Suppose the set consists of four elements: , and two operations and are defined on , with the following multiplication tables:
| + | 0 | 1 | x | y |
|---|---|---|---|---|
| 0 | 0 | 1 | x | y |
| 1 | 1 | 0 | y | x |
| x | x | y | 0 | 1 |
| y | y | x | 1 | 0 |
| 0 | 1 | x | y | |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 0 |
| 1 | 0 | 1 | x | y |
| x | 0 | x | y | 1 |
| y | 0 | y | 1 | x |
| Prove that is a field. (It would take too long to verify completely the associativity of and and the distributivity axiom. They happen to work in this case; please illustrate this fact by choosing an example for each of these properties. On the other hand, make sure you give a complete verification for the other axioms needed to prove that is a field.) |
Answer
The addition and multiplication are clearly closed based on the table.
- Commutativity of addition. Since the table is symmetric, it is commutative.
- Additive identity. 0 is the additive identity, can be seen from the first row and column.
- Associativity of addition. (0 + 1) + x = 1 + x = y = 0 + y = 0 + (1 + x)
- Additive inverse. Each element has an additive inverse since each row has a 0.
- Commutativity of multiplication. Since the table is symmetric, it is commutative.
- Associativity of multiplication. 1(xy) = 1(1) = 1 = xy = (1x)y
- Multiplicative identity. 1 is the multiplicative identity, can be seen from the second row and column.
- Distributivity. x(y + y) = x0 = 0 = 1 + 1 = xy + xy
- Multiplicative inverse. Each nonzero element has a multiplicative inverse since each nonzero row has a 1.