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floating-pointnumber-theory

Proof that the last non-zero digit to the right of the decimal point in a non whole number float is always 5?

Is it true that the last non-zero digit to the right of the decimal point in a non whole number float is always 5? Can/has someone prove[ed] this formally? Also, how many 0s in a row to the right of the decimal point are needed to conclude that there are no more non-zero digits farther to the right--I have seen #s such as 1.23450678500000 (an example, probably not a real float), so I know it is at least two.

Solution

  • Is it true that the last non-zero digit to the right of the decimal point in a non whole number float is always 5?

    In a binary floating-point format, if a number is not an integer, the last non-zero digit of its decimal numeral is 5.

    Generally, a floating-point format represents numbers as some integer M multiplied by power e of a fixed-base b: M•be, with various limits on M and e. If b is two, we call this a binary format. Binary floating-point formats are currently the most common. Some platforms provide decimal formats, where b is 10. Other bases are mathematically possible but rarely used, although 16 was somewhat used in the past.

    To see the last digit is 5, consider the form M•be. The only way this can be a non-integer is for e to be negative. If M is even, we can reduce the form to (M/2)•be−1, and, if M/2 is also even, we can reduce further and continue reducing until we have an odd integer multiplied by a negative power of 2. This is equivalent to dividing by a positive power of 2. Then we can see that the last digit is 5:

    • If we are dividing by 21, the number is .5, 1.5, 2.5, 3.5, and so on.
    • If we are dividing by 22, the number is half of one of those, so it is .25, .75, 1.25, 1.75, and so on.
    • If we are dividing by any greater power of 2, we are always dividing some number that ends in 5 by 2, and that means we carry the ½ fraction from the position where the 5 is to the next position, producing half of 10 there, and half of 10 is 5.

    Alternately, we can see that multiplying a numeral by 2, even repeatedly, can never make a digit position 0 unless it is 0 or 5. Multiplying 1, 2, 3, 4, 6, 7, 8, or 9, will produce 2, 4, 6, 8, 2, 4, 6, and 8, respectively, and multiplying those by 2 produces a digit from the same set. So, if a non-integer multiplied by a power of 2 produces an integer, its last non-zero digit must be 5.

    Also, how many 0s in a row to the right of the decimal point are needed to conclude that there are no more non-zero digits farther to the right…

    This depends on the specific floating-point format. A high-precision format can represent numbers that are very close to some number .xxxx500 but not equal to it. For any given floating-point format, there must be some limit to the number of zeros there can be before a non-zero digit, but it may require some work to figure it out. If you have concerns about finding the number of digits required to represent a floating-point number accurately, there are other ways to approach it.