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The discussion focuses on proving properties of the inverse of a relation, specifically showing that the domain of the inverse equals the range of the original relation, and vice versa. Participants confirm that the first three identities are correct, while also discussing the proof for the fourth identity, which states that the inverse of the inverse relation returns the original relation. A detailed explanation is provided for each identity, including the use of logical implications to establish equivalences. The conversation concludes with an affirmation of understanding regarding the proofs presented.
evinda
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Hi! (Wave)

Let $R$ be a relation.

Show the following sentences:

  1. $dom(R^{-1})=rng(R)$
  2. $rng(R^{-1})=dom(R)$
  3. $fld(R^{-1})=fld(R)$
  4. $(R^{-1})^{-1}=R$

That's what I have tried:

  1. Let $x \in dom(R^{-1})$. Then $\exists y$ such that $<x,y> \in R^{-1} \Rightarrow <y,x> \in R \Rightarrow x \in rng(R)$.

    So, $dom(R^{-1}) \subset rng(R)$.

    Let $y \in rng(R)$. Then $\exists x$ such that $<x,y> \in R \Rightarrow <y,x> \in R^{-1} \Rightarrow y \in dom(R^{-1})$.

    So, $rng(R) \subset dom(R^{-1})$.

    Therefore, $rng(R)=dom(R^{-1})$.
  2. Let $y \in rng(R^{-1})$. Then $\exists x$ such that $<x,y> \in R^{-1} \Rightarrow <y,x> \in R \Rightarrow y \in dom R$.

    So, $rng(R^{-1}) \subset dom R$

    Let $y \in dom R$. Then $\exists y$ such that $<x,y> ain R \Rightarrow <y,x> \in R^{-1} \Rightarrow x \in rng(R^{-1})$

    So, $dom R \subset rng(R^{-1})$.

    Therefore, $dom R = rng(R^{-1})$.
  3. $$fld(R^{-1})=dom(R^{-1}) \cup rng (R^{-1})=rng(R) \cup dom R$$
    $$fld(R)=dom R \cup rng(R)$$

Is that what I have tried right or have I done something wrong? (Thinking)

Also, how could we prove the fourth identity $(R^{-1})^{-1}=R$ ? :confused:
 
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Number 1-3 are correct. Number 4 is even easier than 1 or 2.
 
Evgeny.Makarov said:
Number 1-3 are correct. Number 4 is even easier than 1 or 2.

Is it maybe like that? (Thinking)

Let $<y,x> \in (R^{-1})^{-1}$. Then: $<x,y> \in R^{-1} \Rightarrow <y,x> \in R$.

So, $(R^{-1})^{-1} \subset R$.

Let $<x,y> \in R$. Then: $<y,x> \in R^{-1} \Rightarrow <x,y> \in (R^{-1})^{-1}$.

So, $R \subset (R^{-1})^{-1}$.

Therefore, $(R^{-1})^{-1}=R$.

(Thinking)
 
Yes. I would write it as a chain of equivalences:
\[
\langle x,y\rangle\in R\iff \langle y,x\rangle\in R^{-1}\iff \langle x,y\rangle\in(R^{-1})^{-1}.
\]
 
Evgeny.Makarov said:
Yes. I would write it as a chain of equivalences:
\[
\langle x,y\rangle\in R\iff \langle y,x\rangle\in R^{-1}\iff \langle x,y\rangle\in(R^{-1})^{-1}.
\]

Nice! I understand! (Nod) Thanks a lot! (Smile)
 

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