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title: Arithmetic modulo 3

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# Arithmetic modulo 3

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In the last lesson we learned about [Arithmetic modulo 2](https://hackmd.io/@alexhkurz/ByKQ3EGiU).

Now, what about fields with 3 elements? We know that we will have elements $0,1$ anyway. So let us call the third element $2$. But keep in mind that for now $2$ is just a name. It might be safer to call it something else, such as $\alpha$ (pronounced "alpha").

**Activity:** Fill in the following tables.


|+|0| 1| 2| 
|:---:|:---:|:---:|:---:|
| 0 | | | |
| 1 | | | |
| 2 | | | |

and

|$\cdot$|0| 1| 2|
|:---:|:---:|:---:|:---:|
| 0 | | | |
| 1 | | | |
| 2 | | | |

**Hints:** We proceeded by first filling in everything that follows from $0$ being the neutral element for addition and $1$ being the neutral element for multiplication. Then we guessed that $1+1=2$ (recall that $2$ for now is just the name of the third element, we don't know yet how to compute with it). After that, we saw that the rest of the tables is determined.

**Homework:** Write out the reasoning that showed that after defining $1+1=2$ all the operations on numbers are determined.

**Homework:** What happens if we define $1+1=0$? Is that compatible with the axioms?

**Homework:** Can you write out tables for arithmetic modulo 4?

## Takeaway

If you are stuck just try all possibilities. **Make case distinctions. Derive contradictions. Backtrack.** If you are lucky, as in the example above, you will find that, after all, there was only possibility.

For example, in the situation above, we expect that we must have $1+1=2$, but before we can assert this we need to show that $1+1=0$ is impossible. We do this by assuming that $1+1=0$ and deriving a contradiction:

- First, we compute $2+2=0$ using the axioms.
- It only remains to determine $1+2$. There are three possibilities:
   - If $1+2=0$, then $2=1$, which is a contradiction.
   - If $1+2=1$, then $2=0$, which is a contradiction.
   - If $1+2=2$, then $1=0$, which is a contradiction.

It follows that $1+1=0$ was impossible after all. So we proved $1+1=0$ (after having checked that the two tables then satisfy the axioms).

**Exercise:** Make sure that you understand how to derive the contradictions above from the axioms. We will need this in what follows.