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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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 ...

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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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