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Numerical Multiplication in MATLAB

BM 님이 질문을 제출함. 24 Oct 2019
최근 활동 BM 님이 편집함. 24 Oct 2019
Using MATLAB as a simple calculator, we get the results
2 * 0.155 * 100 = 31,
2 * 0.165 * 100 = 33,
2 * 0.135 * 100 = 27,
2 * 0.125 * 100 = 25
but the moment I compute the following
2 * 0.145 * 100 = 29.0000
I didn't have any variables in the memory, nor any settings enabled. These were on a blank canvas, but this particular value of 0.145 yields 29.0000 with four decimal places. I am a little unsure of why this value of 0.145 yields an answer with decimals, whereas the other calculations yield integers. Does anyone have an answer?

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John D'Errico 님의 답변 24 Oct 2019
John D'Errico 님이 편집함. 24 Oct 2019
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So what is your problem? If you think the result should always be an exact integer, that just means you don't appreciate floating point arithmetic.
Perhaps you wonder why the last of those computations does give an exact integer. That happens because 0.125 is EXACTLY representable as a binary number, thus 2^-3. We could write it in the form of
0.00100000000000000000000000...
in binary bits.
So when you multiply it by 2*100, you turn the result into the exact integer 25.
The other results are NOT exactly representable in binary. For example, the decimal number 0.145 looks like
0.0010010100011110101110000101000111101011100001010001111...
as a repeating binary form, where each 1 in that expansion represents a negative power of 2.
This is no different from the fact that you cannot write the fraction 1/3 as a terminating decimal number.
That means you cannot expect all of those results to always be exact integers.
sprintf('%0.55f',0.145)
ans =
'0.1449999999999999900079927783735911361873149871826171875'
sprintf('%0.55f',2*0.145*100)
ans =
'28.9999999999999964472863211994990706443786621093750000000'
29 == 2*0.145*100
ans =
logical
0
In a binary form, we might write what MATLAB generates for 2*0.145*100 using this expansion:
11100.111111111111111111111111111111111111111111111111
Whereas we know that 29 is:
dec2bin(29)
ans =
'11101'
So the result is off by one bit down at the least significant bit of the number.
All of this is due, not to MATLAB, but to the IEEE representation of floating point numbers, used by most computing languages.

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BM 24 Oct 2019
Hi John,
Thanks for your answer. This part of the code serves as a numerical check for my program. I guess I had been staring at analytical calculations so long that when this checking error was thrown up in the numerical calculation, its reconciliation through binary representation had completely escaped me at that moment.

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