This post is rather out of order, I know. I used MS OneNote to create the tutorial, and book im putting together on Modular Mathematics. So excuse the disordliness of the post as it was posted from text boxes in book form. Some of the formatting is also incorrect, AO's fault, not mine. This is just the beginnings of the Binary chapter. I am also working on chapters for hex, octal, trinary, and some imaginary system for the more advanced at this type of thing. Just remember one thing: Binary is Base-2. It contains only 0's and 1's.

A Few Things To Note:

*** If a number has a 'd' following it, it is in decimal form (base-10...i.e. normal). If it has a 'b' following it, it is in binary form (base-2, 0 or 1)

*** Notice that counting in Binary is as follows: 0, 1, 10, 11. So, when you add two ones together, they equal 10b.

*** You must always carry the left over ones to the next place, even if it is non-existent.

*** The rightmost 1 is 1d, the next is 8d, then 16d, 32d, 64d. The zeros are merely place-holders. Add the decimal numbers: 121d. So 111001b = 121d.

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If you say 11+11=110, people will think you're nuts. So, in this book you will see 11b+11b=110b or 11b+11b=6d. The 'd' is for 'decimal', normal counting numbers here. Let's start off with some basic addition and subtraction, shall we?

First you must know a few things. In decimal, the number 657, could be said to be seven ones, five tens, and six hundreds. In binary, the number 100, could be said to have zero twos, zero fours, and one eight. Notice the pattern? 2's, 4's, 8's, 16's….they are all powers of 2. In decimal, they are all powers of 10. To convert these, simply do the following:

1 0 0 1

^ ^ ^ ^

| | | |_________> 1 x 2^0 = 1 x 1 = 1

| | |___________> 0 x 2^1 = 0 x 2 = 0

| |_____________> 0 x 2^2 = 0 x 4 = 0

|_______________> 1 x 2^3 = 1 x 8 = 8

That means that the rightmost one, has a value of 1d. The leftmost one, has a value of 8d. 8d+1d=9d. So 1001b = 9d.

111001b = ?d. Find the value of 111001b in decimal:

Example:

1111b

+1111b

11110b

110110

+1111101

10110011

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1+11 = 10

Write 0, carry the one. If it all adds to 1, then bring the one down. If it adds to 2 again, (which is 10b) then bring down the zero and add the one to the next left digits. Always put the right most digit of the addition down into the answer, then what ever is left is added to the next set of digits. For example, if the total is 4d (100b) drop the zero down, and put the one and zero (10b which = 2d) in the mext column to the left and add it as 2 to the other ones and zeros.

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For these types of problems you must be able to calculate binary digits equivalents in your head.

1b = 1d

10b = 2d

11b = 3d

100b = 4d

Example #1:

101b

+1010b

1111b

Example #2:

10101010b

+100001101b

110110111b

Example #3:

1101110b

+10001111b

11111101b

Example #4:

101b

1010b

1110111b

+ 101b

10001011b

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Well, as I said, that post is a mess, but useful and interesting to the Math majors out there.

A_T

P.S. I attached a ZIP file containing both a .one file (from MS OneNote 2003) and a .mht file (web archive). The second was exported from OneNote, and is much better looking than this text version, and is viewed via IE.