1/0 = 'inf' why?

1/0 is unable. why did you write inf. Why is not Nan?

[Rik]

It just is. I suspect it has something to do with finding the limit for a/b with 'b' approaching 0 and 'a' a non-zero value.

[KSSV]

Read wiki article it explains. https://en.wikipedia.org/wiki/Division_by_zero

Answer 1

[Jan]

This is the IEEE754 standard:

 *Exception handling*

Division by zero (an operation on finite operands gives an exact infinite
result, e.g., 1/0 or log(0)) (returns ±infinity by default).

Answer 2

[Walter Roberson]

1/0 is the limit of 1/(1/N) as N approaches infinity. But for any finite N, 1/(1/N) is N. So as N approaches infinity, the value approaches infinity.

You could start from the other side of 0, in which case you could conclude that the results should be negative infinity.

There are thus two possible results, infinity and negative infinity. You might say that is enough to declare the result should be nan. But there are only the two possible results and infinity is pretty useful, so IEEE declared that a positive value decided by 0 would be positive infinity and a negative value divided by 0 would be negative infinity. It is useful.

The other cases such as 0/0 or infinity minus infinity that produce nan have stronger reasons for being indeterminate.

[Stefan H]

Yet, this explanation is mathematically incorrect, because 0 in the denominator is not necessarily the sequence 1/N or -1/N or anything with decreasing absolute value, but also the limit of the sequence 0, 0, 0, 0. In this case every element in the sequence 1/0 is undefined and no limit exists.

[Jan]

darova wrote on 28 Jun 2019 at 20:31:

Only MVP can debate with Walter

[Jan]

@darova: All users of this forum are potential MVP in the future. But all discussions are welcome. Sometimes a questions for clarifications helps to improve an explanation or to reveal a missunderstanding.

Do not overestimate the MVP status. I got it by posting many answers over the decades, but it is still an efficient idea to think twice instead of debating with Walter :-)

[Bruno Luong]

There is no rigourous consistent mathemathical justtification so such result.

Jan is correct, because IEEE standard states it like this.

The result is actually

Inf * (bit-sign of numerator) * (bit-sign of 0-denominator).

>> z=0

z =

     0

>> z==-z

ans =

  logical

   1

>> 1/z==1/(-z) % compare Inf and -Inf

ans =

  logical

   0

>>