Show that the relation $R = \{(m,n) : \dfrac{m}{n} ext{ is a power of } 5\}$ on the set $Z - \{0\}$ is an equivalence relation.

Question

Show that the relation $R = \{(m,n) : \dfrac{m}{n} 	ext{ is a power of } 5\}$ on the set $Z - \{0\}$ is an equivalence relation.

Accepted Answer

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> [!TLDR]
> Checking all three properties -- reflexivity, symmetry, transitivity -- confirms R is an equivalence relation.
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> Let $R=\{(m,n): m/n 	ext{ is a power of } 5\}$ on $\mathbb{Z}-\{0\}$ (powers of 5 taken as $5^k$, $k\in\mathbb{Z}$).
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> **Reflexive:** For any $m$, $\dfrac{m}{m}=1=5^0$, a power of 5. So $(m,m)\in R$ for all $m$.
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> **Symmetric:** Suppose $(m,n)\in R$, so $\dfrac{m}{n}=5^k$ for some integer $k$. Then $\dfrac{n}{m}=5^{-k}$, which is also a power of 5. So $(n,m)\in R$.
> …

Q.Show that the relation $R = \{(m,n) : \dfrac{m}{n} \text{ is a power of } 5\}$ on the set $Z - \{0\}$ is an equivalence relation.

Concept understanding — Equivalence Relation Proof

Proving a Relation is an Equivalence Relation

The three checks

A worked template

Q.Show that the relation R={(m,n):mn is a power of 5}R = \{(m,n) : \dfrac{m}{n} \text{ is a power of } 5\}R={(m,n):nm​ is a power of 5} on the set Z−{0}Z - \{0\}Z−{0} is an equivalence relation.

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Q.Show that the relation $R = \{(m,n) : \dfrac{m}{n} \text{ is a power of } 5\}$ on the set $Z - \{0\}$ is an equivalence relation.

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