To prove that the relation $R$ on the set $A$ is complete if and only if $R \cup R^{-1} = A \times A$, we proceed as follows:

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Definitions:

1. Complete Relation: A relation $R$ on $A$ is complete if, for all $a, b \in A$, either $(a, b) \in R$ or $(b, a) \in R$ (or both).

2. Inverse Relation $R^{-1}$: $R^{-1}$ consists of all pairs $(b, a)$ such that $(a, b) \in R$.

3. Cartesian Product $A \times A$: This is the set of all possible ordered pairs $(a, b)$ where $a, b \in A$.

4. Union of Relations: $R \cup R^{-1}$ is the set of pairs that belong to $R$ or $R^{-1}$.

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Proof:

(⇒) If $R$ is complete, then $R \cup R^{-1} = A \times A$:

• Assume $R$ is complete. By definition, for every $a, b \in A$, at least one of $(a, b) \in R$ or $(b, a) \in R$ must hold.
• This implies that every pair $(a, b)$ or $(b, a)$ (or both) is covered by $R \cup R^{-1}$.
• Since $A \times A$ is the set of all pairs $(a, b)$ where $a, b \in A$, the above means $R \cup R^{-1} \supseteq A \times A$.
• Conversely, any pair $(a, b) \in A \times A$ must belong to either $R$ or $R^{-1}$, so $R \cup R^{-1} = A \times A$.

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(⇐) If $R \cup R^{-1} = A \times A$, then $R$ is complete:

• Assume $R \cup R^{-1} = A \times A$. By definition of union, for any $a, b \in A$, $(a, b) \in R \cup R^{-1}$.
• This implies that either $(a, b) \in R$ or $(a, b) \in R^{-1}$. By the definition of $R^{-1}$, $(a, b) \in R^{-1}$ means $(b, a) \in R$.
• Thus, for any $a, b \in A$, either $(a, b) \in R$ or $(b, a) \in R$, which is the definition of completeness.

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Conclusion:

We have shown that $R$ is complete if and only if $R \cup R^{-1} = A \times A$.

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Do you have any questions, or would you like further clarification?

Related Questions:

1. What is an example of a complete relation on a specific set $A$?
2. How can we visualize the union $R \cup R^{-1}$ in a graph representation of a relation?
3. What is the significance of completeness in practical applications of relations?
4. How does the concept of $R^{-1}$ relate to symmetry in relations?
5. What happens if $R \cup R^{-1} \neq A \times A$?

Tip: Always check the symmetry and coverage of a relation to quickly determine if it satisfies completeness.