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In "A Transition to Advanced Mathematics", 8th edition, by Smith, Eggen and St. Andre, exercise 1(f) in chapter 1.6 on page 60 says:

Prove that
 if there exist integers $m$ and $n$ such that $12m + 15n = 1,$
 then $m$ and $n$ are both positive.

This statement has the form $X \implies Y.$ Since the scope of the variables $m$ and $n$ is $X,$ the variables are out of scope when used in $Y$ later. So, the given statement is actually incorrectly worded, isnt it? I think it should have been

  • for all integers $m$ and $n$,
     if $12m + 15n = 1,$ then $m$ and $n$ are both positive.
ryang
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Ben
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    It's not an error, it's just a style of writing. I agree that your way of writing the statement looks better. – azif00 Jul 02 '25 at 00:01
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    maybe it should be called “a transition to super advanced mathematics” because once people get the hang of things they get more relaxed :) – peek-a-boo Jul 02 '25 at 00:03
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    @azif00 How is it not an error if they are using variables that are out of scope? – Ben Jul 02 '25 at 00:05
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    $\exists m,n \in \mathbb{Z}, 12m+15n=1$ implies everything. – Henry Jul 02 '25 at 00:09
  • @Henry My question has nothing to do with the truth value of the statement. Anyway, it doesn't even make sense to assign a truth value to a statement that isn't written correctly. Or would you also say that a nonsense statement like $(\exists m,n \in \mathbb{Z}, 12m+15n=1) \implies (3+\forall \neg = \iff)$ is true? I certainly would not. – Ben Jul 02 '25 at 00:14
  • I don't think it's an error, just confusing to translate into predicate logic. I would translate it as $\forall m,n, (12m + 15n = 1 \Rightarrow m,n > 0)$ or something along those lines. I wouldn't translate "there exists" literally to $\exists$ since the statement isn't conveying an existential statement: I'm not reading it as the author stating the implication holds for at least one $m$, $n$. It's subtle, and I probably wouldn't write that way for students new to logic. – Trevor Norton Jul 02 '25 at 00:33
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    @QiaochuYuan The statement is vacuously true. $12m + 15n \neq 1$ for any integers $m,n$ (the LHS is always a multiple of $3$ and so can not equal $1$). Thus the implication always evaluates to true. – Trevor Norton Jul 02 '25 at 00:58
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    I suppose most people don't consider it an error because this sort of imprecision is common when writing math using words (rather than symbols), and it is obvious what statement the authors are trying to make. – Kenanski Bowspleefi Jul 02 '25 at 01:06
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    You're meant to read the second part as "then that $m$ and $n$ are both positive". It's assumed that by this point in your mathematical education, you know to make such insertions wherever necessary. – Gerry Myerson Jul 02 '25 at 03:24
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    @GerryMyerson I know that that is a way to make it correct. My question is whether it is wrong as written. – Ben Jul 02 '25 at 05:33
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    And my answer is as I wrote: It's assumed that by this point in your mathematical education, you know to make such insertions wherever necessary. It's not wrong – it just assumes the reader is willing and able to do some of the work. – Gerry Myerson Jul 02 '25 at 06:43
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    @GerryMyerson I am going to have to disagree with your answer, then. Yes, any mathematician will understand what the writer meant. That does not make it correct; it is a wrong sentence and is in fact poor writing. It's not a major error, but it's not not wrong: the author wrote one thing, and meant another. if I found something like it in a draft from a coauthor, I would definitely point it out, and I'd expect them to do the same if I made this sort of mistake in a paragraph I wrote. The same applies here. – Z. A. K. Jul 02 '25 at 06:51
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    (1) It's wrong. (2) Every mathematician would instinctively understand what's meant. (3) Students who study "A Transition to Advanced Mathematics" do not yet have that instinct and will either be confused by such an incorrect formulation or decide that it is indeed wrong. (4) Therefore, if I notice such a statement in a research paper, I think "Well, they're being sloppy," but if I see it in a book purporting to bring students into advanced (therefore precise) mathematics, I think "Yikes." – Andreas Blass Jul 02 '25 at 14:32
  • Isn't the statement true? $p \implies q \iff q \lor \lnot p$. $p$ is never true, so $\lnot p$is always true then so is the disjunction. – TurlocTheRed Jul 02 '25 at 15:14
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    @TurlocTheRed This question has nothing to do with the truth value of the statement. See my comment reply to Henry above. – Ben Jul 02 '25 at 16:51
  • This can actually be justified if you use dependent type theory. If you interpret the antecedent as $T = \sum_{m,n : \mathbb Z} 12m + 15n = 1$, then you can recover $m$ and $n$ even after they go out of scope, by projection functions. (Implication is the same thing as a function set in dependent type theory.) – Trebor Jul 04 '25 at 06:15

1 Answers1

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  1. if there exist integers $\color\red m$ and $\color{cyan}n$ such that $12\color\red m + 15\color{cyan}n = 1,$
    then $\color\red m$ and $\color{cyan}n$ are both positive.

the given statement is incorrectly worded, isnt it? I think it should have been

  1. for all integers $m$ and $n$,
      if $12m + 15n = 1,$ then $m$ and $n$ are both positive.

As hinted by Gerry Myerson in the comments above, mathematical writing is not formal logic. We'd read the two sentences in "Mary.... She...." as referring to the same entity. Similarly, with the (justifiable) understanding that variables are not being recycled (as if the given statement is just "if integers $m$ and $n$ are such that $12m + 15n = 1,$ then $m$ and $n$ are both positive", with no visible quantification), we can demonstrate that Statement 1 is logically worded, and in fact equivalent to Statement 2. We begin by making explicit Statement 1's implicit universal quantification.

\begin{align}&\text{Statement 1}\\\equiv{}&\forall x{,}y {\in}\mathbb Z \:\bigg(\exists m{,}n{\in}\mathbb Z \,\Big(12m+15n=1 \;\text{ and }\;(x,y)=(m,n)\Big)\implies x,y>0\bigg)\\ \equiv{}&\forall m{,}n {\in}\mathbb Z \:\bigg(\exists x{,}y{\in}\mathbb Z \,\Big(12x+15y=1 \;\text{ and }\;(m,n)=(x,y)\Big)\implies m,n>0\bigg)\\ \equiv{}&\forall m{,}n {\in}\mathbb Z \:\Big(12m+15n=1\implies m,n>0\Big)\\ \equiv{}&\text{Statement 2}.\end{align}

While at first glance Statement 1 and its common variant

  1. if any integers $\color\red m$ and $\color{cyan}n$ are such that $12\color\red m + 15\color{cyan}n = 1,$
    then $\color\red m$ and $\color{cyan}n$ are both positive.

—which as a matter of fact is also equivalent to it and thus to Statement 2—appear to be incorrectly worded (and, to be clear, I do agree that wordings 1 and 3 are, echoing Andreas Blass above, "Yikes"), consider the fact that the following assertions, corresponding to the above, are idiomatically understood as mutually equivalent:

 (2) Every solution is positive.
 (1) If there exists a $\color\red{\text{solution}},$ $\color\red{\text{it}}$ must be positive.
 (3) If there is any $\color\red{\text{solution}},$ $\color\red{\text{it}}$ must be positive.

Reply to the OP

This isn't an answer to the question. You're just showing that two rewritten versions of the statement are equivalent, not answering the question of whether the original statement as written is syntactically correct.ㅤ- Ben

You're asking whether Statement 1, the verbal statement given by your textbook, is "syntactically correct" (in the formal-logic sense). This depends on its formalisation (i.e., translation from natural langauge to formal logic), which is a function of the idioms and conventions of the source language—here mathematical English or Mathlish—and which is not always a rote, mechanical procedure (consider the non-obvious translation of "P(x) holds for at least two objects" as $$∀a\:∃x\:\big(x\ne a\,∧\,P(x)\big)\quad).$$

To answer your question requires formalising Statement 1, which I did by adopting the convention that the second and third occurrences of $\color\red m$ are meant to be pointing at the same object (likewise, $\color{cyan}n).$ With this understanding, Statement 1 is indeed syntactically correct and, furthermore, is semantically equivalent to your Statement 2.

If we were to treat Statement 1 not as part of mathematical exposition, but as a word-for-word transcription of a logic formula, then yes—rather than being equivalent to Statement 2, it indeed contains a free occurrence of each of the variables $m$ and $n,$ and so is an open formula and technically not even a statement.

ryang
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    This isn't an answer to the question. You're just showing that two rewritten versions of the statement are equivalent, not answering the question of whether the original statement as written is syntactically correct. – Ben Jul 02 '25 at 16:59
  • @Ben Please see the addendum. I understand that mine is not the Answer you're seeking; I too initially would have reacted that Statement 1 is atrociously written. – ryang Jul 02 '25 at 17:46
  • I still think you're not answering the question because you've still changed the statement before addressing the syntactic correctness. I am not asking whether your statement 1 is syntactically correct, I am asking whether "if there exist integers $m$ and $n$ such that $12m+15n=1$, then $m$ and $n$ are both positive" is correct, i.e. $(\exists m,n\in\mathbb{Z}, 12m+15n=1) \implies (m>0 \wedge n>0)$. – Ben Jul 02 '25 at 17:52
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    I didn't miss it, I just think your claim that $\forall x,y \in \mathbb{Z}, \dots$ is a direct translation of the original statement into formal logic is objectively not correct. You didn't translate exactly the content of the original statement, you translated it and fixed the mistake, and only then pointed out that there's no mistake. I disagree that the original statement has an implicit universal quantifier - it has two variables, both under an existential quantifier in the wrong scope. – Ben Jul 02 '25 at 18:10
  • @Ben Again: what I did is premised on the convention that I highlighted, which we don't all have to agree with (in sympathy with this point, I've just added a last paragraph to the addendum). A convention can be good or dumb, but unless it leads to a contradiction, it probably cannot be "incorrect". – – ryang Jul 02 '25 at 18:42
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    To quote from someone else's comment, "Every mathematician would instinctively understand what's meant" and "Yikes!" You've done a good job of formalizing what was almost certainly intended, but that doesn't automatically mean you have answered the question. By the way, the ability to make a compact logical statement equivalent to a verbal statement (as in $∀a:∀b:∃s:\big(s\ne a,∧,s\ne b,∧,P(s)\big)$) is not the same thing as translating the verbal statement to logical syntax. – David K Jul 02 '25 at 23:08
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    @DavidK I've reorganised the Answer to replace that hyperbolic, hastily-included example (the point being that formalising natural language, as is required to evaluate the OP's claim, contains an interpretive aspect), and to affirm Andreas's "Yikes", haha. To be completely explicit: the answer to the OP's query is, "Yes and No, depending on that verbal sentence's agreed reading." – ryang Jul 03 '25 at 04:38