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The problem I have to solve is:

Prove an equivalence relation given $\;a\mathcal R b\;$ iff there exists $\;x\in \{1,4,16\}\;$ such that $\;ax\equiv b\pmod{63}\;$

I understand the definitions of reflexive, symmetry and transitive, but i'm not sure how to prove this with the given problem. Could someone please give a hint as to where to start?

DonAntonio
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Chloe
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  • Ok, so have you already tried reflexivity? Observe that $;16=4^{-1}\pmod{63};$ ....how can this help you with symmetry, say? What about Transitivity: have you done something yet? Show your work in your question... – DonAntonio Sep 22 '19 at 10:25
  • well 1 = $4^0 $, 4 = $4^1$ and 16=$4^2 $and 63 = $4^3$-1 so can I use this somehow as well? – Chloe Sep 22 '19 at 11:19
  • $4^{-1}$ means the multiplicative inverse of the element $[4]$ in the ring $\mathbb{Z}/63\mathbb{Z}$. We have $[4]^{-1} = [16]$ since $[16] \cdot [4] = [4] \cdot [16] = [64] = [1]$ -- and $[1]$ is the multiplicative neutral element. – ComFreek Sep 22 '19 at 11:33

2 Answers2

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Hint $\,\ a\,R\,b \iff b\equiv 4^{\large n} a\pmod{\!63}\, $ for some $\,n\in\Bbb N\ $ (note $\,4^{\large 3}\equiv 1\,\Rightarrow\, 4^{\large -1}\equiv 4^{\large 2}$)

Remark $\ $ This is a special case of oribit equivalence.

Bill Dubuque
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    This is interesting! So we have a group action of the group ${1,4,16}$ on the group $\mathbb{Z}/63\mathbb{Z}$, right? And the equivalence just states $a\ R\ b$ iff. $b$ is "reachable" via a group action from $a$ and the equivalence classes are the orbits. – ComFreek Sep 22 '19 at 15:21
  • @ComFreek Yes, the classes are the cosets of the multiplicative subgroup $4^n \equiv { 1, 4, 16 }.,$ For another more interesting example see this group-theoretic proof of Wilson's Theorem. – Bill Dubuque Sep 22 '19 at 15:56
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Well, start as always: do definition unfolding. I'll show you how to start proving reflexivity:

  • Proof goal: $\forall a\in\mathbb{N}. a\ R\ a$

  • Proof:

    • Let $a \in \mathbb{N}$.
    • Proof goal now: $a\ R\ a \Leftrightarrow \exists x\in \{1,4,16\}.\quad ax \equiv a$

    • Which $x$ can we take?

      Just take $x = 1$ and we have $a \equiv a$ since $\equiv$ is a eqv. relation and thus reflexive.

ComFreek
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  • thank you that makes sense. I have done the same with symmetry and got to a point where i have 63 divides (ax -b) and 63 divides (bx-a). Using the same logic as above can i take x = 1 and then say 63 divides 0? – Chloe Sep 22 '19 at 11:26
  • @Chloe It's easier if you do case analysis. So you know $a\ R\ b$, then you know either $x=1$ or $x=4$ or $x=16$ such that $ax \equiv b$. Divide by $x$ (in the ring $\mathbb{Z}/63\mathbb{Z}$ (i.e. multiply by its inverse) to get $a \equiv bx$ in each case, which implies $b\ R\ a$. – ComFreek Sep 22 '19 at 11:36
  • I'm sorry Im unfamiliar with the ring terminology could you please just clarify for me what i multiply by? is it the multiplicative inverse? – Chloe Sep 22 '19 at 11:38
  • @Chloe Yes exactly, it's the multiplicative inverse. – ComFreek Sep 22 '19 at 11:39