# Tag Info

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

Yes, you just have to add an axiom stating that the relation is a function: $$\forall x\forall y\forall z(R(x,y)\land R(x,z)\rightarrow y=z)$$ And you may want to require that the domain is the entire universe.

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

Yes, indeed, you are correct. The relation is not symmetric, and for the reason you post. $$(X\subseteq Y\; \text{ AND }\;Y\subseteq X) \iff X=Y$$ Hence, since $(X, Y)\in R \implies (X\subseteq Y$ and $X\neq Y$), it cannot be the case that $Y \subseteq X$.

4

The relation cannot be reflexive. $(X, X)\in R$ if and only if both the following statements are true: $$X\subseteq X\tag{1}$$ $$X \neq X\tag{2}$$ $(1)$ is true for all $X \in \mathcal P(A)$. $(2)$ is false for all $X \in \mathcal P(A)$. Hence, for all $X \in \mathcal P(A)$, $(X, X) \notin R$.

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 $\bar{4}^2=\bar{2}^2, \bar{5}^2 = \bar{1}^2, \bar{6}^2=\bar{0}^2, ... 3 This means that if you consider the collection of equivalence classes, and this order defined$\leq$, then it is a partial order. Namely, reflexive, antisymmetric and transitive. The problem is that usually,$|A|$is not a set, and the collection of different equivalence classes is too big to be a set as well (even if we managed to choose a unique ... 3 Hint Let$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$is called a symmetric relation. 3 If$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)$; ... 3 The relation of interest is not transitive, because$R_1(1,2)$and$R_1(2,4)$are true, but$R_1(1,4)$is false. By the way, a good specification of$R_1$would be explicit about the values that the arguments can take. For example, you could use set-builder notation: $$R_1 = \{(a,b) \in \mathbb{Z} \times \mathbb{Z} \mid 2a^2+b^2−3ab=0\}$$ Also, if you're ... 3 Its not a good idea to use subtraction, for by default that relation is not symmetric. However you are correct. But I would suggest the following :$(a,b)R(c,d) \iff a+d =b+c \equiv (a,b)R(a,b) \iff a+b =a+b$therefore R is trivially reflexive. $$(c,d)R(a,b) \iff c+b=d+a$$*and so a+d=b+c, which means (a,b)R(c,d)*. Therefore R is symmetric. ... 2 The axiom explains itself, does it not? For each$n$, and for each point$P$on$x=n$, and point$Q$on$y=n$, there must be a fourth point$Y$in the relation such that$P-(n,n)-Q-Y$is a rectangle. (You could say that if$P$or$Q$coincided with$(n,n)$, you'd be looking for a degenerate rectangle that's either a line or a point.) 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 Yes, a function from$A$to$B$is a relation on A and B such that any element x of the domain cannot be related to two different element$y$and$z$of$B$. Also every element of the domain$A$should be related to some element of the target$B$. In one sentence,every element of the domain should be related exactly one element of the target. 2 Great Question. Yes! In fact, all functions are relations: the set of all functions is a subset of the set of all relations. Think about it: take two sets$A = \{a,b,c\},B = \{1,2,3\}$. Create a function between them. Remember the formal definition of a function: a function is just a set of ordered pairs$(x,y)$, for which we like to say that$x \in X$is ... 2 Without knowing exactly what is meant by "classifying" the relation, I would just answer that the relation is the equality relation over the domain$\{2, 3\}$. You can throw in the terms reflexive, symmetric, and transitive, as well. But I think "equality" over a domain is pretty precise. 2 The relation$R$sends each real number$x$to$x$and$-x$. In other words, $$R(x)=\pm x.$$ From here it is clear that the domain and the range of$R$are all real numbers, and that each$y$in the range is sent to$-y$and$y$in the domain. Thus,$R^{-1}=R$. I leave it to you to find$R\circ R^{-1}$. 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$(b,b)\in f$. 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 ... 1 (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 ... 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 Instead of writing$(a,b)R(c,d)$, we will write that$(c,d)$is equivalent to$(a,b)$. We have$(c,d)$is equivalent to$(3,3)$precisely if$3-2d=c-6$. (I am just reading this from the definition, replacing$a$by$3$and$b$by$3$.) For clarity rewrite the equation$3-2d=c-6$as$c+2d=9$. Now recall that we are working in$\mathbb{Z}^+$, which in your ... 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 Since$a \in [a]$, we must have$A = \cup_{a \in A} [a]$. This is true whether or not the number of equivalence classes is finite or not. If we let$A/{\sim} $denote the equivalence classes, we have$A = \cup_{B \in A /{\sim}} B$. If it happens that$A /{\sim} = \{ A_1,...,A_k \}$, you can write$A = \cup_{i=1}^k A_i$. 1 Your statement$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.

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