To prove this statement, we need to establish two directions of the equivalence:

1. If $R/I$ is commutative, then $ab - ba \in I$ for all $a, b \in R$.
2. If $ab - ba \in I$ for all $a, b \in R$, then $R/I$ is commutative.

Proof:

1. If $R/I$ is commutative, then $ab - ba \in I$ for all $a, b \in R$.

Assume that the quotient ring $R/I$ is commutative. This means that for any $a, b \in R$, we have:

$(a + I)(b + I) = (b + I)(a + I)$

Expanding both sides:

$(a + I)(b + I) = ab + I \quad \text{and} \quad (b + I)(a + I) = ba + I$

Since $R/I$ is commutative, $ab + I = ba + I$. This implies that:

$ab - ba \in I$

Thus, if $R/I$ is commutative, we have $ab - ba \in I$ for all $a, b \in R$.

2. If $ab - ba \in I$ for all $a, b \in R$, then $R/I$ is commutative.

Now, assume that $ab - ba \in I$ for all $a, b \in R$. We want to show that the quotient ring $R/I$ is commutative, meaning for any $a, b \in R$, we have:

$(a + I)(b + I) = (b + I)(a + I)$

Expanding both sides again, we get:

$(a + I)(b + I) = ab + I \quad \text{and} \quad (b + I)(a + I) = ba + I$

Since by assumption $ab - ba \in I$, we can write $ab = ba + i$ for some $i \in I$. Thus, in $R/I$, we have:

$ab + I = ba + I$

Therefore, $(a + I)(b + I) = (b + I)(a + I)$, which shows that $R/I$ is commutative.

Conclusion:

The quotient ring $R/I$ is commutative if and only if $ab - ba \in I$ for all $a, b \in R$.

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Do you need further clarifications or details on any part of the proof? Here are some related questions to deepen understanding:

1. What is the definition of a quotient ring $R/I$?
2. How does the ideal $I$ affect the structure of the quotient ring?
3. Can $ab - ba \in I$ imply any other algebraic properties for the elements of $R$?
4. How does the commutativity of $R/I$ relate to other algebraic structures like fields or modules?
5. How can we generalize this result to non-commutative rings?

Tip: Always verify both directions when working with "if and only if" proofs to ensure completeness.