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{$(a, b) | a∈Z^{+}, b∈Z^{+}, a = 3^{n}b, where n∈N$} (N is the set of natural numbers)

{$(a, b) | a∈Z^{+}, b∈Z^{+}$, a >2b or b >2a}

{$(a, b) | a∈Z^{+}, b∈Z^{+}$, a ≡ 0 (mod b) or b ≡ 0 (mod a)}

Terabyte
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  • Well, I know it's considered an equivalence relation if it's reflexive, symmetric and transitive and a partial ordering relation if it's reflexive, antisymmetric and transitive. But I'm getting confused when it comes to proving these. – Terabyte Apr 12 '13 at 13:05
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    Well, give it a shot --- can you decide whether the first one is reflexive? It really helps if we know what you can do on your own. Also, I think you're missing something in the 3rd one, maybe $b\equiv0\pmod a$. – Gerry Myerson Apr 12 '13 at 13:11
  • Oops, yes. I was missing a 0. – Terabyte Apr 12 '13 at 13:12
  • I don't think the first one is reflexive (which would mean it's neither) but I'm not 100% sure. The first one actually confuses me the most. – Terabyte Apr 12 '13 at 13:17
  • I think the 3rd one is an equivalence relation because it's reflexive and symmetric (still working on the transitive). – Terabyte Apr 12 '13 at 13:25
  • @Terabyte, I'm not sure in what context these relations appeared, but in that particular course/book/paper/whatever are natural numbers defined to start at 0 or 1? – in_mathematica_we_trust Apr 12 '13 at 13:52
  • N = {0, 1, 2, 3,...} so 0. – Terabyte Apr 12 '13 at 13:55
  • For reflexivity of the first relation: you have to check whether $(a,a)$ is always in the set, i.e. whether it is true, for all $a \in \mathbb{Z}^+$, that $a = 3^n a$ for some $n \in \mathbb{N}$. In order for $a = 3^n a$, since $a \neq 0$, we must have $3^n = 1$. Since it is possible to find such an $n$, namely $n = 0$, we conclude that $(a,a)$ is in the set for all $a \in \mathbb{Z}^+$. Thus, the first relation is reflexive. Try to use similar reasoning to see if the other two relations are reflexive. – Michael Joyce Apr 12 '13 at 14:08

1 Answers1

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All three relations are of the form $$aRb\quad \text{ if }\frac{a}{b}\in S$$ where $S$ is some subset of $[0,+\infty]$. Namely,

  1. $S=\{3^n: n\ge 0\}$
  2. $S=[0,1/2)\cup (2,\infty]$
  3. $S=\mathbb N\cup \{1/x:x\in \mathbb N\}$

(One can observe that formally setting $0/0=1$ fits in all three cases.)

  • Reflexivity amounts to $1\in S$.
  • Symmetry amounts to $S=\{1/x:s\in S\}$.
  • Antisymmetry (weak) amounts to $S\cap \{1/x:s\in S\}=\{1\}$
  • Transitivity amounts to $S$ being closed under multiplication.
40 votes
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