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When talking about quantities that vary, does $\neq$, by default mean "is always not equal to" or "is not always equal to"?

The context of this is the following question:

enter link description here

We're having a dispute over the meaning of the $\neq$ in the absence of a $\forall$ or "in general" statement here. Does $\mathbb R^n \neq \mathcal C(A)+\mathcal N(A)$ mean "$\mathbb R^n$ is never equal to $\mathcal C(A)+\mathcal N(A)$", or "$\mathbb R^n$ is not always equal to $\mathcal C(A)+\mathcal N(A)$"

I'd prefer a source for the definition, if possible.

Manishearth
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  • $A$ is clearly meant to be an arbitrary matrix in this problem. –  Feb 07 '13 at 08:39
  • I think this $\ne$ for sets. So $A\ne B$ means that thiere ia an aelment $a\in A$ such that $a\notin B$ or there is an element $b\in B$ such that $b\notin A$. – Nikita Evseev Feb 07 '13 at 08:39
  • There is a a problem here, the assumption is that each bit of the question is assumed to either be a true of false statement for any general matrix $A$...however some parts are dependent on the matrix itself. – fretty Feb 07 '13 at 09:45

2 Answers2

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The question is not well formulated for exactly the reason you give (the quantifier over $A$ is not written). However the mere presence of alternative (e) suggests that the interpretation must be such that the statement "$\Bbb R^n=\mathcal C(A)+\mathcal N(A)$ or $\Bbb R^n\neq\mathcal C(A)+\mathcal N(A)$" can conceivably be false. That is the case if one reads the statement as "$(\forall A:\Bbb R^n=\mathcal C(A)+\mathcal N(A))$ or $(\forall A:\Bbb R^n\neq\mathcal C(A)+\mathcal N(A))$" (and indeed this is false), but not if one reads it as "$(\forall A:\Bbb R^n=\mathcal C(A)+\mathcal N(A))$ or $(\exists A:\Bbb R^n\neq\mathcal C(A)+\mathcal N(A))$" (which is therefore trivially true, by the law of the excluded middle). While implicit universal quantification of statements is quite common, I've personally never seen anybody advocate "implicit existential quantification of negative statements" (which would raise the question of precisely delimiting what a negative statement is; I suppose $x\not\leq y$ would be negative whereas $x>y$ is positive). To summarize, I think the second reading is a far stretch, and I would say (e) is the right answer. But that makes this a bad question for yet another reason, because giving answer (e) does not permit expressing that one has recognised the fact that $\forall A:\Bbb R^n=\mathcal R(A)+\mathcal N(A)$ is indeed true.

A correct formulation would have been "Let $P$ be the statement $\Bbb R^n=\mathcal R(A)+\mathcal N(A)$ and $Q$ the statement $\Bbb R^n=\mathcal C(A)+\mathcal N(A)$, then (a) $P$ and $Q$ are both true, (b) $P$ is true but $Q$ is false, (c) $P$ is false but $Q$ is true, (d) $P$ and $Q$ are both false." This is just to show that one can give a clear formulation even without using quantifiers. And not that adding "(e) none of the above" would be a ridiculous alternative.

Just a final remark: while false in general, $\Bbb R^n=\mathcal C(A)+\mathcal N(A)$ holds for almost all matrices. In order for it to fail, $A$ must have an eigenvalue $0$ with multiplicity at least $2$ in its minimal polynomial (and a fortiori in its characteristic polynomial).

Marc van Leeuwen
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It means "not always equal to". You will likely not find a very good source for this as it is simply a convention of language.

Edit: Here's the first thing that came up via google: http://www.math.hawaii.edu/~ramsey/Logic/ForAll.html

Jim
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  • The link you came up with says the exact opposite: a phrase with unquantified variables is always assumed to be quantified by "for all", not by "there exists" (even if the phrase is negative). So "not $P(x)$" is to be read as "for all $x$ not $P(x)$" (and not as "not for all $x$ $P(x)$" which means "there exists $x$ such that not $P(x)$"). – Marc van Leeuwen Feb 07 '13 at 09:35
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    It depends on the order you apply the symbols. I would say that $A(x) \neq B(x)$ means the opposite of $A(x) = B(x)$ and $A(x) = B(x)$ means $\forall x, \ A(x) = B(x)$. I do see that it's ambiguous though... – Jim Feb 07 '13 at 17:09