If 0 / 0 is indeterminate, are all clauses "0 / 0 != x" true 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.
 A: I had a question like this one a while back. I asked two question about it and got some answers. Take a look at


*

*True or false or not-defined statements

*$[\frac{1}{x} = 1$ for all real numbers $x]$ is not true or false
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.
A: 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).
