Let $\mathbb{R}$ be the relation on $\mathbb{Z} \times \mathbb{Z}$, that is elements of this relation are pairs of pairs of integers, such that $((a,b),(c,d)) \in \mathbb{R}$ if and only if $a-d = c-b$. Can anyone give me a start on how to solve it to be transitive, reflexive and symmetric?
[Math] Equivalence Relations on Set of Ordered Pairs
discrete mathematicsequivalence-relations
Related Solutions
You seem to be overlooking the last three words of the question. The question doesn't want any old equivalence relation on $A$ you can come up with -- it want the particular equivalence relation "induced by $\pi_1$".
Perhaps you have missed that "the equivalence relation that such-and-such partition induces" has a particular definition? The exercise is asking you to apply that definition to find which one of the many possible equivalence relations on $A$ it is speaking about.
There are various equivalent ways to define this concept -- we can either say
We say that the equivalence relation $R$ is induced by the partition $\pi$ if the elements of $\pi$ are exactly the equivalence classes under $R$.
or
Given a partition $\pi$, the equivalence relation induced by this partition is the relation $R_\pi$ defined by $$ x\mathrel{R_\pi}y \iff \exists P\in\pi: \{x,y\}\subseteq P $$
- If $(6,6)\notin R_1$, then $R_1$ would not be reflexive simply because $R_1$ being reflexive means that $(\forall n\in\{1,2,3,4,5,6\}):(n,n)\in R_1$.
- It is transitive because if $3\mid x-y$ and $3\mid y-z$, then $3\mid(x-y)+(y-z)$. But this means that $3\mid x-z$. So, this proves that$$x\mathrel{R_1}y\text{ and }y\mathrel{R_1}z\implies x\mathrel{R_1}z.$$
- Consider the number $1$. For which numbers $n\in\{1,2,3,4,5,6\})$ do we have $1\mathrel{R_1}n$? It is easy to see that this occurs if and only if $n=1$ or $n=4$. So, the equivalence class of $1$ is $\{1,4\}$. Now take some element of $\{1,2,3,4,5,6\}\setminus\{1,4\}$. Suppose that you have taken $5$. For which numbers $n\in\{1,2,3,4,5,6\})$ do we have $5\mathrel{R_1}n$? It is easy to see that this occurs if and only if $n=2$ or $n=5$. So, the equivalence class of $5$ is $\{2,5\}$. And now you start all over again, taking some element from $\{1,2,3,4,5,6\}\setminus\bigl(\{1,4\}\cup\{2,5\}\bigr)$…
Best Answer
Reflexive: $\forall (a,b):\Bigl[(a,b)\in \mathbb{Z\times Z} \to \bigl((a,b),(a,b)\bigr)\in R\Bigr]$
Symmetric: $\forall (a,b,c,d): \Bigl[\bigl((a,b),(c,d)\bigr)\in R \leftrightarrow \bigl((c,d),(a,b)\bigr)\in R\Bigr]$
Transitive: $\forall (a,b,c,d)\exists (e,f): \Bigl[\bigl((a,b),(e,f)\bigr)\in R\land \bigl((e,f),(c,d)\bigr)\in R \leftrightarrow \bigl((a,b),(c,d)\bigr)\in R\Bigr]$
Show that these properties hold (or not) when $\Bigl[\bigl((a,b),(c,d)\bigr)\in R \Bigr]\iff\Bigl[ a-d=c-b\Bigr]$
Hint: $[a-d=c-b] \iff [a+b=c+d]$