everyone. It is my first time here.

I have always had a doubt in deciding whether a proposition can be true, false or neither, depending on the general hypothesis of the definitions in the context. It will be clearer with some examples:

1) Supose we are working in $\mathbb{Z}$ and define for $a, b \in \mathbb{Z}, a \neq 0$, $a|b$ if there exists $k \in \mathbb{Z}$ such that $b = k \cdot a$.

What can I say about the following sentence: $0|1$?

One could say that it is false because there not exists an integer $k$ such that $1 = k \cdot 0$, but I think it wouldn´t be correcte because $a=0$ was excluded in the definition of divisibility. Does this mean that the sentence $0|1$ has no sense? It can be neither true nor false? Is it even a proposition?

2) In real analysis, we know that, in the definition of limit of a real function of real variable, in the expression $\lim_{x \rightarrow a} f(x)$, we assume that $a$ is a limit point (or accumulation point) of $Dom(f)$. Then, if we consider a function $f \colon D \rightarrow \mathbb{R}$, with $Dom(f)=\{2\} \cup (3, +\infty)$, what can I say about the sentence: $\lim_{x \rightarrow 2} f(x)=f(2)$? Is it true, false, or none? Has it any sens? Is it a proposition?

I hope my question to be clear.

Thank you very much and sorry for my English.

  • 1
    $\begingroup$ As you've suggested, both formulae are ill-defined; thus, neither is a proposition or even an open formula that has a varying truth value. $\endgroup$
    – ryang
    May 14 at 8:13
  • 1
    $\begingroup$ Thanks for your answer, ryang. I am new here and I tried to vote your comment. Accidentally I have undone my vote and now I cannot vote again. But I appreciate a lot your comment. $\endgroup$
    – Luks
    May 14 at 15:59


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