How does binary work

binary counting is just like decimal counting.

If you chop off all but your thumbs.

1 ... Ummm lessee... next thumb would make it... 10

10 binary = 2 decimal. Lets call that 10b = 2d

Add another thumb... 10 + 1 = 11. 11b = 3d

One more... oops, ran out of thumbs. We used 10, so the next one has to be 100.

100b = 4d

101, 110, 111, what...

1000. 1000b = 8d (you counting along with me?)

Now here's a stupid thing to remember it all by.

10b has one '0' in it, = 2d
100b has two '0's... = 4d = 2d*2d
1000b has three... =8d = 2d*2d*2d

The number of zeroes is the number of twos...

Have fun.

Since he posted in Music, I think he might be talking about the new binary music system to download from newsgroups, but not sure since he wasn't specific.

If so, this tutorial may be informative. Mp3 binary files - A 56k newbie's guide to downloading music from binary newsgroups

Fellow with 2 posts (at current time) could easily put this in the wrong place...

Fellow with 2 posts could easily never return.

Oh well.

There are 10 kinds of people... those who understand binary and those who don't.

OH man soory i out it in the wrong place

I Remeber in High school My Dean had a big poster behind his desk that said that

It is sometimes a pleasure to fill a page with numbers.
I have enough (Ich habe genug).

0 ..0
00000001 ..1
00000010 ..2
00000011 ..3
00000100 ..4
00000101 ..5
00000110 ..6
00000111 ..7
00001000 ..8
00001001 ..9
00001010 ..10
00001011 ..11
00001100 ..12
00001101 ..13
00001110 ..14
00001111 ..15
00010000 ..16
00010001 ..17
00010010 ..18
00010011 ..19
00010100 ..20
00010101 ..21
00010110 ..22
00010111 ..23
00011000 ..24
00011001 ..25
00011010 ..26
00011011 ..27
00011100 ..28
00011101 ..29
00011110 ..30
00011111 ..31
00100000 ..32
00100001 ..33
00100010 ..34
00100011 ..35
00100100 ..36
00100101 ..37
00100110 ..38
00100111 ..39
00101000 ..40
00101001 ..41
00101010 ..42
00101011 ..43
00101100 ..44
00101101 ..45
00101110 ..46
00101111 ..47
00110000 ..48
00110001 ..49
00110010 ..50
00110011 ..51
00110100 ..52
00110101 ..53
00110110 ..54
00110111 ..55
00111000 ..56
00111001 ..57
00111010 ..58
00111011 ..59
00111100 ..60
00111101 ..61
00111110 ..62
00111111 ..63
01000000 ..64
01000001 ..65
01000010 ..66
01000011 ..67
01000100 ..68
01000101 ..69
01000110 ..70
01000111 ..71
01001000 ..72
01001001 ..73
01001010 ..74
01001011 ..75
01001100 ..76
01001101 ..77
01001110 ..78
01001111 ..79
01010000 ..80
01010001 ..81
01010010 ..82
01010011 ..83
01010100 ..84
01010101 ..85
01010110 ..86
01010111 ..87
01011000 ..88
01011001 ..89
01011010 ..90
01011011 ..91
01011100 ..92
01011101 ..93
01011110 ..94
01011111 ..95
01100000 ..96
01100001 ..97
01100010 ..98
01100011 ..99
01100100 ..100

The binary system works like all the rest with one critical difference its numerical base is 2.

Lets use something you are familiar with.
The decimal system we normally use is base 10. Each digital position in a number represents a power of 10 and the digit currently in its place tells us the number of times we must add that quantity to itself. We then must add up all these position's values in the number to arrive at the quantity or value the number represents...its not as confusing as it sounds. Lets take the number 213 (Two hundred and thirteen) in base 10(ten), the system we normally use . It possesses 3 digits.

By convention the right most digit (in this case 3) holds the number of times we add 10 to the 0 (zero) power expressed as 10^0. The second digit from the right represents 10 to the 1st power (10^1) and the third represents 10 to the 2nd power (10^2) and so on. If we had a fourth digit its position would represent 10 to the third power) so... (note that any number to the zero power=1)

position 3rd 2nd 1st
psitional value (10^2=100) (10^1=10) (10^0=1)

digits 2 1 3

First we calculate the positional values then we simply add them together to obtain the total value of the number so represented.

The digit in the 1st position is three so we add three of the value of the position. Since the 1st positional value is 1 we add up 3 one's to get the value 3. The 2nd position value is 10 and the digit is one which gives a value of 10. The 3rd positional value is 100 so we add up 2 of these to get a value of 200. To get the total value of the number represented by the digits 213 we add the seperate values of all the positions.

position 3 position 2 position 1
200 10 3

200 + 10 + 3 = 213

Works the same for binary but the base is two (thus the preffix "bi")

so the first or right most digit represents two to the zero power or one (2^0=1). The second position two to the first power or 2 (2^1=2). The third position represents 2 to the 2nd power or 4 (2^2=4) or 2 squared. Looking at satt-focusable's above conversion chart let's take 0101 = 5

Right most digit is one and its positional value one so we add it to itself one time getting a value of one. The second is zero and thus evaluates to zero. The 3rd positional value is 2 to the 2nd power or 2 squared or 4 (2^2=4). How many 4's do we have? Looking at the third position we see a one so the total vaule of that position is 4. So we merely add up all the values.

1st or right most value = 1
2nd value = 0 (zero)
3rd value = 4

1 + 0 + 4 = 5

Notice that each position to the left of any given position represents the next power of the base (in this case 2) so in the above example the number's left most position (we find a zero in this position) represents 2^3=8 since there is a zero in this position no value is added.

The binary system shows the importance of the digit zero as a place holder, a concept invented in the Arab world. This system is ideal for computers, one merely has to envision tiny electrical switches as being on or off. An on switch being a one (1) and an off switch representing a zero (0). This is how data is represented in a computer...all data, even text.

Next time we learn how to add in binary! Anybody game?

JM

01010011011101010111001001100101

Convert to binary from decimal by repeatedly dividing by 2 and using the remainder.

33/2=16 1 remaining
16/2=8 0 remaining
8/2=4 0 remaining
4/2=2 0 remaining
2/2=1 0 remaining
1/2=0 1 remaining

Start from the BOTTOM
33 dec = 100001 bin