1 KB = 1024 bytes
1 MB = 1024 KB
1 GB = 1024 MB
etc....
2006-09-13 11:51:21
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answer #1
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answered by Anonymous
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it has to do with binary...
1 bit is the smallest form of info your computer reads
1 byte = 8 bits
1 kilobyte = 1024 bytes or 8192 bits
1 megabyte = 1024 kilobytes or 8388608 bytes or 67108864 bits
1 gigabyte = 1024 megabytes or 1048576 kilobytes or 1073741824 bytes or 8589934592 bits
so the memory you are looking at tells you how many Bytes it can hold, 1073741824 is the number of charectors it can hold
(in binary each number or charector is stored in 8 digits of 1's and 0's...)
i dont know where you got 1024000000 from but it depends on what you look at a gigabyte as because it can be any of the following:
1024 megabytes
1048576 kilobytes
1073741824 bytes
8589934592 bits
this may be confusing but a gigabyte is 1073741824 bytes but also 1024 megabytes.
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for those of you who dont know why 1k is really 1024 its because it has to be in a form of 8...
when a computer transmits data it is all done in 1's and 0's and like i said each number, charector, or letter is represented by 8
and 1024 is a power of 2...,
2^10(2*2*2*2*2*2*2*2*2*2)=1024 where as to get 1000 you would need to do 2^9.965784285 which is alot more complicated to do than 2^10
lol
2006-09-13 11:55:56
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answer #2
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answered by Anonymous
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Because the advertised size is based on Base 10 numbering, whereas the computer uses Base 2 numbering. In base 10, 1 Giga is 10 to the 9th power. In base 2, 1 Giga is 2 to the 30 power.
2006-09-13 11:51:42
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answer #3
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answered by Chris J 6
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yeah it's somewhat misleading....
Scientifice notation for numbers has defined the following
Kilo - 1000
Mega - 1000000
Giga - 1000000000
but in computer terminology, 1 K is really 1024 and not 1000.
So 1G(Giga) = 1K x 1K x 1K = (1024) x (1024) x (1024)
So you get 1073741824
2006-09-13 11:55:09
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answer #4
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answered by sleepysheep 2
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Memory modules are never manufactured precisely with 1024 MB of memory and they will always have a small amount of extra memory.
http://www.cybertopcops.com
2006-09-13 11:53:10
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answer #5
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answered by cppgenius 4
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