Definition of transitivity. 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 ? 
 A: 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.  
A: 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.
