Why is the statement $(\forall x\in\mathbb R)(x^2\ge0)$ true even when $x$ is not a real number?

Question

It seems obvious that the statement $(\forall x\in\mathbb R)(x^2\ge0)$ is true. To prove it, however, we need to show that the implication $x\in\mathbb R\to x^2\ge0$ holds for all $x$. But suppose that $x$ is not a real number. Then, $x^2$ is not defined, and so presumably $x^2\ge0$ does not have a truth value (and nor does $x\in\mathbb R\to x^2\ge0$). Therefore, it is unclear to me how the implication $x\in\mathbb R\to x^2\ge0$ can be true for all $x$. What am I missing?

Answer 1

If $x$ is not a real number then $x \in \mathbb{R}$ is false and the implication $x \in \mathbb{R} \to A(x)$ therefore true, no matter what $A(x)$ is.

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Albeit the above is sufficient as an answer, I feel like I also need to point out what might be incorrectly seen as the problem about $x \in \mathbb{R} \to x^2 \geq 0$. In the end it's all about so-called junk theorems.

Intuition might tells us that $x^2$ is not defined for sets and therefore there seems to be a problem. But then we are not taking the set theory serious on which we base mathematics. If we take it serious, then we need to internalize that (1) everything is a set (2) the only relation between sets we have is $\in$ and everything can be boiled down to this.

This means: $\mathbb{R}$ is a set, two real numbers $x, y \in \mathbb{R}$ are sets, the multiplication function is a set $F_\times$ and $x \cdot y = z$ really means $(x,y,z) \in F_\times$, the relation $\leq$ is a set $R_\leq$ and $x \leq y$ really means $(x,y) \in R_\leq$. It is difficult to make all of these sets explicit for the particular case of these operations on the real numbers, but we can have some look at it in the case of the natural numbers.

There is an axiom in set theory guaranteeing the existence of the uniquely determined set of natural numbers $\mathbb{N}$ and one possible representation of numbers in terms of sets is $$ 0 = \emptyset \hspace{2em} 1 = \{ \emptyset\} \hspace{2em} 2 = \{ \{ \emptyset \}\} \hspace{2em} 3 = \{ \{ \{ \emptyset \} \} \} \hspace{2em} \dots $$ leading to our first junk theorem $2 \in 3$. We might want to define a function $S$, which always adds $1$ to a number, and here this would be achieved by $S(n) := \{ n \}$. But functions in set theory are sets as well. $S$ really corresponds to the set $$ R_S = \{ \, (n, \{ n \}) \mid n \in \mathbb{N} \} $$ and $S(n) = m$ is really just notation for $(n, m) \in R_S$. Addition similarly has a corresponding set, such that $$ x + y = z ~:\iff (x,y,z) \in F_+ $$ and we could then define $\leq$ as $$ x \leq y ~:\iff (x, y) \in \{ \, (a, b) \mid \exists k \in \mathbb{N}: \, (a, k, b) \in F_+ \, \} $$ This leads to a second junk theorem we could probably disprove: $\mathbb{R} \leq \mathbb{N}$. Note that the statement is fine, as it is just asking whether some set (namely $(\mathbb{R}, \mathbb{N})$) is an element of some other set.

Now if you think, after reading this, "I don't like this, I don't want everything to be a set" then you might like the idea of type theories. In a type theory we could actaully prohibit the multiplication making sense on anything but real numbers; in spirit to the intuition we have about the situation.

Answer 2

“ Why is the statement $$\forall x\in\mathbb R\;\;x^2\ge0\tag{*}$$ true even when $x$ is not a real number? ”

The statement $(*),$ written without abbreviation, is $$\forall x\;(x\in\mathbb R\implies x^2\ge0)\tag{*},$$ and is an assertion about some property of $x$ over its entire domain of discourse, which is possibly $\mathbb C.$

$x$ is not a free variable; the claim is regardless of whether $x$ is $7,$ $-28,$ or $5+9i$ (if indeed the domain of discourse contains the latter).

$(*)$ is true because it is claiming exactly this: $$\text{as $x$'s value varies, }\textbf{ whenever}\text{ it is real, its square is non-negative}$$ or, more succinctly, $$\text{as $x$'s value varies over the reals, its square is non-negative;}$$ no claim is being made regarding $x$ being non-real.

For non-real $x$ (if indeed the domain of discourse is a proper superset of the reals), the implication is vacuously true.

Answer 3

There are two ways to remedy this issue (namely the well-formedness of the statement):

Locally: Restrict the language (or domain of discourse) so that this statement is only well formed with respect to real numbers, that is quantification is over real numbers. In this case there is no need to include the hypothesis $(x\in \mathbb R)$ to a quantifier. This would result in the statement

"In the theory of real numbers, $\forall x. x^2 \geq 0$."

Globally: If we choose to keep your domain of discourse to be say, set theory, then we must generalize all real function and relation symbols so that they are total. For example, multiplication and ordering on real numbers produces a value for any set. Since we add the hypothesis that $x\in \mathbb R$ whenever quantifying over $x$, it does not matter how we define these operations on sets. This results in the statement

"In the theory of sets, $\forall x. (x\in \mathbb R)\implies (x^2\geq 0)$, and whenever $x\not\in \mathbb R$ or $y\not\in\mathbb R$, we define $x * y := 0$ and $x \geq y$ to be true."

There is also a third option if you view your statement in type theory, in which multiplication and ordering is defined only for real numbers, but quantification is now indexed by a type. This guarantees that the following is well formed

"$\forall_{\mathbb R} x. x^2 \geq 0$."