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## Verify the set with addition and multiplication is a field

Solution to Linear Algebra Hoffman & Kunze Chapter 1.2 Exercise 1.2.5

Solution: We must check the nine conditions on pages 1-2:

1. An operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right. This is true for the addition table so addition is commutative.
2. There are eight cases. But if $x=y=z=0$ or $x=y=z=1$ then it is obvious. So there are six non-trivial cases. If there's exactly one $1$ and two $0$'s then both sides equal $1$. If there are exactly two $1$'s and one $0$ then both sides equal $0$. So addition is associative.
3. By inspection of the addition table, the element called $0$ indeed acts like a zero, it has no effect when added to another element.
4. $1+1=0$ so the additive inverse of $1$ is $1$. And $0+0=0$ so the additive inverse of $0$ is $0$. In other words $-1=1$ and $-0=0$. So every element has an additive inverse.
5. As stated in 1, an operation is commutative if the table is symmetric across the diagonal that goes from the top left to the bottom right. This is true for the multiplication table so multiplication is commutative.
6. As with addition, there are eight cases. If $x=y=z=1$ then it is obvious. Otherwise at least one of $x$, $y$ or $z$ must equal $0$. In this case both $x(yz)$ and $(xy)z$ equal zero. Thus multiplication is associative.
7. By inspection of the multiplication table, the element called $1$ indeed acts like a one, it has no effect when multiplied to another element.
8. There is only one non-zero element, $1$. And $1\cdot1=1$. So $1$ has a multiplicative inverse. In other words $1^{-1}=1$.
9. There are eight cases. If $x=0$ then clearly both sides equal zero. That takes care of four cases. If all three $x=y=z=1$ then it is obvious. So we are down to three cases. If $x=1$ and $y=z=0$ then both sides are zero. So we're down to the two cases where $x=1$ and one of $y$ or $z$ equals $1$ and the other equals $0$. In this case both sides equal $1$. So $x(y+z)=(x+y)z$ in all eight cases.