[proofplan] We prove equality by double inclusion. First, every element of $I$ annihilates every coset in $R/I$ because ideals are closed under multiplication by elements of $R$. Conversely, if an element of $R$ annihilates all of $R/I$, then applying it to the coset $1_R+I$ forces the element itself to lie in $I$. [/proofplan] [step:Show that every element of $I$ annihilates every coset in $R/I$] Let $r \in I$. To prove $r \in \operatorname{Ann}_R(R/I)$, let $a+I \in R/I$ be an arbitrary coset, with $a \in R$. By the $R$-module structure on $R/I$, \begin{align*} r \cdot (a+I)=ra+I. \end{align*} Since $I$ is an ideal of the commutative ring $R$ and $r \in I$, we have $ra \in I$. Therefore $ra+I=I$, where $I$ is the zero element of the quotient module $R/I$. Hence \begin{align*} r \cdot (a+I)=I=0_{R/I}. \end{align*} Because $a+I$ was arbitrary, $r$ annihilates every element of $R/I$. Thus \begin{align*} I \subset \operatorname{Ann}_R(R/I). \end{align*} [guided] We want to show that an arbitrary element of $I$ acts as zero on the whole quotient module. Let $r \in I$, and let $a+I \in R/I$ be any coset, where $a \in R$. The given $R$-module action on the quotient is multiplication followed by passage to the quotient: \begin{align*} r \cdot (a+I)=ra+I. \end{align*} The ideal property is exactly what is needed here. Since $r \in I$ and $a \in R$, closure of an ideal under multiplication by arbitrary ring elements gives $ra \in I$. A coset $b+I$ is the zero coset precisely when $b \in I$, so $ra+I=I$. Therefore \begin{align*} r \cdot (a+I)=I=0_{R/I}. \end{align*} Since the coset $a+I$ was arbitrary, $r$ annihilates every element of $R/I$. Thus $r \in \operatorname{Ann}_R(R/I)$, and hence \begin{align*} I \subset \operatorname{Ann}_R(R/I). \end{align*} [/guided] [/step] [step:Test an annihilator element on the coset of the identity] Let $r \in \operatorname{Ann}_R(R/I)$. By definition of the annihilator, $r$ acts as zero on every element of the $R$-module $R/I$. In particular, applying this to the coset $1_R+I \in R/I$ gives \begin{align*} r \cdot (1_R+I)=I. \end{align*} Using the quotient module action and the identity property in $R$, \begin{align*} r \cdot (1_R+I)=r1_R+I=r+I. \end{align*} Therefore $r+I=I$, which is equivalent to $r \in I$. Hence \begin{align*} \operatorname{Ann}_R(R/I) \subset I. \end{align*} [/step] [step:Conclude equality by double inclusion] The first inclusion gives \begin{align*} I \subset \operatorname{Ann}_R(R/I), \end{align*} and the second inclusion gives \begin{align*} \operatorname{Ann}_R(R/I) \subset I. \end{align*} Therefore the two subsets of $R$ are equal: \begin{align*} \operatorname{Ann}_R(R/I)=I. \end{align*} This proves the theorem. [/step]