For each of the following relations, determine whether it is reflexive, symmetric or transitive a) $X$ is the set of real numbers, $n$ is a natural number
 $$R =  \{(x, y) \mid x, y \in X, x^n = y^n\}.$$
b) $X$ is the set of people in the world
$$R = \{ (x, y) \mid x, y \in X, x\text{ and }y\text{ share a parent}\}$$
I believe both relations are reflexive, symmetric and transitive, aren't they? 
I mean, for a) you can have $x^n = x^n \rightarrow (x,x)$, you can have $(x,y)$ and $(y,x) \rightarrow$ symmetric and you can have transitivity by $\leq.$
And for b) it's pretty much the same. As it doesn't say $x$ mustn't be equal to $y$, then you can have $(x,x)$ which is true. You can also have $(y,x) \rightarrow$ symmetry and you can also get transitivity for "if $(x,y)$ is true and $(y,z)$ is true, then $(x,z)$ is true," right?
 A: Relation (b) isn't transitive: consider half-siblings.
In detail: let $M(x)$ and $F(x)$ be the mother and father of person $x$. If $M(a)=M(b)$ and $F(b)=F(c)$ but $M(a)\ne M(c)$ and $F(a)\ne F(c)$ then we have $(a,b),(b,c)\in R$ but $(a,c)\not\in R$.
A: a) $X$ is the set of real numbers, $n$ is a natural number

$R$ =  {($x$, $y$) | $x, y \in X, x^n = y^n$ }

You need to be careful, as was pointed out, with your phrasing of "can have" which implies "there exists", and your invocation of the $\leq$ relation to address problem (a).  


*

*First, reflexivity, symmetry, and transitivity of a relation requires that the properties are true for all elements of the set in question. 

*And secondly, to establish the transitivity of the relation $R$ in (a), you must demonstrate that the given defined relation is transitive (as was pointed out above, the $\leq$ relation is not the relation for which you need to establish transitivity.)  


So, if you are going to argue that the relation given in (a) is transitive, you need to show that for all $x, y, z \in N,$ if $x^2 = y^2$ and $y^2 = z^2$, then it must follow that $x^2 = z^2$.  To prove transitivity, you assume that the "if" conditions hold: Let $x, y, z$ be arbitrary elements of the set of natural numbers. Assume $x^2 = y^2$ and $y^2 = z^2$.  Let $X = x^2, Y = y^2, Z = z^2$.  Then our assumption can be expressed as $X = Y$ and $Y = Z$. Since equality is an equivalence relation, it is transitive, and hence $X = Z$, or $x^2 = z^2.$  Since $x, y, z$ were chosen arbitrarily, it follows that for all $x, y, z \in N,$ we have that $R$ is indeed transitive.
