Tag Info

Hot answers tagged

4

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 ...


3

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 ...



Only top voted, non community-wiki answers of a minimum length are eligible