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I'm a bit confused about how to prove that R is reflexive.

By definition, R, a relation in a set S, is reflexive if and only if ∀x∈S, xRx.

Since (a, α)R(b, β), we know that aβ = bα.

Then to prove that this is reflexive, based on the definition, we would have to show that ((a, α)R(b, β)) R ((a, α)R(b, β). After this, I'm not sure as to how to prove why this is reflexive.

Could we possibly do something like (aβ = bα) R (aβ = bα) is reflexive? Or does ((a, α)R(b, β)) R ((a, α)R(b, β)) already show that it is reflexive itself?

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To prove that $R$ is reflexive is to prove that, for each $(a,\alpha)\in\Bbb Z\times\Bbb N$, $(a,\alpha)\mathrel R(a,\alpha)$. But $(a,\alpha)\mathrel R(a,\alpha)$ means that $a\alpha=a\alpha$, and it is clear that this holds.

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  • $\begingroup$ Oh, I see so I don't use it as ((a, α)R(b, β)) R ((a, α)R(b, β). Could I also prove that this is reflexive by using (b, β) R (b, β)? So in this case, to prove that R is reflexive is to prove that, for each (b, β) ∈ Z×N, (b, β) R (b, β). Since (b, β) R (b, β) means that bβ = bβ, R is reflexive. @José $\endgroup$ – Han May 28 at 11:07
  • $\begingroup$ Yes, you can use $(b\,beta)$. Or $(h,\theta)$. Or $(\zeta,\aleph)$. Or any pair of letters you want. $\endgroup$ – José Carlos Santos May 28 at 11:20
  • $\begingroup$ But in this problem, (h,θ) or (ζ,ℵ)would be irrelevant to this problem right? Thank you for all the help! @José $\endgroup$ – Han May 28 at 11:21
  • $\begingroup$ @JoséCarlosSantos Hi professor Santos, could I ask your assistance here? $\endgroup$ – Antonio Maria Di Mauro May 28 at 16:24

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