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Elsewhere arose a discussion about logical clauses that can be made from indeterminate forms, in this case, namely $0 / 0$. Since $0 / 0$ is indeterminate form, can we make these logical clauses:

  • $0 / 0 = 1$ is false?
  • $0 / 0 \neq 1$ is true?

Or in more general form, if $x$ is not just undefined, but indeterminate form, and $y$ is defined and $y \in \mathbb{R}$, can we say that:

  • $x = y$ is false?
  • $x \neq y$ is true?

My thinking goes, that since $x$ cannot be determined in any way, first one is intuitively correct: any clause that tries to say that $x$ is something must be false. Based on simple logics, it follows that also all clauses that $x$ is something else than something that is defined, are true.

Counterargument is, that if $x$ is indeterminate form, we cannot say that it is not $y$, for that it would make a clause that $x$ is something, because it is at least not $y$, which cannot be true if $x$ is truly indeterminate (i.e. if $x$ is indeterminate, we cannot say that it isn't $y$. To my thinking, this leads to also back to that $x = y$ is at least not true. But with the same argumentation, it cannot be false either, and hence both logical clauses are themselves undefined in answer, not true but not false either.

And counter-counterargument is that, if we make a clause that says $x \neq y$, it does not make $x$ any less indeterminate, because it only says that $x$ is at least not that particular defined form, but still leaves open possibilities that $x$ is something else than $y$ or still completely indeterminate.

My thinking goes that if $x$ is indeterminate form, it means that $x \notin \mathbb{R}$ (or $\mathbb{R}$, for that matter), and hence all clauses that $x \neq y \in \mathbb{R}$, are true, because what we can say about indeterminate forms is that they are not in real number space. But that might be untrue as well, if it is also indeterminate whether indeterminate forms exist in real numbers or not.

What comes to undefined numbers, e.g. if instead of $0 / 0$ we talk about $a / 0$, where $a \neq 0$, it is more clear that they do not exist in real numbers, so if $x$ was only undefined, I believe $x \neq y$ is without doubt true. I.e. clauses that undefined numbers are always not equal to any or all of defined numbers are necessarily true, because it is basically the same clause that $x \notin \mathbb{R}$, which is true if $x$ is undefined.

The answer to this is probably very much not indeterminate, but with my basic knowledge of algebra a definitely correctly-argumented answer cannot be made.

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    $\begingroup$ My view: if $0/0$ is undefined, then $0/0=1$ is undefined and has no meaning. You can't say if something that you haven't defined is true or not. $\endgroup$
    – user236182
    Sep 18, 2015 at 18:23
  • $\begingroup$ @user236182, first of all, 0/0 is not undefined, it is indeterminate form. 1/0 is undefined. But when something is undefined, does not necessarily mean that all logical clauses containing it are undefined. If something is undefined, we know that is is and will be undefined, hence we can make a clause "x is undefined", which is true. And hence "x is defined" is false, not undefined. And if we know that x is not defined (and will never be), we should also know that x cannot be something that is defined, which would lead to that we know that "x = 1" is false. $\endgroup$ Sep 18, 2015 at 18:41
  • $\begingroup$ Not to argue, but $0/0$ is undefined (and an indeterminate form when dealing with limits). If it were defined, then you could say:$0/0=...$ $\endgroup$
    – TravisJ
    Sep 19, 2015 at 1:22

3 Answers 3

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I had a question like this one a while back. I asked two question about it and got some answers. Take a look at

Here the question is basically: If a statement isn't welldefined, does it make sense to say that it is true or false. I think there are some disagreement and there are different points of view, but here is what I cam away with.

We only assign the value of true or false to a statement that is well defined / well formed. So saying that

fsdfsdf

is not a true of false statement because it isn't a (well defined) statement. So the question is now: Is $0/0 = 1$ a well defined statement? I don't think it is since $0/0$ isn't defined. $0/0$ doesn't make sense. So saying that something is equal to something else assumes that the somethings are well defined.

Some will complain about this because we can say that $\lim_{n\to 0} \frac{1}{x} = 1$ is a false statement (so it is a well defined statement) even though the limit doesn't exist. I think you can solve this problem by (1) simply not saying that the statement is false, or (2) by still saying that while $\lim_{x\to 0}]\frac{1}{x}$ doesn't exist, it is still a valid expression, or (3) simply definition the statement $\lim_{x\to 0}\frac{1}{x} = 1$ as an equivalent form of $\forall \epsilon > 0 \exists \delta > 0 : 0<\lvert x \rvert < \delta \Rightarrow \lvert \frac{1}{x} - 1\rvert < \epsilon$. In this case you simply consider the statement as a piece of notation that stands in the place of something else that has a perfect fine meaning. I would tend to go with option (1) (or (3)) because of the same problem with $1/0$.

Now, we do sometimes talk about certain limits as being indeterminate. We do this, for example, when we have a fraction where the numerator and denominator both approach zero. In this case we say that we have an/the indeterminate form $0/0$. But this doesn't mean that $0/0$ is well defined.

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A good way to interpret $0/0$ is that it is a sign that an error has been made.

Your question in this context: Is $x = y$ true? Think of it this way: if you are certain that an error has already been made, why try to draw a conclusion from it? It would be unreliable either way.

$0/0$ = "an error has already been made" = "everything from this point on is unreliable" = "go back and fix the error so that you no longer run into $0/0$".

What if you encounter $0/0$ as a limit? That means the same as it always means: "an error was made, go back and fix it". The entry "$0/0$" in the table of indeterminate forms indicates how to correct an error of that type (use l'Hôpital's rule).

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  • $\begingroup$ Why do you think it is a sign of error? For instance, if we have a function f(x, y) = x / y, it forms 0 / 0, if x goes to 0 and y goes to 0. It does not indicate that any error would have been made, it is just undefined if x = 0 and y = 0. You cannot "fix" f(x, y) = x / y so that we wouldn't have to determine its behaviour at (0, 0). We have to, and we have determined it's an indeterminate form. And my question was that does it lead to being not equal to all real numbers or also indeterminate regarding its truth value. $\endgroup$ Feb 18, 2017 at 20:24
  • $\begingroup$ Yes, it would be an error to try to determine a defined value for something that cannot be determined. But it is not an error that we have a function that cannot have a defined value at all points. And my question is that if such point would be not equal to all real numbers, as it definitely cannot be said that it would be equal to any real number. And, in my question 0 / 0 was just an example. $\endgroup$ Feb 18, 2017 at 20:27
  • $\begingroup$ In computing there are various protocols to deal with $0/0$. The most common is to stop the computation and return an error message (which is what I meant with my answer) but there is also the NaN mechanism (Not A Number). Here, comparisons always return "false" so even something like $x=x$ evaluates to "false" when $x = $NaN. You could of course propose a different protocol for NaN. $\endgroup$
    – Mark
    Feb 19, 2017 at 12:42
  • $\begingroup$ My question was not about computing at all. $\endgroup$ Feb 20, 2017 at 13:26
  • $\begingroup$ The point is that mathematics is completely reliable and consistent as long as one follows the rules. Break one rule, such as dividing by $0$, and reliability can no longer be guaranteed; some proven facts may no longer hold if you do this. On your question "can we say that $x=y$ is false?". The answer is: you can choose to allow division by $0$ and still assign true/false values to statements like that, in whichever way you want. However, you can't always trust the results that come from this. $\endgroup$
    – Mark
    Feb 20, 2017 at 15:23
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Indeterminate forms arise in analysis in the context of limits, but it does not mean they cannot be assigned values algebraically. Thus, indeterminate form is not necessarily undefined. Here is a list of values that create the least troubles algebraically and break the least rules:

  • $0^0=1$ - this is more or less universally accepted, follows from algebraic property of empty product.
  • $\infty^0=1$ - from the same property as the previous one.
  • $1^\infty=1$ - this follows from the algebraic properties of multiplicative unity (it is adsorbing element of exponentiation).
  • $0\cdot\infty=0$ - this follows from the absorbing property of zero.
  • $0/0=0$ - follows from the previous one and associativity.

The remaining indeternimate forms cannot be assigned a value algebraically:

  • $\infty-\infty$
  • $\infty/\infty$
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