Math Error with Primitive Operators - c++

Question

I am having an issue with primitive types using built in operators. All of my operators work for all datatypes except for float and (un)signed long long int.
Why is it wrong even when multiplying by one? Also, why does +10 and -10 give the same number as +1, -1, /1, and *1.
The number 461168601 was chosen because it fits within the max float and max signed long long int.
Ran the following code and got the following output:
fmax : 340282346638528859811704183484516925440
imax : 9223372036854775807
i : 461168601
f : 10
f2 : 1
461168601 / 10 = 46116860
461168601 + 10 = 461168608
461168601 - 10 = 461168608
461168601 * 1 = 461168608
461168601 / 1 = 461168608
461168601 + 1 = 461168608
461168601 - 1 = 461168608
The following code can be ran here.
#include <iostream>
#include <sstream>
#include <iomanip>
#include <limits>
#define fmax std::numeric_limits<float>::max()
#define imax std::numeric_limits<signed long long int>::max()
int main()
{
signed long long int i = 461168601;
float f = 10;
float f2 = 1;
std::cout << std::setprecision(40);
std::cout <<"fmax : " << fmax << std::endl;
std::cout <<"imax : " << imax << std::endl;
std::cout <<"i : " << i << std::endl;
std::cout <<"f : " << f << std::endl;
std::cout <<"f2 : " << f2 << std::endl;
std::cout <<std::endl;
std::cout << i << " / " << f << " = " << i / f << std::endl;
std::cout << i << " + " << f << " = " << i + f << std::endl;
std::cout << i << " - " << f << " = " << i - f << std::endl;
std::cout <<std::endl;
std::cout << i << " * " << f2 << " = " <<i * f2 << std::endl;
std::cout << i << " / " << f2 << " = " << i / f2 << std::endl;
std::cout << i << " + " << f2 << " = " << i + f2 << std::endl;
std::cout << i << " - " << f2 << " = " << i - f2 << std::endl;
}

Answer 1

The error is caused by the too big difference between 4611686018427387904 and 1 or 10. You should never sum numbers with a such difference, because actual difference between two closest floating point numbers grows with exponent value.
When two floating point numbers are added, the first of all they are aligned to the same exponent value (the bigger one), so before operation you have e.g. 1e10 and 1e-10 and after alignment you have 1e10 and 0e10 the result is 1e10.

Answer 2

Dug around some and found this article.
Casting opens up its own can of worms. You have to be careful, because your float might not have enough precision to preserve an entire integer. A 32-bit integer can represent any 9-digit decimal number, but a 32-bit float only offers about 7 digits of precision. So if you have large integers, making this conversion will clobber them. Thankfully, doubles have enough precision to preserve a whole 32-bit integer (notice, again, the analogy between floating point precision and integer dynamic range). Also, there is some overhead associated with converting between numeric types, going from float to int or between float and double.
So, essentially once the whole part of a number reaches about more than seven digits, the float begins to shift the number to keep the whole part of the number about seven digits. When this shifting of the decimal place occurs, the number begins to reach the floating point inaccuracy.