1
$\begingroup$

I want to prove this statement: "there exists a real number $x<0$ whose square is 2". This could mean $$ \exists x\in \mathbb{R}, x < 0 \text{ and } x^2=2$$ or $$ \exists x\in \mathbb{R}^-, x^2=2$$ I'm having trouble dissecting where the "domain of discourse" ends and where the "predicate" begins. Is the domain of discourse $\mathbb{R}$, or $\mathbb{R}^-$? Or can it even be taken to be "everything", so the statement becomes $\exists x, P(x)$, where $P(x)$ says "$x$ is a negative real whose square is 2"? (Doesn't this have something to do with Russel's paradox?)

Now, suppose I want to attempt a proof by contradiction and assume the negation, should I use $$\forall x \in \mathbb{R}, x \geq 0 \text{ or } x^2\neq 2 $$ or $$\forall x \in \mathbb{R}^-, x^2\neq 2 $$

Any help on the proof is also appreciated (I'm aware of the direct proof of letting $c = \inf \{z \in \mathbb{R}| z^2 =2\}$ and showing $c^2=2$, but I want to see if it's possible to do it without constructing $\sqrt 2$).

$\endgroup$
1
  • 1
    $\begingroup$ In fact, in the usual set-theoretic foundations approach, the domain of discourse is all sets and the real numbers are defined as a particular sets of sets, and order and arithmetic can be defined as relations and functions thereon. So $\exists x\in \mathbb R(x<-\land x^2=2)$ is just an abbreviation for $\exists x (x\in \mathbb R\land x<0\land x^2=2)$ (where the abbreviation is often called a "bounded quantifier"). And, as always, the unqualified existential quantifier in the latter ranges over the domain of discourse (sets). $\endgroup$ – spaceisdarkgreen Jul 4 '18 at 19:41
3
$\begingroup$

In first order logic, the phrase $\exists x\in \mathbb R$ is not actually an official phrase of the language. All actual statements involve just bare quantifiers $\exists x,$ and the domain of discourse is a property of the interpretation. For instance in a first order language with relation symbol $<$ and constant symbol $0$ we could write $\exists x(x<0)$ and then we could interpret it as a statement about real numbers, or about natural numbers, or whatever else where we interpret $<$ and $0$ as the usual relation and number in whichever domain.

This does not precluded us from informally writing $\exists x \in \mathbb R(x<0)$ to emphasize that we are interpreting the statement in the reals. More importantly, 'bounded quantifiers' like this are often used in an official capacity as abbreviations. For instance, say we're working in the reals and by some means we can define a predicate identifying the naturals within the reals. Then we can write such a predicate $x\in \mathbb N$. Then $$\exists x\in \mathbb N(x<0)$$ can be regarded as an abbreviation for the actual formal sentence $$\exists x(x\in \mathbb N \land x <0)$$ where again, we're interpreting in the reals here.

So in other words, you're right. It's a fuzzy line. For your purposes here, I wouldn't worry too much about the difference between $\exists x\in \mathbb R(x<0\land x^2=2)$ and $\exists x\in \mathbb R^-(x^2=2)$ or what the ultimate domain of discourse is. (As I mentioned in a comment, it could be set theory, so that $\mathbb R$ is a defined set and thus $x\in \mathbb R$ is a defined property and we're using bounded quantifiers as an abbreviation). It's much more important that you understand what the negation says logically than niceties about notation. It's clear that, whatever the formality, $\forall x\in \mathbb R( x\ge 0 \lor x^2\ne 2)$ and $\forall x \in \mathbb R^- (x^2\ne 2)$ mean the same thing and are logically the negation of the first statement. So this is what you need to prove.

(By the way, for a bounded universal quantifier, $\forall x \in A (P(x))$ abbreviates $\forall x (x\in A \to P(x))$ or equivalently $\forall x (x\notin A \lor P(x)),$ so you can see accords with what you wrote down.)

$\endgroup$
5
  • $\begingroup$ Thanks! In the last paragraph, did you mean "$\forall x\in A(P(x))$ abbreviates $\forall x (x\in A \to P(x))$" for bounded universal quantifier? Also, isn't $\forall$ in some sense optional in the implication? $\endgroup$ – ladiesman Jul 4 '18 at 20:39
  • $\begingroup$ @ladiesman yes, I meant that, sorry. I'm not sure what you mean by 'isn't $\forall $ in some sense optional in the implication?' $\endgroup$ – spaceisdarkgreen Jul 4 '18 at 21:15
  • $\begingroup$ Sorry I meant the unbounded universal quantifier $\forall x$, and that $x\in A \to P(x)$ ("if $x\in A$, then $P(x)$") intuitively says the same thing as $\forall x ( x\in A \to P(x))$ (aren't they logically equivalent?), but I don't know if omitting the $\forall x$ is grammatically legal. $\endgroup$ – ladiesman Jul 4 '18 at 22:38
  • 1
    $\begingroup$ @ladiesman Yeah, that makes sense, but it really applies to any formula. It's certainly natural to think of a formula $Q(x)$ with a free variable as meaning the same thing as the sentence $\forall x Q(x)$ (a sentence is a formula with no free variables). We call $\forall x Q(x)$ the universal closure of the formula $Q(x).$ Some treatments do give a formula the same meaning as its universal closure. Other approaches say only sentences have meaning, or to have an interpretation also assign a domain value to each free variable so $Q(x)$ means different things in different interpretations. $\endgroup$ – spaceisdarkgreen Jul 4 '18 at 23:06
  • $\begingroup$ @ladiesman (note also when I say 'give it a meaning', I mainly mean 'assign it a truth value', (though strictly speaking, it can be broader than that). So your question 'aren't they logically equivalent?' makes sense rigorously here, and the answer's yes if we interpret open formulae as their universal closures.) $\endgroup$ – spaceisdarkgreen Jul 4 '18 at 23:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.