If $R=\{(x,y): x\text{ is wife of } y\}$, then is $R$ transitive? 
If $R=\{(x,y): x\text{ is wife of } y\}$, determine whether the relation $R$ is transitive or not.

My Try: For Transitivity, If $(a,b) \in R$ and $(b,c)\in R\;,$ Then $(a,c)\in R.$.
Here If $x$ is a wife of $y$, then $y$ is not a wife of $z$. Therefore $x$ is not a wife of $z$.
Means $(x,y)\in R$ and $(y,z)\notin R$, then $(x,z)\notin R$
So $R$ is not a Transitive Relation.
But in Book, It is Transitive Relation.
Plz explain me with example. How this can be possible? Thanks
 A: The best explanation is probably that the possibility of same-sex marriages were not on the author's mind when he/she came up with the exercise.
If we suppose that marriages is always between a man and a woman (and genders are binary, bla bla bla), then the relation is indeed transitive, but vacuously so.
The condition it has to satisfy is that if $x$ is the wife of $y$ and $y$ is the wife of $z$, then $x$ is the wife of $z$. However this is true because the premise (i.e. the part after "if") is impossible to satisfy (under our assumed assumptions): $x$ can only be the wife of $y$ if $y$ is a man, and $y$ can only be the wife of $z$ if $y$ is a woman. Since nothing is both a man and a woman, both of these cannot be true at the same time.
In other words, the only way for a relation not to be transitive is if there is some $x$, $y$, and $z$ such that $xRy$ and $yRz$ and not $xRz$. If there is no such example, the relation is transitive -- no matter why there is no example.
Thus, the relation is transitive for the same reason that the empty relation is transitive: Because there are no triples $(x,y,z)$ such that $xRy$ and $yRz$, the condition for being transitive is not automatically met.
On the other hand, if $R$ includes information about at least one lesbian marriage, then it fails to be transitive: Suppose Alice and Betty are married. Then, in the case $(x,y,z)=(\mathit{Alice},\mathit{Betty},\mathit{Alice})$ we have that $\mathit{Alice}\mathrel{R}\mathit{Betty}$ and $\mathit{Betty}\mathrel{R}\mathit{Alice}$, yet not $\mathit{Alice}\mathrel{R}\mathit{Alice}$, since Alice is not her own wife.
A: It is transitive. In logic the statement $p \rightarrow q$ is true when $p$ is false irrespective of $q$.
Here $p$ can be seen as $(x\mathbf{R}y) \land (y\mathbf{R}z)$ which is not a valid boolean since if $(x\mathbf{R}y)$ is true, then $(y\mathbf{R}z)$ is false. In addition, $q$ can be viewd as $(x\mathbf{R}z)$.
In fact, for proving transitivity you need to prove that $p \rightarrow q$ is true which is the case here since $p$ is false.
A: The answer seems to depend on jurisdiction (and maybe subscription to certain formulations of gender identity).
We'll make some common assumptions for the sake of producing an answer: Let $X$ be a set (whose elements are people), each of which is labeled with exactly one of two genders, which we'll denote $M$ and $W$ (if you like, this is a function $X \to \{M, W\}$). Furthermore, suppose we have a irreflexive, symmetric relation on $X$, which we'll regard as a simple graph with vertex set $X$ and some set of edges. We say that two vertices $x, y \in X$ are married to one another iff there is an edge between them. Moreover, we'll say that $x$ is a wife of $y$ iff (1) $x$ and $y$ are married to one another, and (2) $x$ has gender $W$.
In this formulation, the answer depends on constraints on the marriage relation:


*

*If marriages between two people of gender $W$ are allowed, then the relation is not in general transitive: Given two people $x, y$ of gender $W$ married to one another, then by definition $x$ is a wife of $y$ and $y$ is a wife of $x$, but (as the graph of the marriage relation is simple) $x$ is not a wife of $x$. (If polyamory is also allowed, transitivity can fail in another way: Given $x, y, z$ of gender $W$ such that $x$ is a wife of $y$ and $y$ is a wife of $z$, $x$ need not be married to $z$ and so $x$ need not be a wife of $z$.)

*If marriages between two people of gender $W$ are not allowed, then the relation is vacuously transitive: If $x$ is a wife of $y$, then since $x$ has gender $W$, $y$ must have gender $M$, and hence $M$ is not a wife. So, under such a restriction there are never two pairs of related elements that satisfy the hypothesis of the transitivity criterion that there are elements $x, y, z \in X$ such that $(x, y), (y, z) \in R$.

A: Let $R_1$ be a relation on a set $A$ and if we are required to show that it is not transitive relation then we have to show that there exists $a,b,c\in A$ such that $(a,b),(b,c)\in R_1 \not \Rightarrow (a,c)\in R_1$  
$R=\{(x,y):x\text{ is wife of }y\}$
In case of $R$ if $(x,y)\in R$ then there is no $z$ such that $(y,z)\in R$. Since we are unable to show that $(x,y),(y,z)\in R \not \Rightarrow (x,z)\in R$. Therefore $R$ is transitive relation. 
A: It is transitive, considering that a set (y,z) (implying that y who is wife of x, is husband of z) does not exist.
To prove transitive condition you need to have the set (x,y) and also (y,z).Then you find out whether (y,z) is in the relation to figure out the transitive condition.
And if the condition (y,z) does not exist (like in the above case) you don't need test for transitiveness. It is trasitive by default.
