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6

You could draw the relations on a piece of paper to make it clearer: R $$ 1\to 2\to 3\to 4\to 5 $$ and S $$ \begin{array}{cc} &&3\\ &\nearrow&\\ 2&&\downarrow\\ &\searrow&\\ &&4 \end{array} $$ and then squaring them means to form every relation that can be achieved by two consecutive arrows, so $$ ...


3

Your proof does indeed show that $R$ is not a non-strict partial order. Perhaps what you read elsewhere claimed that $R' = \{(x,y) ~|~ \text{the first letter of }x\text{ occurs later than the first letter of }y\}$ is a strict partial order, which is of course true.


3

HINT: Show that the relation has only finitely many equivalence classes, say $C_1,\ldots,C_m$, and use the subscripts on the equivalence classes to define your function $f$.


2

This answer starts where you stopped. We focus on the $4$ sets $[1],[0],[4],[9]$. Any equivalence relation with the mentioned properties will induce a partition such that each of its elements is a non-empty union of these sets. We have $[1]\cup[0]\cup[4]\cup[9]=\mathbb N$, but what can be said about mutually disjointness? On this the third property applies ...


2

HINT:Denote by $[x], [y]$ the equivalence classes. To test that the addition is well defined you must see that $[x+y]=[x'+y']$, for all $x'\in[x],y'\in[y]$. In other words: $$x+y\sim x'+y'.$$


2

You only have two pairs of the form $\;(a,b),\,\,(b,c)\in R\;$ , and also $\;(a,c)\in R\;$ , so it is true that whenever $\;(x,y),\,(y,z)\in R\;$ , also $\;(x,z)\in R\;$, and that's transitivity


2

The reflexive closure of some relation $R$ over $A$ is the smallest subset of $A\times A$ that (a) contains $R$ and (b) is reflexive. In this case $R$ is the empty set, so every subset of $A\times A$ satisfies condition (a). We're left with looking for the smallest subset of $A\times A$ that is a reflexive relation on $A$. This smallest subset is evidently ...


1

Hints: (a) Try solutions of the form $u_n=a \lambda^n$. (b) Try solutions of the form $u_n = a 3^n + b n^2 + c n + d$ (c) Try solutions of the form $u_n = a 3^n + b 4^n$ The main principle is to combine solutions of the homogenous equation with a particular solution. (a) is a homogenous equation and you can just try $u_n=a \lambda^n$ and see which ...


1

Because no matter how you select elements $x,y,z \in \{a, b, c\}$ such that $xRy$ and $yRz$ (hint: there's only one way), you have $xRz$, which is what it means to be transitive.


1

Let $R$ be a relation in $A \times B$. If you want $R^{-1}$ to be function on $B$, then you need $R$ to be injective and surjective. Indeed, if $R$ is injective, then $R^{-1}$ is a function on the range of $R$. If $R$ is surjective, then its range is $B$. (A relation is injective iff $(a_1,b) \in R$ and $(a_2,b) \in R$ imply $a_1=a_2$. A relation is ...


1

If I understand your picture correctly, you want to know, given the value of the blue curve at some $x$-value between -8 and 0, the value of the corresponding red curve. So the input to your function will be a number $s$ between $200$ and perhaps $300$, and the output will be between about $0$ and $60$. Here goes. Suppose we call the blue value $t$. Then ...


1

Apparently, the formula tries to express the existence of a supreme and it's unicity (hence the $y=z$ consequent). Since that kind of number does not exist neither in $\mathbb{N}$ nor in $\mathbb{Q}$, then you can be sure that neither $S_1$ nor $S_2$ model the formula.


1

You have done most of the work, you just need to prove transitivity. If $a R b$ and $b R c$, $\frac{a}{b}=2^{m_1}$ and $\frac{b}{c}=2^{m_2}$, multiply them together, you have $$\frac{a}{c}=2^{m_1+m_2}$$ Since $m_1+m_2 \in \mathbb{Z}$, $\frac{a}{c} \in H$, i.e. $a Rc$.


1

It depends on the context. If there is no ambiguity, "less than or equal to" works. In a lecture, you might pronounce it "curly less than" to help people who are taking notes. If you want a short way to pronounce it, you might vocally label it "r" or "rel" (short for "relation"), as in "Suppose that rel is a partial order", but this is less standard and I ...


1

Your proof is right. I would clarify the steps in the transitivity part by saying something like this: Assume $aRb$ and $bRc$. Now $2a+2b\equiv 0 \mod 4$ and $2b+2c\equiv 0 \mod 4$, which means $2a+2b=4m$ and $2b+2c=4n$ for some integers $m$ and $n$. Now by adding the two equations together we have $2a+2b+2b+2c=4m+4n$, i.e. $2a+2c=4(m+n-b)$, and thus ...


1

You seem to be fundamentally misunderstanding something here. Given a set $S$, a relation $R$ is simply defined as some subset of $S^2$, that is, where $$R=\{\langle s_1, s_2\rangle\in R: s_1\in S, s_2\in S\}$$ $R$ is by definition a relation. The domain of $R$ (admittedly this isn't a term I haven't heard or seen before, but from intuition and Google this ...


1

Basically, to prove a relation $R$ is reflexive on a set $S$, we need to prove $(s, s) \in R$ for all $s \in S$. Since we already know this is true for $R_1$ and $R_2$, it becomes very easy to prove this for their intersection and union. Given an element $s \in S$, we know that $(s, s) \in R_1$ and $(s, s) \in R_2$ because $R_1, R_2$ are reflexive. Since ...


1

The relation $R$ is not reflexive: take $A=\{1\}$; then $|A\cap A|=1<2$. The relation $T$ is indeed a partial order, but is not a well-ordering, because, for instance, $\{(1,2),(2,1)\}$ has no minimum. The relation $S$ is indeed an equivalence relation and it is not antisymmetric (the only equivalence relation that is also a partial ordering is the ...


1

Work it out by the definitions. Reflexivity: means x R x for all x. So: $ARB \iff |A\cup B| \ge 2$. Does $|A \cup A| \ge 2$ for all subsets of the natural numbers? Well... $|A \cup A| = |A|$. Do all subsets have cardinality $\ge 2$? Obviously not as, say {0} or {1} or {k} have cardinality of 1 (not to mention $|\emptyset| = 0$). So R is not ...


1

Honestly, I think that the best way to approach this as a beginning math student is just to try to parse through a few examples. If you find a counterexample, you're done, otherwise, after a few tries, see if you can see why it would be true. Try it out with $R$: Reflexivity: $\forall X\in P(\mathbb{N}), R(X,X)$. Note that $A\cup A=A$, so $R(A,A)$ iff ...


1

It looks like you misread slightly: partial orders and equivalence relations are both reflexive and transitive, but only equivalence relations are symmetric, while partial orders are antisymmetric. A relation $R$ is symmetric if $aRb$ implies $bRa$, while a relation $R$ is antisymmetric if $aRb$ and $bRA$ implies $a=b$. For example, the relation "has the ...


1

R = {(a,c), (c,a), (a,d), (d,a), (d,c), (c,d), (a,a), (c,c), (d,d), (b,b)} Hope that helps. Note: The claim that R should be a subset of {a,b,c,d}x{0,1} is incorrect.



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