Recipe

Add two elements in $\mathbb{Z}_n$

Use when: you need $a+b$ inside $(\mathbb{Z}_n, +)$ , including finding an additive inverse or filling in one cell of a Cayley table

Goal: the single representative in $\{0,\ldots,n-1\}$ that $a+b$ actually equals, with the wraparound step never skipped

Quick recipe

1
Ignore $\mathbb{Z}_n$ for a moment and add $a+b$ the normal way, without any reduction yet.

Example

In $\mathbb{Z}_9$ : $6+4=10$ , computed exactly as ordinary addition.

Why: reduction only makes sense once the raw, unreduced sum is in hand.
2
Compare the raw sum from Step 1 to $n$ .
- the raw sum is already between $0$ and $n-1$ — it's already the final answer, no further step needed.
- the raw sum is $n$ or more (or negative) — it needs the wraparound step before it's a genuine element of $\mathbb{Z}_n$ .
Why: an unreduced sum like $10$ isn't wrong arithmetic, but it isn't one of $\mathbb{Z}_n$ 's actual elements either.
3
Subtract $n$ as many times as needed if the sum is too large, or add $n$ if it's negative, until the result sits in $\{0,\ldots,n-1\}$ .

Example

Continuing $6+4=10$ in $\mathbb{Z}_9$ : $10 \ge 9$ , so subtract once, $10-9=1$ , giving $6+4=1$ in $\mathbb{Z}_9$ .

Why: this wraparound step is exactly the division algorithm from the readiness review, applied to the raw sum with divisor $n$ .
4
Set the target sum to $0$ and solve for $x$ directly: $x=n-a$ works for any nonzero $a$ , since $a+(n-a)=n$ , which reduces to $0$ by the same wraparound step.

Example

Invert $5$ in $\mathbb{Z}_7$ : $x=7-5=2$ , and checking confirms $5+2=7\equiv0 \pmod 7$ .

Why: an inverse is the special case of Steps 1-3 where the target sum is the identity, $0$ , instead of some other value.

Common mistakes

Reporting the raw, unreduced sum (like $10$ in $\mathbb{Z}_9$ ) as the final answer.
Subtracting $n$ the wrong number of times, instead of just enough to land the result in $\{0,\ldots,n-1\}$ .
Forgetting the wraparound step entirely when the sum happens to already look like a reasonable small number.
Computing an inverse with ordinary subtraction, $-a$ , instead of $n-a$ , and getting a negative representative instead of one in range.

Add two elements in \mathbb{Z}_n