To show that $I = \langle n \rangle = n\mathbb{Z}$ is a maximal ideal of $\mathbb{Z}$ if and only if $n$ is prime, we need to prove both directions of the "if and only if" (⇔) statement.

First Direction: If $n$ is prime, then $\langle n \rangle$ is a maximal ideal of $\mathbb{Z}$.

Proof:

1. Prime ideal definition: If $n$ is a prime number, then the ideal $\langle n \rangle$ consists of all multiples of $n$, i.e., $\langle n \rangle = \{ n\cdot k \mid k \in \mathbb{Z} \}$.

2. Maximal ideal definition: To show that $\langle n \rangle$ is maximal, we need to show that the only ideals of $\mathbb{Z}$ that contain $\langle n \rangle$ are either $\langle n \rangle$ itself or $\mathbb{Z}$.

3. Quotient ring: Consider the quotient ring $\mathbb{Z}/\langle n \rangle$. The elements of $\mathbb{Z}/\langle n \rangle$ are the equivalence classes of integers modulo $n$, which means $\mathbb{Z}/\langle n \rangle = \mathbb{Z}/n\mathbb{Z}$.

4. Structure of $\mathbb{Z}/n\mathbb{Z}$: If $n$ is prime, $\mathbb{Z}/n\mathbb{Z}$ is a field. This is a well-known result from number theory: the ring of integers modulo a prime number is a field.

5. Maximal ideal property: Since the quotient ring $\mathbb{Z}/n\mathbb{Z}$ is a field, the ideal ( \langle n \rangle \