Why does rem(a, b) function give different answers for input values' class 'double' and 'uint64'?

I define two values as double,

a1 = 9419588158802421600;    b1 = 44;

I am trying to use built-in remainder function,

v1 = rem (a1,b1)

output>> v1 = 28

Now, I define the same values but as uint64

a2 = uint64(9419588158802421600);    b2 = uint64(44);

and use remainder function

v2 = rem (a2,b2)

output>> v2 = (uint64) 0

Why two different answers are obtained for the same input values?

Answer 1

[Steven Lord]

Let's look at the value stored in a1 converted to a uint64.

>> a1 = 9419588158802421600;    b1 = 44;
>> a2 = uint64(9419588158802421600);    b2 = uint64(44);
>> [uint64(a1); a2; intmax('uint64')]
ans =
  3×1 uint64 column vector
    9419588158802421760
    9419588158802421600
   18446744073709551615

Not all integer values on the magnitude of 9419588158802421600 can be exactly represented by a double. For example, the next representable number after a1 is:

>> uint64(a1 + eps(a1))
ans =
  uint64
   9419588158802423808

and the one immediately before it:

>> uint64(a1 - eps(a1)/2)
ans =
  uint64
   9419588158802419712

You can see this by looking at the hex representation of the numbers:

>> [a1 - eps(a1)/2; a1; a1 + eps(a1)]

ans =

   43e0572330befc9c
   43e0572330befc9d
   43e0572330befc9e

They differ only in the last hex digit. So the computations in double and in uint64 are operating on different data. When you try to compute on numbers that are close to or greater than flintmax be very careful.