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text: 'It takes $\\log_2(8) = 3$ [3p0 bits] of data to [3y1 carry] one message from a set of 8 possible messages. Similarly, it takes $\\log_2(1024) = 10$ bits to carry one message from a set of 1024 possibilities. How many bits does it take to carry one message from a set of 3 possibilities? By the definition of "bit," the answers is $\\log_2(3) \\approx 1.58.$ What does this mean? That it takes about one and a half yes-or-no questions to single out one thing from a set of three? What is "half of a yes-or-no question"?\n\nFractional bits can be interpreted as telling about the [4b5 expected] cost of transmitting information: If you want to single out one thing from a set of 3 using bits (in, e.g., the [+3tz] thought experiment) then you'll have to purchase two bits, but sometimes you'll be able to sell one of them back, which means you can push your expected cost lower than 2. How low can you push it? The lower bound is $\\log_2(3),$ which is the minimum expected cost of adding a [3v9 3-message] to a long message when encoding your message as cleverly as possible. For more on this interpretation, see [+3v2].\n\nFractional bits can also be interpreted as conversion rates between types of information: "A 3-message carries about 1.58 bits" can be interpreted as "one [3ww] is worth about 1.58 bits." To understand this, see [427], or [ How many bits is a trit?]\n\nFractional units of data can also be interpreted as a measure of how much we're using the digits allocated to encoding a number. For example, working with fractional [3wn decits] instead of fractional bits, it only takes about 2.7 decits carry a [3v9 500-message], despite the fact that the number 500 clearly requires 3 decimal digits to write down. What's going on here? Well, we could declare that all numbers that start with a number $n \\ge 5$ will be interpreted as if they start with the number $n - 5.$ Then we have two ways of representing each number (for example, 132 can be represented as both 132 and 632). Thus, if we have 3 decits but we only need to encode a 500-message, we have one bit to spare: We can encode one extra bit in our message according to whether we use the low representation or the high representation of the intended number. Thus, the amount of data it takes to communicate a 500-message is one bit lower than the amount of data it takes to encode a 1000-message — for a total cost of 3 decits minus one bit, which comes out to about 2.70 decits (or just short of 9 bits). For more on this interpretation, see [+44l].',
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