Could not find solution to round a number up with math without if statements, conditionals, relational operators and built-in functions(basically purely arithmetic) so I tried to make one(in python)
How it works: Uses x^(0^x) to act as a conditional, if x(For all values 0 and greater) representing the decimal place, is 0 or equivalent, the function will return 0 else it will return 1. The returned value is then added to the number's floor division by 1.
Example: Number = 7.09 x = 0.09 #decimal value(Number Modulo 1) Plug into function y = x^(0^x): y = 0.09^(0^0.09) = 0.009^(0) = 1 Add y to Number's floor division: Number(rounded up) = y + (7.09//1) = 1 + 7 = 8 (successfully rounded up!)
Now with an integer: Number = 6 x = 0 #decimal value(Number Modulo 1) Plug into function y = x^(0^x): y = 0^(0^0) = 0^(1) = 0 Add y to Number's floor division: Number(rounded up) = y + (6//1) = 0 + 6 = 6 (success, 6 does not round up as it is already a whole number)
With Negative number:
Number = -8.9
x = 0.1 #decimal value(Number Modulo 1)
Plug into function y = x^(0^x):
y = 0.1^(0^0.1)
= 0.1^(0)
= 1
Add y to Number's floor division:
Number(rounded up) = 1 + (-8.9//1)
= 1 + -9.0 #floor division rounds towards negative infinity
= -8 (successfully rounded up!)
Function's Graph:
