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Round-by-chop: The base-expansion of is truncated after the ()-th digit. This rounding rule is biased because it always moves the result toward zero. Round-to-nearest: () is set to the nearest floating-point number to . When there is a tie, the floating-point number whose last stored digit is even (also, the last digit, in binary form, is equal ...
For example, the decimal number 123456789 cannot be exactly represented if only eight decimal digits of precision are available (it would be rounded to one of the two straddling representable values, 12345678 × 10 1 or 12345679 × 10 1), the same applies to non-terminating digits (. 5 to be rounded to either .55555555 or .55555556).
In the example from "Double rounding" section, rounding 9.46 to one decimal gives 9.4, which rounding to integer in turn gives 9. With binary arithmetic, this rounding is also called "round to odd" (not to be confused with "round half to odd"). For example, when rounding to 1/4 (0.01 in binary), x = 2.0 ⇒ result is 2 (10.00 in binary)
However, it is quite likely that it is not safe to round off the intermediate steps in the calculation to the same number of digits. Be aware that roundoff errors can accumulate. If M decimal places are used in the intermediate calculation, we say there are M−N guard digits.
For example, while a fixed-point representation that allocates 8 decimal digits and 2 decimal places can represent the numbers 123456.78, 8765.43, 123.00, and so on, a floating-point representation with 8 decimal digits could also represent 1.2345678, 1234567.8, 0.000012345678, 12345678000000000, and so on.
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What you are saying is simply that a stop watch rounds down to one decimal place so one can have the problem of double rounding. You really need a citation before sticking in your own thoughts on things like that. Dmcq 18:13, 12 April 2016 (UTC) There was the same issue about the bias, which is also due to double rounding.
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