kbw wrote: |
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But you could extend that statement to any base, base 3, 10 or 16 for example. There's nothing special about using base 2. |
Don't confuse OP. He's working with bit representation on binary computer systems.
Thought experiments:
(1)
unary: digits = { 0 }, of which there is exactly one.
There is no number of bits that can hold exactly one value.
(One bit holds two values, which is too many.)
(2)
binary: digits = { 0, 1 }, of which there are exactly two.
A single bit can hold exactly two values.
(3)
ternary: digits = { 0, 1, 2 }, of which there are exactly three.
There is no number of bits that can hold exactly three values.
(One bit holds two values, which is too few; two bits holds four values, which is too many.)
(4)
quaternary: digits = { 0, 1, 2, 3 }, of which there are exactly four.
Two bits can hold exactly four values.
(5)
quinary: digits = { 0, 1, 2, 3, 4 }, of which there are exactly five.
There is no number of bits that can hold exactly five values.
(Two bits holds four values, which is too few; three bits holds eight values, which is too many.)
etc.
Consider:
n bits holds exactly 2
n values.
n-ary has exactly n digits.
You should be able to put this together.
Hope this helps.