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Check reflexivity: Is it the case that for all $(a, b)\in \mathbb N\times \mathbb N$, it is true that $(a, b) R (a, b)$? That is, is it true that for all such $(a, b)$, $ab = ba$? Check symmetry: Is it the case that for all $(a, b), (c, d) \in \mathbb N\times \mathbb N,$ that If $(a, b) R (c, d)$, then $(c, d) R (a, b)$? This means that if $ad = bc,$ is ...
To check if a relation @ is an equivalence relation, you need to verify three things: @ is reflexive @ is symmetric @ is transitive For your particular relation, reflexivity is apparent as $f(x) = f(x)$, (so $x@x$). Symmetry is also not very difficult. If $x@y$, do we have $y@x$? Of course, $x@y$ means $f(x)=f(y)\leftrightarrow f(y)=f(x) \leftrightarrow ... 2 Yes. 1) Reflexivity. Clearly all elements are in their own group of the partition, so we have$aRa$for all$a$. 2) Symmetry. If$aRb$then$a$and$b$are in the same group of the partition, so$b$and$a$are in the same group, so$bRa$. 3) Transitivity. If$aRb$and$bRc$then$a,b,c$are all in the same group, so$aRc$. Also note that the converse ... 2 You don't prove the partitions are equivlaence relations: you prove that there is a one-to-one correspondence between partitions and equivalence relations. That is, Given an equivalence relation on a set, you can use it to construct a partition of the set. Given a partition of a set, you can use it to construct an equivalence relation on the set and ... 1$(a,b) \in RZ \iff (a,x)\in R \land (x,b) \in Z$This means:$a$and$b$are related by the product$RZ$if, and only if, there exists an$x$such that$a$and$x$are related by$R$, and$x$and$b$are related by$Z$.$(a,b) \in S\cup T \iff (a,b) \in S \lor (a,b) \in T$This means:$a$and$b$are related by the union of$S$and$T$if, and only if, ... 1 A total order relation requires 4 things: 1)refxivity(it is reflexive in this case) 2)anti-symmetricity(it is anti-symmetric) 3)transitivity(it is transitive) 4)comparibility Now compatibility means that if you choose any two elements say a,b then either aRb or bRa But in this case if we take for example b and c, neither bRc nor cRb so the relation is not ... 1 Some suggestions: If you have antisymmetry, then your preorder is actually a partial order and it's easy to construct a corresponding lattice. If you don't have antisymmetry, then there exists a pair$a,b$such that$a \preceq b$and$b \preceq a$. How would you define$\{a,b\}$? In fact$\preceq$could be the full relation$Q \times Q$(i.e. it is ... 1 Here is a nice way to proceed - one that works quite generally. Notice that$\rm\,x\sim y\,$iff$\rm\,x\,$and$\rm\,y\,$have equal$ $magnitude:$\rm\ \ x\sim y {\overset{\ def}{\color{#c00}\iff}} f(x) = f(y)\ $for$\rm\:f(x) = \rm\,|x|.\$ Now it is straightforward to prove that any relation of the above form is an equivalence relation. More ...