Why 2.4 mod 0.8 returns 0.8

Question

Why does calculating a module of 2.4 by 0.8 returns 0.8? If it does not return 0?

This happens both in PHP and through fmod and JavaScript through % .

Even google research with this operation fails: http://www.google.com/search?q=2.4+mod+0.8

Answer 1

If you look at what is happening in this particular example, the influence of the binary format becomes more noticeable.

Input is saved as a binary floating-point approximation. This can then be printed as a decimal number that uniquely identifies the binary,

 In [1]: print "%.25f"%2.4 2.3999999999999999111821580 In [2]: print "%.25f"%0.8 0.8000000000000000444089210 

The first step is to calculate the quotient and its integer part. Obviously, since the stored numbers are less than 2.4 and greater than 0.8 , the coefficient is less than 3 ,

 In [3]: print "%.25f"%(2.4/0.8) 2.9999999999999995559107901 

The result modulo - the remainder - is, finally, the difference after removing the integer,

 In [4]: print "%.25f"%(2.4-2*0.8) 0.7999999999999998223643161 

Answer 2

Simply put, to represent a number in decimal, we do something like this:

 1550 = 1000 + 5·100 + 5·10 = 1·10³ + 5·10² + 5·10¹ 

To represent a binary number from a decimal number, you must change the bases:

 106 = 2^6 + 2^5 + 2^3 + 2^1 = 0110 1010 

Where 0 will be equal to 0 · 2 ^ 7, 0 · 2 ^ 4, etc.

As you can see, the pattern is clear: just sum the values ​​of 2 as needed until you get the number you need, until you reach 0 (2 ^ 0) for number 1. If you work a little, the same goes for decimals, you just continue subtract 1 from the authority:

 0.5 = 2^-1 = 1/2^1 

Now for the other numbers:

 0.6875 = 0.1011 = 1/2^1 + 1/2^3 + 1/2^4 

However, there are numbers that you cannot imagine as sums of inverse powers of two, something as simple as 0.1 should require you to add very small numbers to approximate this value and it will look like 0.0001100110011... which actually is rounded to 0.0999755859375... , and since the language must change the bases every time you have to do arithmetic, these problems begin to appear.