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Definition :

A relation $\rm R$ in a set $\rm A$ called transitive, if $(a_1, a_2) \in \mathrm R$ and $(a_2, a_3) \in \mathrm R \implies (a_1, a_3) \in \mathrm R \quad \forall a_1, a_2, a_3 \in \mathrm A$

Problem : (source)

Let $\mathrm A$ be finite set of human beings.

Let $\mathrm R$ be a relation on the set $\mathrm A$ defined as $$\mathrm R = \{ (x,y) : \text{$x$ is wife of $y$}\}$$

Determine whether it is transitive or not.

I would say it is not transitive because if $x$ is wife of $y$ then $y$ can't be wife of $z$ and certainly $x$ can't be wife of $z$ assuming no same sex marriage or extramarital affairs by the people of set $\mathrm A$.

Here if we define $p : (x,y) \in \mathrm R \ \land \ (y,z) \in \mathrm R $ and $q : (x, z) \in \mathrm R$,

Then clearly both $p,q$ is false here and so $p \implies q$ should be false.

By the definition of transitivity ,$\text{if $(p \implies q)$ then transitive}$, the relation $\mathrm R$ is not transitive because $p \implies q$ is false.

Here is the part I don't understand, in the source of this problem the answers suggest that the relation is transitive and it is so because $p \implies q$ is false, provided I understand them properly.

I don't understand why if both $p,q$ is false then the relation is transitive and how does this follows from the definition of transitivity ?

J. M. ain't a mathematician
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user8277998
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    This is why I hate most "real world exercises" about basic set theory. There is so much left ambiguous, and mathematics is exactly about clarify. So can only women be wives? Can women marry other women? Can men be the wives of other men, or better yet, of other women? Do you assume monogamy? Do you at least assume that every woman is married at most to one person, if monogamy is not assumed? – Asaf Karagila Aug 14 '17 at 09:07
  • @AsafKaragila For the sake of an answer assume monogamy and no homosexuality. – user8277998 Aug 14 '17 at 09:16
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    The relation is transitive because it's not non-transitive (there are no counterexample $x, y, z$ to the statement "$R$ is transitive"), and we assume the law of excluded middle. – Arthur Aug 14 '17 at 12:56

2 Answers2

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I write $xRy$ for $(x,y)\in R$.

A relation $R$ is transitive if and only if, for all $x$, $y$, and $z$, if $xRy$ and $yRz$, then $xRz$.

Note that a statement of the form `if $p$, then $q$' is true if $p$ is false.

The point of the exercises is to notice that there are no $x$, $y$, and $z$ such that $xRy$ and $yRz$. So, it is (trivially) true that, for all $x$, $y$, and $z$, if $xRy$ and $yRz$, then $xRz$. So, the relation $R$ with which you are dealing in this exercise is transitive.

MatteoBianchetti
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  • Why "if $p$ then $q$" is true if $p$ is false ? – user8277998 Aug 14 '17 at 09:56
  • There is a long literature on this, since it is actually a complicated issue. In short, I say the following. The negation of "if $p$, then $q$" is: "$p$ and non-$q$". Therefore, the negation of the negation of "if $p$, then $q$" is: "either non-$p$ or $q$". But the negation of the negation of "if $p$, then $q$" is true exactly when "if $p$, then $q$" is true. So, "if $p$, then $q$" is true when non-$p$ is true, i.e. when $p$ is false. – MatteoBianchetti Aug 14 '17 at 10:03
  • So, if we let $r : \text{if $p$ then $q$}$ then $r$ is true when $p$ and $q$ is true, if $p$ is false and if $q$ is true, right ? – user8277998 Aug 14 '17 at 10:16
  • "If $p$, then $q$" is false exactly in one case: when $p$ is true and $q$ is false. It is true in all the other cases, i.e. when $p$ is false or $q$ is true – MatteoBianchetti Aug 14 '17 at 10:20
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A relation is transitive if whenever $(x,y)\in R$ and $(y,z)\in R$, then also $(x,z)\in R$.

In the case of "$x$ is the wife of $y$", it is necessarily the case that $x$ is a female, $y$ is a male, and that $y$ is not married to anyone else than $x$ (assuming the conditions in the comments).

In particular, only women can be wives, so if $(x,y)\in R$, it automatically means that there is no $z$ such that $(y,z)\in R$. To paraphrase Anouk, $y$ is nobody's wife.

So it is never the case that you have both $(x,y)\in R$ and $(y,z)\in R$. So now we fall back to the truth definition of an implication, $p\implies q$. And the truth definition says that if $p$ is false, then the implication as a whole is true.

In our case, $p$ is the conjunction $(x,y)\in R$ and $(y,z)\in R$. And as remarked above, it is necessarily false. So the implication is true, and therefore the relation is transitive.

Asaf Karagila
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    You highlight the wrong IF. You highlight the IF joining the definiendum (R is transitive) and the definiens (for all x, y, z, if xRy and yRz, then xRz). The IF-THEN statement whose truth value is relevant to the solution of the exercise is the second one (which you start with WHENEVER). – MatteoBianchetti Aug 14 '17 at 09:55
  • In the definiens, you should also add: for all x, y, z. – MatteoBianchetti Aug 14 '17 at 09:55
  • So now we fall back to the truth definition of an implication, $p\implies q$. And the truth definition says that if $p$ is false, then the implication as a whole is true. Why ? – user8277998 Aug 14 '17 at 09:59
  • @123: I explain that in the sentence. – Asaf Karagila Aug 14 '17 at 10:04
  • @AsafKaragila I mean to ask the reason behind the truth table. – user8277998 Aug 14 '17 at 10:18
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    @123: https://math.stackexchange.com/questions/70736/in-classical-logic-why-is-p-rightarrow-q-true-if-p-is-false-and-q-is-tr and many other threads on the site covered this fairly extensively. – Asaf Karagila Aug 14 '17 at 10:23
  • Admittedly, I am confused about this downvote. Especially since the other answer is not significantly different from a mathematical point if view... – Asaf Karagila Aug 14 '17 at 12:15
  • "To paraphrase Anouk, $y$ is nobody's wife." - Actually, $y$ is $x$'s wife... and I'm not sure "nobody is $y$'s wife" is a song. The whole thing is a stone hanging 'round 123's neck though :) – CompuChip Aug 14 '17 at 18:49
  • @CompuChip: As far as I understand it, you're absolutely wrong. $y$ is $x$'s husband. And again we return to me hating these kind of "real world examples" which are full of the inherent ambiguity of natural language. – Asaf Karagila Aug 16 '17 at 14:27