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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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