Analysis I (Terence Tao)—Free and bound variables

Question

In Terence Tao's Analysis I (page 321), he says that the statement "$x+3=5$ does not have a definite truth value if $x$ is a free real variable". On the other hand, the statement "let $x=2$" binds the variable $x$. Tao's remarks left me with a few questions:

If $x$ is not a free real variable, can the equation $x+3=5$ be understood to bind the variable $x$? If so, then is the equation $x+3=5$ a statement? What about the equation $x=x+1$, which has no real solutions? Does this bind the variable $x$?
The words "let $x=2$" are said to bind the variable $x$. Is it simply a convention that when we use the word "let", the variable $x$ becomes bound? Can the equation $x=2$ also be used in the case where $x$ is a free real variable, in which case $x=2$ does not have a definite truth value? And when we write "let $x=2$", is this a statement?
The identity $(x+1)^2=x^2+2x+1$ is true for all real values of $x$. In other words, the following proposition is true: $$ \forall x:(x+1)^2=x^2+2x+1 \, . $$ According to Tao, this means that "the statement $(x+1)^2=x^2+2x+1$ ... is true even when $x$ is a free variable". What I find paradoxical about this is that in order to know that $(x+1)^2=x^2+2x+1$ is true if even if $x$ is a free variable, we have to show that $\forall x:(x+1)^2=x^2+2x+1 $, where $x$ becomes bound. In other words, the statement $(x+1)^2=x^2+2x+1$ for any choice of free variable $x$ seems to bind the variable $x$. Is there something I'm missing here?

I'm not very knowledgeable about free and bound variables, but this question reminds me of how in computer programming "assignment" uses a different syntax (often x = c) than "checking for equality" (often x == c). — Joe, Jun 30 '21 at 12:39
If $x$ is in the scope of a quantifier, like e.g. $\forall x (x \ge 0)$, then it is bound. If not, like e.g. in $(x \ge 0)$ it is free. — Mauro ALLEGRANZA, Jun 30 '21 at 12:45
3: "The identity..." it means that we have an implicit leading universal quantifier: $\forall x [(x+1)^2=x^2+2x+1]$ — Mauro ALLEGRANZA, Jun 30 '21 at 12:46
See the post Definition for free and bound variables in logic — Mauro ALLEGRANZA, Jun 30 '21 at 12:49
@MauroALLEGRANZA, I (not the OP) probably confused the situation by commenting on a post by someone with the same username. Sorry. The OP is probably more knowledgeable about free and bound variables than I am. — Joe, Jun 30 '21 at 12:52
@MauroALLEGRANZA: OP here. Thanks for responding. I'm still not sure if the equation $x+3=5$ be understood to bind the variable $x$ though, just as the expression "let $x=2$" does. — Joe, Jun 30 '21 at 13:02
The formula $\forall x (x+3=5)$ is false while $\exists x (x+3=5)$ is true; thus, when we write the equation, we usually understand to state a "problem": to find the solution (if any) of the equation. — Mauro ALLEGRANZA, Jun 30 '21 at 13:05
@MauroALLEGRANZA: Is it possible if you explain the answer to my third question? — Joe, Jun 30 '21 at 13:57
Already answered above: the formula is true for every choice of value for $x$, because the universal quantification is true. See Universal Instantiation $\forall x \varphi \to \varphi[a/x]$ — Mauro ALLEGRANZA, Jun 30 '21 at 14:00
@MauroALLEGRANZA: Tao says that the statement $(x+1)^2 = x^2 + 2x + 1$ can be considered true even if $x$ is a free variable, and that's what I find confusing. — Joe, Jun 30 '21 at 14:02
Very few to add... The def of free and bound are related to a formal language with quantifiers and they are clear and simple: in $\forall x [(x+1)^2=x^2+2x+1]$ $x$ is bound while in $(x+1)^2=x^2+2x+1$ $x$ is free. That's all. In current math jargon, we sometimes omit quantifiers because the correct reading is dictated by the context. We have that $(x+1)^2=x^2+2x+1$ is called an identity because it is always true, i.e. it is true for every value assigned to variable $x$. 1/2 — Mauro ALLEGRANZA, Jun 30 '21 at 14:10
In this sense (this IMO is the meaning of the convoluted discussion by Tao) we have a formula with free variable - and thus without a definite truth value - that in fact "works like" a sentence (i.e. a formula without free variables) because we know that every possible assignment of values to the variable will output the same truth value, i.e. TRUE. 2/2 — Mauro ALLEGRANZA, Jun 30 '21 at 14:11
@MauroALLEGRANZA: Okay, that clears things up. I have one other question: Tao says that $x = 2 \rightarrow x^2=4$ is a true statement. However, I believe that if we are being formal, then we should write $\forall x : x=2 \rightarrow x^2=4$. Strictly speaking, the assertion that $x=2 \rightarrow x^2=4$ is neither true nor false because $x$ is free, and only statements with no free variables are true or false. Is that correct? — Joe, Jun 30 '21 at 15:32
See this question: https://math.stackexchange.com/questions/2995394/terminology-for-free-variables . There are 2 very common opposing mechanics to free variables, both used in many logics. — DanielV, Jun 30 '21 at 16:46
Elementary algebraic formulas such as f(x)=x+1 we learned in middle school is a well formed formula with free variable x, since x may be range over the domain of natural number, or real number or even complex number. So we cannot be sure what value f(x) got evaluated unless we introduce domain and quantifier, say, natural numbers, then x is bound and more importantly we can have truth value for a predicate such as ∀x(f(x)=2) which of course it's false, but ∃x(f(x)=2) is true. — cinch, Aug 17 '21 at 22:48

Troposphere · Answer 1 · 2021-06-30T14:11:31.203

It is not particularly useful to think of "free" versus "bound" as an absolute distinction where every variable is either one or the other.

Rather, "free" or "bound" is something a variable can be in relation to a certain amount of context. The classification is really a function of both the variable and how much context you're looking at, and if you switch to looking and a larger context, the variable might change from free to bound.

The real concept is not "$x$ is a free variable", but "$x$ is free in (some text or formula hat has the name $x$ in it)". Sometimes we omit stating what the context is, if we trust the reader can figure out which context we're talking about, but the technical concept is not complete without it.

For example, suppose we say:

Let $b=5$.

Does $x^2=b+3$ have a solution for $x$?

If we take the equation "$x^2=b+3$" as our context, both $x$ and $b$ are free in that context.

If the context is just the second line in the example, the $x$ in $x^2$ is now bound -- namely, the phrasing "solution for $x$" binds it in the sense of telling us what the point of that variable is. (It does not make sense to ask "does $x^2=b+3$ have a solution for $x$ if $x=2$?" because speaking about a "solution for $x$" assumes that $x$ we're free to vary the value of $x$ -- that is why "solution for $x$" binds the variable). But $b$ is still free when that line is our context.

When the entire example is our context, both variables in $x^2=b+3$ are bound.

Another way to say this is, the actual question is not whether a occurrence of a variable is bound, but where it is bound -- and in particular whether the binding happens within some particular context we're interested in.

Identities such as $(x+1)^2 = x^2 + 2x + 1$ are an interesting case.

Strictly speaking, the question "is $(x+1)^2$ the same number as $x^2+2x+1$?" does not get an answer before $x$ is a number such that we can compute the two sides and compare them. However, we do know that when we get an answer, that answer is sure to be "yes".

When we're doing actual calculations, the expectation is always that every variable will eventually be bound by something if we look far enough for context -- if nothing else, then the entire book we find the text in always comes with an implicit convention that every variable that is not bound explicitly can have an arbitrary value and what the book claims about those variables is supposed to be true no matter which actual value we give them. And that convention itself then counts as a binding for the variable. (In practice this convention almost always really applies to smaller pieces of text than an entire book: chapters, sections, individual proofs or paragraphs).

So given that we're expecting the $x$ in $(x+1)^2 = x^2 + 2x + 1$ is bound somewhere, that expectation is what allows us to rewrite the entire thing to "true" -- because it will be true when we view he rewriting in the broadest possible context.

Kevin Carlson · Answer 2 · 2021-06-30T14:01:20.457

No, the equation $x+2=5$ cannot be understood to bind $x.$ it has no clear meaning on its own. The sentence “Let $x$ be the unique real solution to $x+2=5$” is a fair binding of the variable $x$, though. You’ve bound a variable when it makes sense to then say something like “then in particular, $x\ne 4$.”

None of this is too closely connected to the main occurrence of an equation like $x+2=5$ in middle and high school math courses, where the goal is generally to find such an $x,$ that is, give it in some more explicit form. However, you can imagine in this context that’s what being said, though often only implicitly, is something like “Let $x$ be such that $x+3=5.$ What are the possible values of $x$?” The algebra student then proceeds with manipulations that depend on $x$ representing a particular number, which is what it means to be a bound variable.

It is implied by the meaning of “to bind a variable” that “Let $x=2$” binds $x.$ To bind a variable is to set it to a value. So I would not say it’s a matter of convention, except to the extent that the meaning of any string of letters is a matter of convention. As to whether this variable binding sentence is a statement, you should be able to decide for yourself by remembering that a statement is a sentence that is either true or false.
All that’s happening here is that we often state identities with the universal quantifier left implicit. You can clarify any confusion on this matter by putting the implicit quantifier back in. Don’t worry about what it means for a statement containing a free variable to be “true.”

Thanks very much for clarifying this, Kevin. Is it possible if you address questions 2 and 3, as well, please? — Joe, Jun 30 '21 at 13:33
Okay, that makes sense, thanks. I don't believe that "let $x=2$" is a statement because it simply defines what the variable $x$ means, rather than makes an assertion. Sorry to bother you, but I think there are still an "edge cases" that I need to clarify: suppose we wrote "let $x$ satisfy $x^2=9$". Does this constitute a binding of the variable $x$, given that the above equation has two real solutions? — Joe, Jun 30 '21 at 14:51

ryang · Answer 3 · 2021-08-28T05:48:18.513

OP: I think there are still an "edge case" that I need to clarify: suppose we wrote "let $x$ satisfy $x^2=9$". Does this constitute a binding of the variable $x,$ given that the above equation has two real solutions?

“Let $x$ satisfy $x^2=9$”, i.e., “let $x\in\{-3,3\}$” declares $x$ be an arbitrary element of $\{-3,3\}$, so yes $x$ is certainly being bound. In the context of a proof, this formally translates as $$\forall x\,\big(x\in\{-3,3\}\implies\ldots\big),$$ or, slightly less formally, $$\forall\, x\in\{-3,3\}\ldots.$$
These two answers that I recently wrote—and the english.stackexchange link contained within—pertain to your various queries and complement the other responses on this page: open vs. closed formulae and how to interpret “let”.

Analysis I (Terence Tao)—Free and bound variables

3 Answers3