Say we're back in the 1950s and our clunky computer has a $1$ kilobyte memory. $1$ kilobyte is $1024$ bytes, which is $8192$ bits. When we consider the entire available storage space, there are $2^{8192}$ states we can harness (each bit can be on or off), but in actuality we only utilize $1024$ size-$8$-bit states. In short, we store only $1024$ items, each item being allotted a byte ($8$ bits) of memory when it seems possible to use a $8192$ bit system to store $2^{8192}$ items.

I have very little knowledge of computers. Gracias, muchas.

My question is, for a $1$ kilobyte memory, wouldn't it be better to use an $8192$-bit system (each word or unit of memory being $8192$ bits long) rather than a $8$-bit system (the unit of memory being $1$ byte or $8$ bits). There are more combinations in the former than the latter, which in my book would mean I could store more information. What's all the Sturm und Drang about $32$-bit and $64$-bit "systems"?

  • 2
    $\begingroup$ "1 kilobyte is 1024 bytes, which is 8192 bits" – In the 1950s, that would be a bold assumption. There were computers with 6-bit, 7-bit, 9-bit, and even 12-bit bytes. Even today, there are computers with byte sizes other than 8. $\endgroup$ Commented Feb 14 at 16:52
  • $\begingroup$ @JörgWMittag, thank you for correcting me. As I said, my knowledge of computers is limited to $2$ years of basics. $\endgroup$
    – Hudjefa
    Commented Feb 14 at 20:06
  • $\begingroup$ Don't confuse "storing more information" with "the number of combinations". In your example, using a single 8192 bit "word", you can only store one item, which is not very practical for most real world applications. Programmers generally decide on a practical "word" size for the data they manage (generally somewhere between 8 and 64 bits for numeric items). That way, they can store many more individual items, which is a requirement in most applications. $\endgroup$ Commented Feb 15 at 14:46
  • $\begingroup$ @SteveMathwig $2^{8192}$ different permutations of $1$s and $0$s. $\endgroup$
    – Hudjefa
    Commented Feb 16 at 1:15

2 Answers 2


You are thinking about it wrong. 8192 bits can only store 8192 single bit items. Now, the number of combinations that the entire collection of 8192 bits can store is indeed 2 to the 8192.

  • $\begingroup$ I'm sorry, but isn't each combination a valid memory address/unit? The reason $8$ bites or $1$ byte was selected is because it enables $256 = 2^8$ combinations (ASCII ?). $\endgroup$
    – Hudjefa
    Commented Feb 14 at 20:04
  • 1
    $\begingroup$ Again, it is how you look at it. The entire 8192 bits, if looked at as a single 8192-bit "word", can have 2 to the 8192 unique values. If looked at as 1024 8-bit "words", it can have 1024 "words", each word can consist of 2 to the 8 unique values. If looked at as 512 16-bit "words", it can have 512 "words", each word can consist of 2 to the 16 unique values. No matter how big the "word" size is, the entire memory, if looked at as consisting of a single 8192-bit "word", would contain up to 2 to the 8192 different values. $\endgroup$ Commented Feb 14 at 21:25
  • $\begingroup$ $8192$ bits = $2^{8192}$ different permutations (on, $1$, and off, $0$). $\endgroup$
    – Hudjefa
    Commented Feb 16 at 1:17

It does not matter whether 8192 bits are subdivided into 1024 items of 8 bits, 512 items of 16 bits, 256 items of 32 bits, or not subdivided at all - the amount of information one can store in that memory stays always the same. To express this fact in simple math counting the combinations:

$2^{8192} = (2^8)^{1024} = (2^{16})^{512} = (2^{32})^{256}$

which shows your idea of

There are more combinations in the former than the latter,

must have a calculation error.

  • $\begingroup$ Maybe, non liquet! Consider how, though, a "word-length" of $8$ bits has only $256$ possibilities, while a word-length of $16$ bits has $65536$ possible permutations (read words). $\endgroup$
    – Hudjefa
    Commented Feb 16 at 1:11
  • $\begingroup$ @AgentSmith that also doesn't change anything "real", if you have 16 bits then there are 65536 possible patterns those bits can have, doesn't matter whether you look at those 16 bits as one 16-bit word or as two independent 8-bit bytes. It's the same either way, for the same reason described in this answer. If you simply add more bits in total, of course there are more possible patterns, but then you're cheating. $\endgroup$
    – user555045
    Commented Feb 16 at 3:26
  • $\begingroup$ @AgentSmith: I have no idea what you are trying to tell me. My math tells me $2^8 \times 2^8 = 2^{16} = 65536$, hence two words of 8 bits can store the same amount of information as one word of 16 bits. If it is still not clear, try to express the unequality you see as a mathematical formula, then we can discuss where your calculation error is. $\endgroup$
    – Doc Brown
    Commented Feb 16 at 5:52
  • $\begingroup$ @DocBrown, A unit of memory consists of a certain number of bits. An $8$ bit memory unit can hold $2^8 = 256$ individual pieces of data, each individual piece of data of size $8$ bits. A $8192$ bit unit of memory can hold $2^{8192}$ individual pieces of data, each one being $8192$ bits long. $1$ kilobyte = $1024$ bytes = $8192$ bits. $\endgroup$
    – Hudjefa
    Commented Feb 17 at 0:14
  • $\begingroup$ @AgentSmith: so your comment in short says 1024 units of 8 bytes = 8192 bits, 1 unit of 8192 bits = 8192 bits, too. So where is there an inequality? $\endgroup$
    – Doc Brown
    Commented Feb 17 at 6:53

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