The number N can be represented using 0 bits if we know in advance that's the only number that could be used in a context. − Woodstone ( talk) 12:16, (UTC) Reply It could be phrased a bit better. You should indicate the rounding up of the value explicitly. Jakob.scholbach ( talk) 11:47, (UTC) Reply However, how do you represent N=1 in zero bits? Perhaps you mean to say that log 2 N bits are needed to encode all integers from 1 to N (inclusive)? Then (1) carries no information, 0 bits, for (1,2) you need 1 bit, for (1,2,3) 2 bits. "so any non-negative integer N needs (log_2 N) + 1 bits to represent it."ĭependent Variable ( talk) 07:53, (UTC) Reply Go ahead! (I would prefer the first of the two).
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"so any positive integer N needs (log_2 N) bits to represent it." While "no more than" covers both cases, wouldn't it be clearer to chose one definition, either positive integers or non-negative integers? Thus: According to that article, it may or may not include 0. "with k bits (each a 0 or a 1) one can represent 2^k distinct values, so any natural number N can be represented in no more than (log_2 N) + 1 bits." contribs) 23:23, 24 March 2010 (UTC) Reply apropos history section.Preceding unsigned comment added by Timeroot ( talk and all the other properties there are easily derived from three basic ones (change of base, exponent->coeff, coeff->addend), and still many of them only work once restricted to positive/integer numbers only.
The Russian article on Logarithms has all the properties as well as many of their proofs.ĭavid Yakubov ( talk) 01:57, 5 March 2010 (UTC) Reply As stated before, the equation: log ab = log a + log b is more generally true than log ab = log |a| + log |b| This more complex and less valid formula does certainly not belong in the lead section.
L o g ( x ) = lim n → ∞ n x 1 / n − x − 1 / n 2 I was bored one day, I was playing around with the ln(x) function on a piece of graph paper and a calculator. :-P I noticed an unusual property:
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