How to solve an equation with FMOD in it

[1plus1is10]

TL;DR Summary

Replace FMOD with algebraic math

I have:
a = FMOD(((x*y)-z), x) / x

When I look at this, I think it could most likely be simplified algebraically if I could only replace FMOD with algebraic math.
After searching the web, I realize that this is more than I would know how to do since FMOD uses FLOOR.

Anyway, I thought I would ask here to see if my function can be simplified somehow (hopefully without FMOD).
Thanks

 

[anuttarasammyak]

You may transform it by function as
$$mod(m,n)=m-n\lfloor \frac{m}{n} \rfloor$$

 

[Mark44]

Summary:: Replace FMOD with algebraic math

I have:
a = FMOD(((x*y)-z), x) / x

In C and C++, fmod is the floating point modulus. Is that how you're using this function?

If so, I don't think the mod function that @anuttarasammyak mentioned will be helpful. fmod(x, y) returns the floating point remainder of x divided by y. For example, fmod(5.0, 4.0) would return 0.25.
Correction: the return value is 1.0

 

[1plus1is10]

Actually fmod(5.0, 4.0) is 1.0 and anuttarasammyak is on the right path with converting it that way.

You can test it here
http://cpp.sh/

#include <iostream>
#include <cmath>
int main(){ std::cout << fmod(5.0, 4.0); }The problem becomes, I have no idea how to do algebra with FLOOR unless it can be "transformed" or replaced somehow also.
Thanks

 

[1plus1is10]

std::cout << 5.0-(4.0*floor(5.0/4.0));

 

[pasmith]

The usual notation for FLOOR(x) is $\lfloor x \rfloor$.

 

[1plus1is10]

Yes, I know. I just do not know math markup code.

 

[Mark44]

Actually fmod(5.0, 4.0) is 1.0

Bad example from me, caused by not checking some code and from not using this function for a long time. The mod function returns an integer, but fmod's return value is a double or a float, depending on the arguments to it, that may or may not be the float/double equivalent of an integer.
Here's a better example to show what I mean:

double rem = fmod(3.5, 3.0);

After execution, rem will be set to 0.5.

 

[Mark44]

The problem becomes, I have no idea how to do algebra with FLOOR unless it can be "transformed" or replaced somehow also.

You can cast a float or double to type int, which will get rid of any fraction part, which is equivalent to what floor() does.
For example:

double dVal = 4.3;
int iVal = static_cast<int>(dVal);

iVal will be set to 4 .

 

[1plus1is10]

Mark44,
I appreciate the fact you know C++ which is what I use to perform calculations. But this for me really is a "converting-to-basic-algebra" problem at the moment.

I'm posting my "Math Notes" for you to get an idea of what I typically use to help me:

   A1       A2        A3        A4
(x/y)*z = x/(y/z) = x*(z/y) = (x*z)/y
(1/2)*3 = 1/(2/3) = 1*(3/2) = (1*3)/2
 0.5*3  = 1/0.666 =  1*1.5  =   3/2
  1.5   =   1.5   =   1.5   =   1.5

   B1       B2        B3         B4:B3      B5:B4,A3,A2   B6:B5,B1,B2   B7:B6,A3,A2
(x/y)/z = x/(y*z) = (x/z)/y = (1/y)*(x/z) = (1/y)/(z/x) = 1/(y*(z/x)) = 1/(y/(x/z))
(1/2)/3 = 1/(2*3) = (1/3)/2 = (1/2)*(1/3) = (1/2)/(3/1) = 1/(2*(3/1)) = 1/(2/(1/3))
 0.5/3  =   1/6   = 0.333/2 =  0.5*0.333  =   0.5/3     =  1/(2*3)    = 1/(2/0.333)
 0.1666 = 0.1666  = 0.1666  =   0.1666    =   0.1666    =  0.1666     =   0.1666

   1=A1      1=A2       1=A3       1=A4
a=(x/y)*z, a=x/(y/z), a=x*(z/y), a=(x*z)/y
a/z=(x/y), a*(y/z)=x, a/(z/y)=x, a*y=(x*z)
1/z=(x/y), 1*(y/z)=x, 1/(z/y)=x, 1*y=(x*z)
    ^        (y/z)=x,    <-x-^     y=(x*z)=(z*x)
    ^          (y/z)=1/(z/y),           y/x=z
1/(y/x)=(x/y)    <--substitute z---------^

   1=B1      1=B2       1=B3          1=B4
a=(x/y)/z, a=x/(y*z), a=(x/z)/y, a=(1/y)/(z/x)
a*z=(x/y), a*(y*z)=x, a*y=(x/z), a*(z/x)=(1/y)
1*z=(x/y), 1*(y*z)=x, 1*y=(x/z), 1*(z/x)=(1/y)
  z=(x/y),   (y*z)=x,   y=(x/z),   (z/x)=(1/y)  sub y  (z/x)=1/(x/z)
   z*y=x,     y=x/z,     y*z=x      z=(1/y)*x   sub z  (x/y)=(1/y)*x

And, of course, I am aware that not everything can be converted to basic algebra (oh well).
But if FLOOR can be, then that would be nice.
Thanks, again.

 

[1plus1is10]

After doing some more Googling "replace floor". I can now remodel my kitchen.
Ha ha, just being funny.

But I did find a couple of related hits:
https://math.stackexchange.com/ques...ctions-applied-to-a-fraction-using-only-the-a

https://math.stackexchange.com/questions/2868605/are-ceiling-and-floor-elementary-functions

To me this means, forget about it.
There is not anything basic about it.
I will just live with what I got.

Thanks everyone for trying to help.