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8

The best explanation is probably that the possibility of same-sex marriages were not on the author's mind when he/she came up with the exercise. If we suppose that marriages is always between a man and a woman (and genders are binary, bla bla bla), then the relation is indeed transitive, but vacuously so. The condition it has to satisfy is that if $x$ is ...


5

Doesn't $(\mathbb R,<)$ have both of these properties?


5

We would like to know how many functions from $\{ 1, 2, 3\}$ to $\{1,2,3\}$ there are for which $f(1)+f(3)\equiv 0 \mod 2$. In other words, we would like to know how many ways we can match the parity of $f(1)$ with $f(3)$. If $f(1) = 1$ then $f(3) = 1$ or $f(3)=3$. If $f(1) = 2$ then $f(3) = 2$. If $f(1) = 3$ then $f(3) = 1$ or $f(3)= 3$. In each of ...


4

Since $R$ is small, you can actually consider all possibilities for $a,b$, and $c$. However, this is easier if you have a systematic arrangement of the ordered pairs. One way is to represent them by a matrix: $$\begin{array}{c|ccc} &t&u&v&w&x&y&z\\ \hline t&1&&1&&&&1\\ ...


4

It is entirely valid to have a relation that relates every element of $A$ to every element of $B$. It can even have fancy properties -- for example if $B$ is a singleton set, the relation $A\times B$ will be a function!


3

They're two different things, there isn't really a strong relationship between the two. Based on the definitions you're using, they both give two different criteria for concluding that $(x, x) \in R$. For any antisymmetric relation $R$, if we're given two pairs, $(x, y)$ and $(y, x)$ both belonging to $R$, then we can conclude that in fact $x = y$, so that ...


3

Here are a few relations on subsets of $\Bbb R$, represented as subsets of $\Bbb R^2$. The dotted line represents $\{(x,y)\in\Bbb R^2\mid y = x\}$. Symmetric, reflexive: Symmetric, not reflexive Antisymmetric, not reflexive Neither antisymmetric, nor symmetric, but reflexive Neither antisymmetric, nor symmetric, nor reflexive


3

Remember that the implication: $$p \rightarrow q$$ Is always true when $p$ is false, regardless of the truth value of $q$. Thus, consider your first relation $R = \{(1,6),(2,7),(3,8)\}$. Note that: $$(x,y)\in R \wedge (y,z)\in R$$ is always false, hence the implication: $$(x,y)\in R \wedge (y,z)\in R \rightarrow (x,z)\in R$$ is always true, and hence ...


3

I assume the undirected edges are meant left-to-right ($\to$). Start with $s_0$ and repeat as many $a$ and $c$ as you want, then transition with $b$: [ac]*b. Then we need a $d$,$a$ and optionally any number of occurrences of $b$,$d$,$a$, i.e. [ac]*bda(bda)* Or shorter [ac]*(bda)+


3

I hope no one minds if I assemble the various comments by Cloudscape and coffeemath (and myself) into an answer. For reflexivity, you want to show that $xSx$, or $x=|x|$ for all $x\in\mathbb{R}$. Clearly this is not true for any negative values of $x$, so $S$ is not reflexive. For symmetry, you want to show that $xSy$ implies $ySx$, or that whenever ...


3

You'd need to prove that if $\frac{a}{b}=\frac{c}{d}$ then $$\frac{\max(a,b)}{\min(a,b)}=\frac{\max(c,d)}{\min(c,d)}$$ Then it depends on what $a,b$ are allowed. We know that $a,b\neq 0$, but can they have a common divisor? Can $b$ be negative? Or are we taking some "usual" representation for each rational - for example, $a,b$ relatively prime with ...


3

$(4,1)$ and $(1,2)$ are both in the relation, but $(4,2)$ is not...


3

One thing that might be useful is to observe that this is a linear function. We can write $f(x,y) = (x + 3y, 3x + y)$ also in the matrix form $$ f(x,y) \;\; \to \;\; \left [ \begin{array}{cc} 1 & 3 \\ 3 & 1 \\ \end{array} \right ] \left [\begin{array}{c} x \\ y \\ \end{array} \right ] \\ $$ Finding the inverse of $f$ should then be equivalent to ...


2

You need to solve $(u,v) = (x+3y, 3x+y)$ for $x,y$. In this case we get $3u-v = 8y$, hence $y = \frac{1}{8}(3u-v)$ and the same way we get $x = \frac{1}{8}(3v-u)$. So $f^{-1}(u,v) = \frac{1}{8}(3v-u, 3u-v)$. You can check that this is true by computing $f\circ f^{-1}$ and $f^{-1}\circ f$.


2

firstly, I am not understanding the function of R,I think that, this is only a relation (according to me) how can it be an equivalence it's not a set. A relation $R$ is a set of pairs, $R\subseteq A\times A$. $aRb$ is just a short form to write $(a,b)\in R$. Now if we define $R=R_1\cap R_2$, we have $$\begin{align} aRb &\iff (a,b)\in R ...


2

The word "equivalent" is built from "equal" and "value" — well, actually from the Latin words aequus (equal) and valere (to be worth). So two elements are equivalent if they are in a sense equal-valued, or interchangeable. Often this is indeed used literally, by defining some quantities as equivalence classes; for example, the fraction $\frac42$ has the same ...


2

If you define the reals with Dedekind cuts, yes: $$\forall a,b\in\mathbb R(a\leq b\iff a\subseteq b)$$ $$\forall a,b\in\mathbb R(a<b\iff a\subsetneq b)$$ It's a little harder to define ordering on the reals if you use Cauchy sequences. (This is because Dedekind cuts are specifically designed to "complete" the ordering of the rationals, while Cauchy ...


2

It all comes down to how we're setting the reals up. If we're using Dedekind cuts, then yep, that's how to do it! If we're using equivalence classes of Cauchy sequences, then we have to say something messier: $a<b$ if, whenever we pick representative sequences $\alpha\in a$ and $\beta\in b$, there is some $n$ such that for all $m>n$, ...


2

Depends on the construction you make of $\mathbb{R}$. Assuming that you already have built the operators $+,\cdot$, a simple way to notice it would be to make a subset $\mathcal{P}\subset\mathbb{R}$ which represent the positive numbers, then $a\leq b \iff b+-a\in \mathcal{P}$, where $-a$ is the additive opposite of $a$ such that $a+-a = 0$.


2

I'm no expert, but a relation can be treated as a function that maps to propositions. The book is written under the propositions = types paradigm, so a binary relation over types $A$ and $B$ is any element of the type $A \times B \rightarrow \mathfrak{U}$ where $\mathfrak{U}$ is the universe.


2

For (b), the first recurrence implies $$ a_{n+2} = 3a_{n+1} + b_{n+1}. $$ Now substitute $a_{n+1} = 3a_n + b_n$ and $b_{n+1} = 5a_n - b_n$ into the right hand side of this equation and regroup to get $$ a_{n+2} = 3(3a_n + b_n) + 5a_n - b_n = 14a_n + 2b_n = 8a_n + 2(3a_n + b_n) = 8a_n + 2a_{n+1}. $$


2

Relations are subsets of products $A\times B$ where $A$ is the domain and $B$ the codomain of the relation. A function $f$ is a relation with a special property: for each $a\in A$ there is a unique $b\in B$ s.t. $\langle a,b\rangle\in f$. This unique $b$ is denoted as $f(a)$ and the 'range' of function $f$ is the set $\{f(a)\mid a\in A\}\subseteq B$. You ...


2

Both relations are transitive by vacuity. In the first case you would need to prove that, for example, if 1 is related to 6 and 6 is related to some $c$ then $1$ is related $c$. Since 6 is not related to anything the condition for transitivity holds. In the same way, for the second case you would need to proe that: if $x$ is wife of $y$ and $y$ is wife of ...


2

In your notation, try $m'' = mm'$ and $n'' = nn'$.


2

Lets recall the definition of transitivity. A relation $R \subseteq A \times A$ on $A$ is called transitive, if we have $$(a,b),(b,c) \in R \Rightarrow (a,c)$$ for all $a,b,c \in A$. Your initial set is $R = \{(1,2),(2,3),(3,4),(4,1)\}$. The way you described your approach is basically the way to go. Just go through the set and if you find some ...


1

The answer seems to depend on jurisdiction (and maybe subscription to certain formulations of gender identity). We'll make some common assumptions for the sake of producing an answer: Let $X$ be a set (whose elements are people), each of which is labeled with exactly one of two genders, which we'll denote $M$ and $W$ (if you like, this is a function $X \to ...


1

Let $a^m = b^n$ and $b^x = c^y$. Then $(a^m)^x = (b^n)^x$ and $(b^x)^n = (c^y)^n$. We have $a^{mx} = b^{nx} = c^{yn} \implies a^{mx} = c^{yn}$


1

Recall that the definition of a relation $R$ between two sets $A$ and $B$ is a subset $R \subseteq A \times B$ so relations are not always a proper subset of $A \times B$. For example, given $A = \{a,b,c\}$ and $B = \{d,e,f\}$, we have $R = \{(a,d),(a,e),(a,f),(b,d),(b,e),(b,f),(c,d),(c,e),(c,f)\}$ which covers the entire set $A \times B$.


1

No, there are plenty of anti-symmetric relations that are not reflexive. For instance, let $R$ be the relation $R=\{(1,2)\}$ on the set $A=\{1,2,3\}$. This relation is certainly not reflexive, but it is in fact anti-symmetric. This is vacuously true, because there are no $x$ and $y$, such that $(x,y)\in R$ and $(y,x)\in R$. Edit: Why is this ...


1

Suppose $R$ is a relation, and let $X$ be the domain of $R.$ I claim that $R$ is the equality relation on $X$--that is, $$R=[=]_X:=\bigl\{\langle x,x\rangle:x\in X\bigr\}.$$ On the one hand, if $\langle x,y\rangle\in R$--that is, $x\:R\:y$--then $x\in X$ by definition. By symmetry, $y\:R\:x,$ and by antisymmetry, $x=y.$ Hence, $R\subseteq [=]_X.$ On the ...



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