Find the different equivalence classes of this relation on A and show their connection to a partition of A. 
Define a relation $R$ on the set $A = \{n \mid n \in \mathbb{N} \textrm{ and } 0 &lt n &lt 14\}$ such that $R$ is an equivalence relation on $A$. (You can either define a property on $A$ or simply list the elements of $R$.) Then find the different equivalence classes of $R$ and show their
  connection to a partition of $A$.

I understand what $A$ is but I don't understand the rest of the question, can anyone point me in the right direction?
 A: Recall that $R$ is a relation on $A$ if $R\subseteq A\times A$, that is the elements of $R$ are ordered pairs that both the elements come from $A$. If $\langle x,y\rangle\in R$ we often write $x R y$.
Let us define some properties of a relation $R$ on $A$:


*

*We say that $R$ is reflexive if for every $x\in A$ the pair $\langle x,x\rangle\in R$. That is $xRx$ is true for every $x\in A$.

*We say that $R$ is symmetric if for every $x,y\in A$ if $\langle x,y\rangle\in R$ then $\langle y,x\rangle\in R$. That is if $xRy$ then $yRx$.

*We say that $R$ is transitive if for every $x,y,z\in A$ if $\langle x,y\rangle,\langle y,z\rangle\in R$ then $\langle x,z\rangle\in R$. That is if $xRy$ and $yRz$ then $xRz$.


If $R$ has all three properties then $R$ is called an equivalence relation. Your question asks you to define an equivalence relation on $A$.
Next we suppose that $R$ is an equivalence relation on $A$, for $a\in A$ we define the equivalence class of $a$ (in $R$) to be the set: $$[a]_R=\{x\in A\mid aRx\}$$

Exercise: Prove that if $R$ is an equivalence relation on $A$ and $xRy$ then $[x]_R=[y]_R$.

The question now asks to take the equivalence relation which you have defined and write down all the equivalence classes.
Last you are require to discuss the relation between the equivalence classes and the characteristics of a partition of $A$. Let us define this as well:
We say that $P$ is a partition of $A$ if:


*

*Every $X\in P$ is a non-empty subset of $A$.

*For every $a\in A$ there is some $X\in P$ such that $a\in X$.

*For every $X,Y\in P$ if $X\neq Y$ then $X\cap Y=\varnothing$.

