$R$ is a relation in a set $A$ (from $A$ to $A$) and $R^{-1}$ is the inverse relation of $R$ prove:

(1) $R ∪ R^{-1}$ is symmetric.

(2) If $R$ reflexive ⇒ $R^{-1}$ is reflexive.

(3) If $R$ symmetric ⇒ $R^{-1}$ is symmetric.

(4) If $R$ transitive ⇒ $R^{-1}$ is transitive.

(5) If $R$ antisymmetric ⇒ $R^{-1}$ is antisymmetric.

I did the (5) as follows:

If $R$ antisymmetric ⇒ $R^{-1}$ is antisymmetric. Demonstration: we must prove that $R^{-1}$ is antisymmetric, that is, we must prove that

$xR^{-1}y$ $∧$ $yR^{-1}x$ ⇒ $x = y.$

We see it,

$xR^{-1}y$ ⇒ $yRx$, (I), by definition of inverse relationship.

$yR^{-1}x$ ⇒ $xRy$, (II), by definition of inverse relationship.

(I) ∧ (II) ⇒ $xRy ∧ yRx$

⇒ $x$ = $y$, and then, by hypothesis, $R$ is antisymmetric.

Is this demonstration correct? How should I demonstrate the other 4 properties between $R$ and $R^{-1}$?


For the purpose of this answer I assume that you define $R^{-1} = \{(y,x)\text{ }|\text{ }(x,y)\in R\}$.

Proof. We examine each of the above propositions in turn.

  • Notice that $(1)$ and $(2)$ follow immedietly from the definition of $R^{-1}.$

  • For $(3)$ you can assume that $R$ is symmetric and let $x,y\in A$ such that $(x,y)\in R^{-1}$ equivalently $(y,x)\in R$ then by symmetry of $R$ we have $(x,y)\in R$ and equivaletly $(y,x)\in R^{-1}$.

  • For $(4)$ assume $R$ is transitive then given arbitrary $x,y,z\in A$ such that $xR^{-1}y$ and $yR^{-1}z$ equivalently we have $yRx$ and $zRy$ which in conjunction with the transitivity of $R$ imply $zRx$ equivalently $xR^{-1}z$.

and yes your proof of proposition $(5)$ is very much correct.

Lastly if you would oblige me may i ask which text is this problem from?

  • 2
    $\begingroup$ Typo: in (3), "by symmetry of $R$, we have $(x,y)\in R^{-1}$": it's $(x,y) \in R$. $\endgroup$ – amrsa Jun 11 '18 at 16:06
  • $\begingroup$ Thank you very much for the reply, @Atif Farooq ! I found the exercise in a partial examination of Algebra for Astronomy of the UNLP. $\endgroup$ – Ayesca Jun 11 '18 at 16:11
  • 1
    $\begingroup$ @amrsa Thankyou for you remark $\endgroup$ – Atif Farooq Jun 11 '18 at 16:30

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