To prove $R \circ I_A = R$, let us proceed step-by-step, where:

• $A$ and $B$ are sets.
• $R$ is a relation from $A$ to $B$, i.e., $R \subseteq A \times B$.
• $I_A$ is the identity relation on $A$, defined as $I_A = \{ (a, a) \mid a \in A \}$.

We aim to show $R \circ I_A = R$.

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Definition of Composition of Relations

If $S$ and $T$ are two relations, then their composition $S \circ T$ is defined as: $S \circ T = \{ (x, z) \mid \exists y \text{ such that } (x, y) \in T \text{ and } (y, z) \in S \}.$

Substituting $R$ and $I_A$

Using the definition, $R \circ I_A$ can be written as: $R \circ I_A = \{ (a, b) \mid \exists x \text{ such that } (a, x) \in I_A \text{ and } (x, b) \in R \}.$

Analysis of $I_A$

The identity relation $I_A$ only relates elements of $A$ to themselves: $(a, x) \in I_A \implies x = a.$

Substitute $x = a$

From the definition of $R \circ I_A$, $(a, b) \in R \circ I_A$ if and only if: $\exists x \text{ such that } (a, x) \in I_A \text{ and } (x, b) \in R.$ Since $(a, x) \in I_A$ implies $x = a$, the condition simplifies to: $(a, b) \in R \circ I_A \iff (a, b) \in R.$

Conclusion

Since every element $(a, b)$ in $R \circ I_A$ corresponds exactly to an element $(a, b) \in R$, we conclude: $R \circ I_A = R.$

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Summary

The composition of $R$ with the identity relation $I_A$ results in $R$ itself, because $I_A$ does not modify any elements in $R$ when composed.

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Would you like further details or another example for clarity?

Here are some related questions:

1. How does this proof extend to $I_B \circ R$?
2. What happens if $R$ is not a relation between $A$ and $B$?
3. Can this property be applied to functions as relations?
4. How does the composition behave for non-identity relations?
5. What is the role of $I_A$ in preserving the structure of $R$?

Tip: Identity relations are foundational in many proofs involving relations and functions; understanding their composition simplifies many concepts in set theory and algebra.