We are given a relation $\sim$ defined on the set of integers $\mathbb{Z}$ by the rule: $a \sim b$ if and only if

$2a + b \equiv 0 \pmod{3}.$

We are asked to prove that this relation is surjective. This means that for every integer $b \in \mathbb{Z}$, there must be an integer $a \in \mathbb{Z}$ such that $a \sim b$, i.e.,

$2a + b \equiv 0 \pmod{3}.$

Step 1: Rewriting the condition

To make the relation clearer, we can rewrite the given condition for $a \sim b$ as:

$2a \equiv -b \pmod{3}.$

Since $-b \pmod{3}$ is just equivalent to $3 - b \pmod{3}$, this can be further rewritten as:

$2a \equiv -b \pmod{3} \quad \text{or equivalently} \quad 2a \equiv (3 - b) \pmod{3}.$

Step 2: Investigating possible values of $b \mod 3$

We will now investigate the three possible cases for $b \mod 3$, which are:

1. $b \equiv 0 \pmod{3}$,
2. $b \equiv 1 \pmod{3}$,
3. $b \equiv 2 \pmod{3}$.

Case 1: $b \equiv 0 \pmod{3}$

If $b \equiv 0 \pmod{3}$, then we need to solve:

$2a \equiv -0 \equiv 0 \pmod{3}.$

This means $2a \equiv 0 \pmod{3}$. Since $2$ and $3$ are coprime, the only solution is:

$a \equiv 0 \pmod{3}.$

Thus, for $b \equiv 0 \pmod{3}$, we can choose $a \equiv 0 \pmod{3}$ (i.e., $a$ is any multiple of 3).

Case 2: $b \equiv 1 \pmod{3}$

If $b \equiv 1 \pmod{3}$, we need to solve:

$2a \equiv -1 \equiv 2 \pmod{3}.$

This simplifies to $2a \equiv 2 \pmod{3}$. Dividing both sides by $2$ (since $2$ and $3$ are coprime), we get:

$a \equiv 1 \pmod{3}.$

Thus, for $b \equiv 1 \pmod{3}$, we can choose $a \equiv 1 \pmod{3}$ (i.e., $a$ is any number of the form $3k + 1$).

Case 3: $b \equiv 2 \pmod{3}$

If $b \equiv 2 \pmod{3}$, we need to solve:

$2a \equiv -2 \equiv 1 \pmod{3}.$

This simplifies to $2a \equiv 1 \pmod{3}$. Dividing both sides by $2$, we get:

$a \equiv 2 \pmod{3}.$

Thus, for $b \equiv 2 \pmod{3}$, we can choose $a \equiv 2 \pmod{3}$ (i.e., $a$ is any number of the form $3k + 2$).

Step 3: Conclusion

In each of the three cases, we have found an integer $a \in \mathbb{Z}$ such that $a \sim b$. Specifically:

• If $b \equiv 0 \pmod{3}$, choose $a \equiv 0 \pmod{3}$,
• If $b \equiv 1 \pmod{3}$, choose $a \equiv 1 \pmod{3}$,
• If $b \equiv 2 \pmod{3}$, choose $a \equiv 2 \pmod{3}$.

Thus, for every $b \in \mathbb{Z}$, there exists an $a \in \mathbb{Z}$ such that $a \sim b$. Therefore, the relation $\sim$ is surjective.

---

Would you like a more detailed explanation of any part of the solution? Here are some related questions to further explore:

1. How can we prove that the relation is also reflexive or symmetric?
2. What would it mean if the relation were injective as well as surjective?
3. How does the divisibility rule for 2 relate to modular arithmetic in this case?
4. Can you derive a more general form for a relation of the form $2a + b \equiv c \pmod{n}$?
5. How would the proof change if the modulus was a different prime number instead of 3?

Tip: When working with modular arithmetic, always check how each term in your congruence can be simplified before solving. This often leads to quicker solutions!