2/2 = 1 + 0

1/ 2 = 0 + 1

0 / 2 = 0 + 0

Our binary conversion is (1111101011100101) Now becuase octal takes 3 bits per number we need to split our binary up into groups of 3 bits.

Since theres 16/ 3 = 5 Remainder 1 Let the one remainder be the first bit. The group the rest.

1 111 101 011 100 101. Now lets see what are octal number is.

101 = 1 + 4 = 5

100 = 4

011 = 1 + 2 = 3

101 = 5

111 = 1 + 2 + 4 = 7

1 = 1

175345

Hope you had fun reading this. Have fun showing off to friends.

this looks suspiciously like my old homework.

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