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I've encountered the following misunderstanding. I pose a question (to undergraduates in the U.S.), for example:

Let $P$ be a polygon of $n$ vertices. Is it true that every triangulation of $P$ has the same number of triangles?

This question depends on what constitutes a "triangulation," but assume the student know that. The answer is Yes: every triangulation of $P$ consists of $n-2$ triangles.

Here is the problem I encounter. The students apparently don't understand that "Let $P$ be a polygon" means, let your mind run over all possible polygons, so $P$ is an "arbitrary" polygon, in that it can be anything that fits the definition of a polygon (which the students also know). They wonder, well, maybe $P$ is a convex polygon, and should I answer specific to convex polygons?

This example doesn't quite illustrate the problem because the answer is always Yes. But when the answer is sometimes Yes, sometimes No, they seem to get confused over the quantifier. I think it may come down to the meaning of the phrase: "Let $A$ be a $B$." Let $p$ be a point in the plane $\mathbb{R}^2$—meaning any point in the plane, an "arbitrary" point in the plane. Let $P$ be a polygon, meaning any polygon. Let $T$ be a triangulation of a set of $n$ points. Does every triangulation of $n$ points have the same number of triangles? (No, not always.)

Have you encountered this confusion in your teaching? If so, how do you circumvent it?

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    $\begingroup$ Reminds me of this question I asked: matheducators.stackexchange.com/questions/13594/… Perhaps we need to be more explicit/verbose with "let"! Would students be less inclined to this confusion if we said, for instance, "Suppose P is any polygon, with no other assumptions about its properties"? $\endgroup$ Feb 24, 2020 at 7:02
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    $\begingroup$ Isn't that exactly what "Let P be a polygon" means outside our classroom? @BrendanW.Sullivan $\endgroup$ Feb 24, 2020 at 13:36
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    $\begingroup$ @ChrisCunningham Yes, exactly. But it seems like maybe the problem is students don't realize that the single word "let" for us means all of that other stuff. I'm recommending that we spell it out for them a lot. Emphasize that idea in class so that, on a written problem like in OP, part of the question is assessing their understanding that "Let P be a polygon" means to consider all possible polygons. $\endgroup$ Feb 24, 2020 at 16:17
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    $\begingroup$ My initial thought was that "convex" was intended, because omissions such as this are common (sometimes on purpose, sometimes just overlooked), and in my opinion to avoid having a gotcha question you'd want to say something like "polygon (convex or not)", unless prior context or reader exposure was that non-convex polygons arose often enough that your intended audience should be thinking about them (which I can't judge based only on what you've posted). So regarding "how do you circumvent it", I would say by anticipating common/expected assumptions unless you're specifically testing that. $\endgroup$ Feb 24, 2020 at 16:55
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    $\begingroup$ Have you tried saying "suppose $P$ is a polygon" or "assume $P$ is a polygon" instead? $\endgroup$ Aug 8, 2022 at 13:51

2 Answers 2

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Many logicians that I have spoken to have concurred with my assessment that this is an issue of the misleading use of "let". Many teachers use this word in two very different and incompatible ways. The first is universal quantification, as in your example. The second is existential instantiation, as in "Let $z = \exp(x+y)$. Then [blah blah] about $z$.".

The solution is simple. Do not use the bare word "let" for universal quantification, because it is not even technically the most precise way of expressing the desired meaning in English. Here are some much preferable alternatives:

  • Take/consider any polygon $P$. ...
  • Let $P$ be any given polygon. Then ...

The key word is "any", which is needed to precisely convey the universal quantification. If you want to express universal quantification in a single sentence, you can use the following:

  • Given any polygon $P$, ...
  • For every polygon $P$, ...

Note that the English word "any" behaves differently from "every". For example, "If for any $x∈S$ we have $f(x)∈T$, then ..." actually means "For every $x∈S$, if $f(x)∈T$ then ..." and not "If for every $x∈S$ we have $f(x)∈T$, then ...". So to minimize confusion it is better to stick with "for every" in single-sentence universal quantification.

I often hear the excuse that people have been using "let" in the confusing way for so long already, and that students have to learn to interpret it anyway. That is an excuse, because we have no reason not to use precise words in our teaching. As you yourself pointed out emphatically, using the word "any" would make things so much clearer. So we should use it! After students have acquired a proper grasp of logical reasoning (and not before that), facilitated by precise teaching, we can then tell students that some textbooks actually use "let" in the confusing manner, and they will have no trouble with it.

Related to this, students understand quantifiers far better when they are explained in terms of game semantics. Similarly for teaching structural induction. These are also perfectly in line with the use of the phrasing "given any ...".

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    $\begingroup$ Agreed. But I would personally lean towards arbitrary rather than any as it is a more mathematically precise word. $\endgroup$ Apr 13, 2020 at 19:19
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    $\begingroup$ @MatthewDaly: Yes, you can use "given an arbitrary polygon $P$", but it is not any more mathematically precise than using "given any". This is standard English. If we do not want to use English, there is always "∀". $\endgroup$
    – user21820
    Apr 14, 2020 at 6:38
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    $\begingroup$ @MatthewDaly: There is also another reason to stick to "any", because it is more concise. Consider: "Given any $a,b∈\mathbb{R}$ with $a≤b$, and any continuous $f : [a,b]→\mathbb{R}$ such that $f(a) ≤ 0 ≤ f(b)$, there is some $c∈[a,b]$ such that $f(c) = 0$. Similarly for theorems involving more universally quantified variables. =) $\endgroup$
    – user21820
    Apr 14, 2020 at 6:57
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    $\begingroup$ “Many teachers use [the] word [“let”] in two very different and incompatible ways.” I don’t see how the example you chose are incompatible. “Let $x$ be a real number” and “let $x$ be the real number $\pi$” are two sides of the same coin: both are universal quantification, just that the latter one has an additional condition to satisfy… $\endgroup$
    – MacRance
    Aug 24, 2022 at 4:18
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    $\begingroup$ @MacRance: Although the example I gave in the post may be viewed in the way you suggest, the ordinary mathematician does not actually view it your way, and it is not the case in general either. For example, when a mathematician says "Let $f : ℕ→ℝ$ be defined by the recursive relation ...", he/she must use ∃elim on the appropriate instantiation of the recursion theorem. There is logically no other possible way. $\endgroup$
    – user21820
    Aug 24, 2022 at 7:42
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Adding to what has been said:

  1. These sentences in a proof are equivalent:

    • Let $P_n$ be a polygon of $n$ vertices.

    • Let $P_n$ be an arbitrary polygon of $n$ vertices.
    • Let $P_n$ be an arbitrary polygon and suppose that $P_n$ has $n$ vertices.
    • Let $P_n$ be an arbitrary polygon such that $P_n$ has $n$ vertices.

    The adjective ‘arbitrary’ (alternatively: ‘any’) is frequently tacit—as in the OP's given sentence—as the intended meaning can be inferred from the context. So, as suggested by user21820, an immediate solution to the issue raised by the OP of clarifying Mathematical English is to not omit it from the sentence.

  2. Concluding the proof and asserting its result $\psi(P_n):$

    • For each polygon $P_n$ of $n$ vertices, $\psi(P_n).$
    • $\forall P_n{\in}\{x \mid x \text{ is a polygon of $n$ vertices}\}\:\:\psi(P_n).$

    This implicitly invokes Universal Introduction to generalise a representative (any $P_n$) to the universal (each $P_n$).

  3. The word ‘let’ per se does not connote universal (or existential) quantification; it merely signifies that a definition/restriction/value is to be assigned.

    However, having answered such questions on Mathematics.Stackexchange

    My issue comes with statements like "Let $x$ be an integer". I really do not know how I should intuitively interpret such a statement. Should I interpret it like how I interpret "Suppose $x$ is an integer"?

    What does "Let $G$ be a group" even mean?

    I agree with user21820 that the more specific phrasing

    • Consider an arbitrary polygon $P_n$ of $n$ vertices

    is probably clearer.

  4. Here's a comparison of the various translations of $$∀x{\in}F\; \psi(x).$$

    • For all elements $x$ in $F,\,\psi(x)$ holds” sometimes sounds like the property $\psi$ might belong to $F$ as a whole rather than to its individual members: “for all members of the family, they have a house” (1 house in total? or 5?). Contrast with “for each member of the family, they have a house” (definitely 5 houses in total). (“For all $x$” is Mathlish.)

    • For every element $x$ in $F,\,\psi(x)$ holds”, despite ‘every’ too having a collective sense, says that the property $\psi$ is common to the members of $F.$

    • For each element $x$ in $F,\,\psi(x)$ holds” directly attributes the property $\psi$ to individual members of $F.$

    • For any element $x$ in $F,\,\psi(x)$ holds” doesn't strongly communicate that the property $\psi$ belongs to each and every member of $F;$ nevertheless, “for any $x$” typically means “given an arbitrary $x$”, which, logically, is synonymous with “for every $x$” and “for each $x$”.

  5. In any case, when writing and teaching mathematics, the word ‘any’ should be used judiciously, because its meaning alternates between ‘every’ and ‘some’, or could be ambiguous.

    • If any intruder enters, the alarm will sound.”

      Here, ‘if any’ means ‘if some’ instead of ‘if every’.

    • “He does not have any pet.”

      Here, ‘not any’ means ‘not some’ instead of ‘not every’.

    • Wikipedia's definition of set disjointedness: “A collection of two or more sets is disjoint if any two distinct sets of the collection are disjoint.”

      Are three sets are disjoint if every pair of distinct sets is disjoint or if some pair of distinct sets is disjoint? It turns out from reading between the lines two sections down that Wikipedia intends the ‘if every’ reading, even as the ‘if some’ reading is probably more idiomatic. No wonder authors disagree between these two contrasting definitions!

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    $\begingroup$ I would like to add to this excellent answer one additional comment: students commonly use the word "random" as a synonym for "arbitrary" (e.g., "Suppose $P$ is some random polygon...") This should be anticipated, because in common language "random" is often used to mean "arbitrary, non-specific, generic". ("This random guy came up to me on the street today and said...") However, this usage should be discouraged whenever it occurs in a mathematical context, as "random" has a technical meaning in mathematical probability. $\endgroup$
    – mweiss
    Feb 14, 2023 at 22:44
  • $\begingroup$ @mweiss Thanks! Allow me to add: Two aspects of randomness and Arbitrary vs. Random. $\endgroup$
    – ryang
    Feb 15, 2023 at 15:07

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