[hamdani yusuf (OP)]

Acording to Mathmetica which I consider the last word on maths 0^0 is inditerminate

What's the consideration behind its answer? Argument from authority is not a strong one in a scientific discussion.

[alancalverd]

Ok. They don't change the result much.

Squeeze the trigger slowly and you'll appreciate the enormous difference between "not much" and "unstable".

[hamdani yusuf (OP)]

1 * 1 is less stable. Perturbing one of the operands with a finite constant does change the result. But multiplying them with a constant infinitesimally close to 1, or adding them with a constant infinitesimally close to 0 doesn't change the result.

Oh yes it does!
(1 * Δ)² = Δ² ≠ 1
(1+Δ) ² = 1 + 2Δ + Δ² ≠ 1.

Ok. They don't change the result much.

Squeeze the trigger slowly and you'll appreciate the enormous difference between "not much" and "unstable".

How can 1*1 unstable?

[alancalverd]

It isn't. Only you suggested it was.

[hamdani yusuf (OP)]

It isn't. Only you suggested it was.

I said 1*1 is less stable than 0*0.

[Eternal Student]

Hi.
   This kind of "stability" idea just isn't worried about too much, to the best of my knowledge.  I think that's why you (Hamdani) aren't getting many good replies about the issue.   The term "stability" is usually applied to other things in mathematics and I'm not sure that you (Hamdani) need to develop a new set of ideas for the stability of basic field operations like addition and multiplication.   There is already a simple idea called continuity which seems to offer all we need.
   However, if you want to develop ideas for this sort of stability, then I don't suppose there's any law against it.

Here's something to think about if you are developing some ideas for stability of basic field operations like + and x on different choices of input numbers:
      You mentioned that 0 x 0  was very stable but that depends on how you measure stability.  Surely you'd want to consider a proportional or percentage change?    A change of  +/- 1 unit in an expected answer of 100 units is nothing to worry about but a change of +/- 1 unit in an answer of  0.01 units   is a    ten-thousand percent error.     Anyway, using this notion of proportional difference,  0 x 0   and   also  0 + 0    isn't stable, it's terrifyingly unstable.  Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded.   No other choice of input numbers is that bad.

Best Wishes.

[alancalverd]

A simple example of a trigger instability is the equation y = 1/x. As x passes through zero its value switches between +∞ and -∞.

[hamdani yusuf (OP)]

You mentioned that 0 x 0  was very stable but that depends on how you measure stability.

0 * 0 is extremely stable. The operands can be perturbed heavily without changing the result (e.g. by multiplying them with some constants). Adding one of the operands with a finite constant doesn't change the result. Adding both operands with a finite constant does change the result.

Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.

Anyway, using this notion of proportional difference,  0 x 0   and   also  0 + 0    isn't stable, it's terrifyingly unstable.  Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded.   No other choice of input numbers is that bad.

Can you show how?

[hamdani yusuf (OP)]

A simple example of a trigger instability is the equation y = 1/x. As x passes through zero its value switches between +∞ and -∞.

Yes. I understand if that instability makes people think that 1/0 is undefined.
https://www.wolframalpha.com/input?i=1%2F0
Wolframalpha says that 1/0 is complex infinity instead of undefined.
1/0=-1/0=i/0=-i/0=complex infinity. In Riemann's sphere, they are located on the same point.

There are seven indeterminate forms involving 0, 1, and infinity:
https://mathworld.wolfram.com/Indeterminate.html

If complex infinity is allowed as well, then six additional indeterminate forms result:

[Eternal Student]

Hi.

    Anyway, using this notion of proportional difference,  0 x 0   and   also  0 + 0    isn't stable, it's terrifyingly unstable.  Perturb the input numbers even a tiny amount and you generate a proportional difference that is unbounded.   No other choice of input numbers is that bad.

Can you show how?

0 + 0   = 0  = the expected answer for 0 + 0.
Perturb the input values slightly and consider  δ + ε  where  δ, ε are arbitrarily small but positive.
Then  δ + ε   ≠ 0, instead δ + ε =  something small but positive.   The percentage change from the expected answer, 0, is then undefined:
 (something positive)  /   0          x  100%       
 -  the division by 0 is a problem.

Best Wishes.

[hamdani yusuf (OP)]

0 + 0   = 0  = the expected answer for 0 + 0.
Perturb the input values slightly and consider  δ + ε  where  δ, ε are arbitrarily small but positive.
Then  δ + ε   ≠ 0, instead δ + ε =  something small but positive.   The percentage change from the expected answer, 0, is then undefined:
 (something positive)  /   0          x  100%       
 -  the division by 0 is a problem.

Best Wishes.

Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.

[Eternal Student]

Hi.

Perturbation by multiplication already implies ratio. Percentage change simply converts perturbation from addition to multiplication.

   ?  I've no idea what you neant there.

Best Wishes.