On Fri, 15 Apr 2016 12:05:18 +0100, John Pote wrote:
> On 15/04/2016 03:38, Christian Gollwitzer wrote:
>> Am 15.04.16 um 02:36 schrieb Dennis Lee Bieber:
>>>>> I should also have said that the square root of integer squares with
>>>>> between 15 and 30 decimal digits will only be correct if the square
>>>>> numbers themselves are exactly representable in 53 bits. So we can
>>>>> expect failures for squares with 16 or more digits.
>>>>
>>>> However, if a number with 31 or less digits is known to be the square
>>>> of an integer, the IEEE754 sqrt function will (I believe) give the
>>>> correct result.
>>>
>>> How could it EXCEPT by having ~15 significant digits and an
>>> exponent --
>>> since that is all the data that is provided by a double precision
>>> floating point. That is, for example,
>>>
>>>>>> 1000000000000000.0 * 1000000000000000.0
>>> 1e+30
>>>>>> import math math.sqrt(1e30)
>>> 1000000000000000.0
>>>>>>
>>>>>>
>>> only has ONE significant digit -- even though it has thirty 0s before
>>> the decimal point.
>>
> As I was taught in school and university the number of significant
> digits are the number of digits written after the point, be that
> decinal, binary or any other base.
Then I would complain to your education regulator
Significant digits are the 1st X digits in the number (with rounding)
after converting to standard form.
examples to 3 significant digits
Number Standard form 3 Significat digits
10 1.0e1 1.00e1 = 10.0
100 1.0e2 1.00e2 = 100
3.14159 3.14159e0 3.14e0 = 3.14
0.01777 1.777e-2 1.78e-2 = 0.0178
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