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9

This is unusual terminology, but legitimate. We usually talk about a relation being transitive, meaning that $aRb,bRc$ implies $aRc$. If we take $R$ to be the relation which has $aRf(a)$, then it will be transitive if $aRf(a)$ and $f(a)Rb$ implies $aRb$. But since $f()$ is a function that means $b$ must equal $f(a)$. In other words, the point $f(a)$ which ...

8

A binary relation (or relation, means the same) from a set $A$ to a set $B$ is any subset $R\subseteq A\times B$. We take any here seriously so in particular, if $A$ contains some element $a$, and $B$ contains some element $b$, then $R=\{(a,b)\}$, being a subset of $A\times B$ is a relation from $A$ to $B$. For that matter, given any two sets $A$ and $B$, ...

6

Let $S=\{a,b,c,d\}$. Let $T$ be the image of $f$. Then $f(t)=t$ for all $t\in T$, and $f:S\setminus T\to T$ can be anything. There are $$|T|^{|S\setminus T|}$$ such functions for each $T$. So it depends only on the size of $T$, we'd have: $$\sum_{\emptyset\neq T\subseteq S} |T|^{|S\setminus T|}=\sum_{t=1}^4 ... 4 just trust the definition! a relation on a set A is any element of \mathfrak{P}(A \times A). a singleton is such an element. don't confuse with a function A \to A, which is a relation which must satisfy two further conditions. what about the empty relation? 4 Suppose that x\sim y and y\sim z. Then 3 divides |x-y|. It follows that 3 divides x-y. Similarly, 3 divides y-z. So 3 divides (x-y)+(y-z). It follows that 3 divides x-z, and therefore 3 divides |x-z|. We conclude that x\sim z. 3 If |a-b| < n \cdot 0.1, then (a,b) lies in H^{\circ n} and hence in the transitive closure. Since \{n \cdot 0.1 : n \in \mathbb{N}\} is unbounded, we are done. 3 If X and Y are sets and f:X \to Y is a function, then R is an equivalence relation on X if "\equiv" is an equivalence relation on Y. We need merely check the axioms: I.) Reflexitivity: for any x \in X, we have f(x) \equiv f(x), so x\ R\ x; II.) Symmetry: for x, y \in X, we have f(x) \equiv f(y) \Leftrightarrow f(y) \equiv f(x); ... 3 Hint Let r = a+b with a,b\in X How many different values can r have? 3 Note that we can remove the negation because this is an if and only if. Therefore this is the same as saying a\mathrel{R}b\iff b\mathrel{R}a. This property is called symmetry, and such R is called a symmetric relation. 3 This means that if you consider the collection of equivalence classes, and this order defined \leq, then it is a partial order. Namely, reflexive, antisymmetric and transitive. The problem is that usually, |A| is not a set, and the collection of different equivalence classes is too big to be a set as well (even if we managed to choose a unique ... 3 Since there are only 12 elements, you can simply check this element by element. Clearly \bar{0},\bar{1},\bar{2},\bar{3} form distinct equivalence classes since their squares are all different. Since \bar{4}^2=\bar{2}^2, \bar{5}^2 = \bar{1}^2, \bar{6}^2=\bar{0}^2, ... 2 (Hurkyl + Ittay Weiss are right) If u define Rel to have sets as objects and binary relations as arrows and you show this makes it a category, then u have (as for any category): Say \mathcal{C} is an arbitrary category (not necessarily small) Define \mathcal{\hat C} to be the category having as objects all \mathcal{C}-arrows and as arrows between ... 2 As this relation's defined, we see that (a,b)\in f and (b,a)\in f so f(\color{red}a)=b and f(b)=\color{red}a respectively. So f(f(b))=f(\color{red}a)=b. This means that (b,b)\in f. 2 There is some subset on which it is the identity function, and all elements not in that subset are mapped into that subset. That subset cannot be empty because some elements are mapped to members of it. So first choose one of the 15 subsets to be the image. Then one must choose where to map the other elements. The number of ways to do that depends on ... 2 Not true because \dfrac{1}{2} = \dfrac{2}{4} but 1-2 = -1 while 2-4 = -2 2 It is not. Your relation will include (\frac 1 2,-1) and (\frac 2 4, -2) = (\frac 1 2, -2) implying that \frac 1 2 is mapped to both -1 and -2, which violates the definition of a function. 2 Let S be the set of all countries on Earth and consider the "is a neighbour of" relation on S. Notice that this relation is not transitive. To see this, observe that Canada is a neighbour of USA and USA is a neighbour of Mexico, but Canada is not a neighbour of Mexico. Furthermore, notice that this relation is not intransitive. To see this, observe ... 2 Actually, to prove that f is injective, you have to show that f(p_1,q_1)=f(p_2,q_2) implies (p_1,q_1)=(p_2,q_2). But since f(p,q)=(r,q), then this immediately implies q_1=q_2. The rest of the argument you gave shows that p_1=p_2, thus (p_1,q_1)=(p_2,q_2). For the surjectivity, let (r,q)\in H. Let's find p such that (p,q)\in G and ... 2 If relation R is a function then (a,b)\in R\wedge (a,c)\in R implies that b=c. This is not the case for R:=\{(a^2,a)|a\in\mathbb R\}. For instance we have (4,2)\in R and (4,-2)\in R while 2\neq-2. 1 Well this works:$$Z - Y = 10X + 11\left\lfloor\frac{X-8}{10}\right\rfloor -2 PARI-GP Script: for(x = 1, 20, printf("X = %d, Z - Y = %d\n", x, 10 * x - 2 + 11 * floor((x - 8) / 10))) Output: X = 1, Z - Y = -3 X = 2, Z - Y = 7 X = 3, Z - Y = 17 X = 4, Z - Y = 27 X = 5, Z - Y = 37 X = 6, Z - Y = 47 X = 7, Z - Y = 57 X = 8, Z - Y = 78 X = 9, Z - Y = 88 ...

1

Order people alphabetically by name. This is reflexive and transitive but obviously not symmetric. It's not anti-symmetric either, because people can have the same name without being the same person (as anyone unlucky enough to have the same name as someone placed on a no-fly list can tell you).

1

Your statement $P = \{X\ :\ a\in X\}$ is not correct. $P$ is the set of all elements of the partition, while $\{X\ :\ a\in X\}$ consists only of that element of the partition containing $a$. The portion of the proof you quoted is saying this: choose $a\in S$; then there is a $X\in P$ such that $a\in X$. By definition of $R$, $[a] = \{b\in S\ |\ b\sim a\} = ... 1 If I understand you right, then you want to calculate how much sb can buy in a state relative more/less to Missouri (in percent). The term equation would be$\left( \frac{X}{113.51}-1\right) \cdot 100 \% $. Thus you divide by 113.51. 1 See Partially ordered set or poset : A (non-strict) partial order is a binary relation "≤" over a set$P$which is reflexive, antisymmetric, and transitive, i.e., which satisfies for all$a, b, c \in P$:$a ≤ a$(reflexivity); if$a ≤ b$and$b ≤ a$then$a = b$(antisymmetry); if$a ≤ b$and$b ≤ c$then$a ≤ c$(transitivity). We have ... 1 To prove that$R \subseteq \mathcal{P}(X) \times \mathcal{P}(X)$defined by$(A,B) \in R \Leftrightarrow A\subseteq B$is a partial order you need to show: Reflexivity: for all$A\in \mathcal{P}(X)$we have$A\subseteq A$; Anti-symmetry: if$A,B \in \mathcal{P}(X)$and$A\subseteq B$and$B\subseteq A$then$A=B$; Transitivity: if$A,B,C \in ...

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A relation is called – mostly in economics terminology – preference relation if is reflexive, transitive, and dichotomous. Az $R$ relation on an $X$ set is called reflexive, if for all $x \in X$: $(x,x) \in R$; transitive, if for all $x,y,z \in X$ if $(x,y) \in R$ and $(y,z) \in R$ then $(x,z) \in R$; dichotomous or total, if for all $x,y \in R$ $(x,y) ... 1 Hint note that$|x-z|=|x-y+y-z|$1 A relation can be trivially transitive, so yes. The condition for transitivity is: Whenever$aRb$and$bRc-$then it must be true that$aRc$. That is, the only time a relation is not transitive is when$\exists \; a,b,c$with$aRb$and$bRc$, but$aRc$does not hold. So the relation corresponding to the graph is trivially transitive. You can learn ... 1 Two sets agree if they agree on the complement of$Y$, so there is a canonical bijection$P(X)/R \to P(X \backslash Y) = P (\{ 1,3,5,7\})$. The bijection identifies$[A] \in P(X)/R$with$A \cap (X \backslash Y)$. Then you use the fact that$|P(\{x_1,\cdots,x_n\})| = 2^n$. Hope that helps, 1 An ordered pair is defined as the following by Kuratowski$(x,y):=\{ \{x\},\{x,y\} \}.$If$X,Y$are sets, we define their Cartesian product as$X\times Y = \{\,(x,y)\mid x\in X, \ y\in Y\,\}.$A set called binary relation if all of its elements are ordered pair. If$R$is a binary relation we say$(x,y) \in R$or$xRy\$. Sometimes we also speak about the ...

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