Why does a division by zero produce an error or undefined message?

AtoMick-u235 said:

A division by zero is a mathematical oddity, , if a computer or calculator etc performs a 1\0 they would crash, the OS prevents the execution, and displays ERROR or UNDEFINED, , , some versions of windows calculator display positive or negative infinity, , try it

AtoMick-u235 said:

Any ideas on this ?

Dale said:

It is pretty straightforward. $$x=a/b$$ means solve the following equation for ##x##:$$bx=a$$If ##b=0## then that equation has no solution. So division by zero is undefined

AtoMick-u235 said:

Hi, i understand the math:-

x = a/b so .5 = 1/2

bx = a so 2 x .5 = 1

But Infinity is not a Natural number, Integer, Rational, or even a Real Number so it cannot be represented with pen and paper, but i think in the real world with a computer the answer is 00000001 / 00000000 = infinity

DrClaude said:

Apologies to my co-mentors @Mark44 and @fresh_42, but I will reopen the thread as some of the answers are incorrect/incomplete.

A floating point following the IEEE Standard for Floating-Point Arithmetic (IEEE 754) can represent a real number, but can also represent positive or negative infinity, and NaN (not a number). Following that standard, since ##\lim_{ x \rightarrow 0} a/x = \pm \infty##, division by zero should return ##\pm \infty## (the sign will depend on the sign of ##a## and the sign of ##0##, which is a signed number in IEEE 754). See https://en.wikipedia.org/wiki/Division_by_zero#Computer_arithmetic

fresh_42 said:

IIRC, then the C++ compiler from MS throws an exception which has to be handled by the programmer or is thrown all the way up to the top, resulting in an automatic error message.

Your "can represent ##\pm \infty ## or ##NaN##" is exactly the exception handling, since the program cannot distinguish whether you tried to compute ##\lim_{n \to \pm\infty}\frac{1}{n}## or ##1/0.##

Hence we have an exception anyway. Makes sense as many IDE are written in C++.

#include <iostream>

using namespace std;

int
main ()
{
  double a = 1.;
  double b = -1.;
  double zero_plus = 0.;
  double zero_minus = -0.;

  cout << a/zero_plus << " " << a/zero_minus << " " << b/zero_plus
       << " " << b/zero_minus << "\n";

  return 0;
}

$ g++ div_zero.cpp
$ ./a.out
inf -inf -inf inf

DrClaude said:

As I said above, the sign of ∞ is determined fro the sign of the numerator and the sign of the zero, which in IEEE 754 is signed.

DrClaude said:

There is no exception thrown when the IEEE 754 standard is followed.

martinbn said:

Is this thread about why the calculator does something when asked to divide by zero or is it about why division by zero is not defined?

martinbn said:

Is this thread about why the calculator does something when asked to divide by zero or is it about why division by zero is not defined?

AtoMick-u235 said:

At the time i thought of both, but it's just a general question about "Division By Zero" errors

fresh_42 said:

Well, C++ has been a while since I last directly used it. And the IDE I use nowadays fortunately throws an exception since anything else is mathematical nonsense; and this thread has landed in GM.

jack action said:

It kind of makes sense to separate inf from undefined.

jack action said:

Playing around with the C++ program of @DrClaude , the only time it returns nan is when you have a form ##\frac{0}{0}## in your equation^[1].

For example, it returns the correct answer ##0## to the equation cout << 1/(2/0.); even if you do it into 2 separate equations like so: a = 2/0.; cout << 1/a;.

jack action said:

----------------------------------------------------
^[1] ##0 \times \infty## and ##\frac{\infty}{\infty}## are just special cases of ##\frac{0}{0}##, namely ##0 \times \frac{a}{0}## and ##\frac{^a/_0}{^b/_0}##.

Mark44 said:

In my experience, computers don't crash if a program attempts to divide by zero. Instead, the program will throw an exception; namely a divide by zero exception.

In Windows 10, both the Scientific mode and Standard mode simple display a message: "Cannot divide by zero". Which versions of the Windows calculator are you talking about?

BTW, I have moved your thread to the General Math section as it has nothing to do with relativity.

fresh_42 said:

I meant the C++ compiler, not the program code.

AtoMick-u235 said:

In my all my programing over the years, i have done a lot of compiling , a "Division by Zero" is extreamly rare, my cousin has been an accountant for 40 years, i asked her, and she never encountered a "Division by Zero" ever, , ha

fresh_42 said:

This does not answer the question of how Throw Exception Catch Exception is realized so that it can catch division by zero or any other exception.

• When the CPU encounters an exception condition it executes a trap with a certain value (a trap is a special kind of interrupt, one that is generated by the CPU rather than external hardware or software).
• The processer handles the trap like other interrupts by looking up the value in an interrupt vector which stores the address of a subroutine to call.
• The addresses stored in the interrupt vector are set by the operating system which will default to terminating the relevant program when an exception occurs.
• An application program may include code to handle exceptions occurring while it is executing itself. On loading it will tell the operating system about this so that instead of terminating the program the OS will call the relevant code.
• Many high level languages include instructions to handle exceptions, typically using a ‘try…catch’ block (‘try…except’ in Python and Pascal). When the compiler encounters the ‘try’ token it inserts code to call the ‘catch’ block when an exception is encountered.
• Most high level languages will ignore floating point divide by zero by default, allowing the program to continue with the value of + or - Infinity (or NaN for 0/0). Among modern languages used for scientific applications (including Matlab and Mathematica), Python is an outlier in this respect: it throws an exception for any division by zero.

fresh_42 said:

It's not always a numerical problem, but those count, too, e.g. the logarithm of a negative number or an invalid string handling.

pbuk said:

SIMD

pasmith said:

removable discontinuities