How to show properties of a given relation? I am given that R is a relation on the given set X, and I have to show if the relation is 


*

*(i) reflexive, 

*(ii) symmetric,

*(iii) transitive,

*(iv) asymmetric, and

*(v) give an example of an element of the relation.


X= the positive $\mathbf{Z}$ (the positive integers)and R is the relation defined by nRm if and only if there is a nonzero $k \in \mathbf{Q}$ for which $n^k=m$. 
I don't even know where to start. Keep in mind that I am in an introductory to higher mathematics course so I may not be entirely familiar with really advanced concepts. 
 A: I will try to explain a approach for part ii) maybe that gives you insight on how to tackle the other parts.
ii) Symmetric: if we have $n\sim m$ then there is a $q\in \mathbb{Q}$ such that $n^q=m$. Now for we want to know if it is symmetric, so then we want to know if $m\sim n$. In other words is there a $p\in \mathbb{Q}$ so that $m^p=n$? (Hint: look at $n^q=m$, then raise both sides to a certain power.)
A: I'll start with the definitions. Note that '$\wedge$' is the logical AND operator.


*

*R is reflexive on a set $\mathbb{X}$ if and only if:
$$\forall a \in \mathbb{X}, aRa$$

*R is symmetric if and only if:
$$aRb \Leftrightarrow bRa$$

*R is transitive if and only if:
$$aRb \wedge bRc \Rightarrow aRc$$

*R is asymmetric if and only if:
$$aRb \wedge bRa \Rightarrow a=b$$



The example relation you defined was:
$$R = \left\{(a,b) :a^{k}=b \textit{ for some } k \in \mathbb{Q},k\ne 0\right\}$$
R is reflexive because $\forall a \in \mathbb{Z_{+}}, a^{1} = a$. Setting k=1 will yield the equality $a=a$ which is true for all elements of $\mathbb{Z_{+}}$. Thus every element is related to itself.
R is symmetric because: $$aRb \Rightarrow a^{k}=b \Rightarrow b^{1/k}=a$$ 
and
$$(k \in \mathbb{Q}) \wedge (k \ne 0) \Rightarrow \frac{1}{k} \in \mathbb{Q}$$
Thus $bRa$ satisfies the definition of the relation and so ultimately, $aRb \Rightarrow bRa$
R is transitive because:
$$aRb \Rightarrow a^{k_{1}}=b$$
$$bRc \Rightarrow b^{k_{2}}=c$$
Substituting $b$ in the second expression for $a^{k_{1}}$ in the first expression yields:
$$(a^{k_{1}})^{k_{2}}=a^{k_{1} k_{2}}=c$$
and so there exists $k =k_{1} k_{2}$ such that $a^{k} = c$.
Note once again that $k$ is a nonzero rational number and so $aRc$ satisfies the definition of the relation.
R is not antisymmetric. A simple counterexample would be to notice that:
$$2R4 \wedge 4R2$$
because
$$2 = 4^{\frac{1}{2}} \textit{ and } 4 = 2^{2}$$
This contradicts the definition of antisymmetry which says that for two elements to be related to eachother, they must be equal. Thus R is not antisymmetric.
