In a field $F=\{0,1,x\}$, how does $1 + 1 = x$? I understand that in a field with two elements $1 + 1 = 0$, but in a field with three I do not understand how $1 + 1 =x$.
The work I have done so far is:
\begin{align}
1 + 1 &= \{ 0 , 1 , x\}\\
1 + 1 &= 1
\end{align}
This cannot be true because of the additive inverse axiom,
     and I get $1 = 0$ which is a contradiction.   
I do not know what to do next.
 A: We're given that $F=\{0,1,x\}$ is a field.
As you already observed, we cannot have $1+1=1$.
Supposing for the sake of contradiction that we have $1+1=0$, then what is $x+1$?


*

*We can't have $x+1=x$, since then we'd get $1=0$

*We can't have $x+1=1$, since then we'd get $x=0$

*We can't have $x+1=0$, since then we'd get $x=1$
But of course $x+1$ is some element of $F$; this is a contradiction.
Therefore the only possibility is $1+1=x$.
A: Note that $\{0,1,x\}$ is a group under addition. By definition of a field, $0$ is the identity for this group. Since $1$ and $x$ are non-identity elements, they each must have order dividing $3$ (the order of the group, by Lagrange), and neither has order $1$. So, they both have order $3$.
In particular, neither has order TWO. So, $-1$, which must be an element of $\{0,1,x\}$, cannot be $0$ (since $0$ is its OWN additive inverse), and cannot be $1$ itself (since $1$ does not have order $2$). The only thing left is $x$:
$-1 = x$.
Now, since $1$ has order $3$:
$0 = 1 + 1 + 1$, so we see that $1+1 = -1$, that is: $1 + 1 = x$.
A: Either $1+1 = 0$ or $1 + 1 = 1$ or $1 + 1 = x$. The first equation would imply that the additive group generated by 1 has order 2, but by Lagrange's thoerem this is not possible. The second equation would imply that 1 is an additive identity, so it is also not possible. We conclude that $1 + 1 = x$.
A: It is very straight forward. If $1+1=1$ then cancelling $1$ you get $1=0$ (not happening). Next $1+1=0$ that means the order of $1$ is $2$ but the order of the field is $3$ and $2$ does not divide $3$. Therefore it must hold $1+1=x$. 
A: Any field with three elements is isomorphic to $\Bbb F_3$, which we can identify with the ring $\Bbb Z / \langle 3 \rangle$. (This is a special case of an important theorem about finite fields, which says that if two finite fields have the same order, then they are isomorphic.) In particular, given that $0$ and $1$ are respectively the additive and multiplicative inverses, $x$ must be the image of $2$ under the canonical quotient $\Bbb Z \to \Bbb Z / \langle 3 \rangle$, i.e., reduction modulo $3$. In particular, reducing both sides of $1 + 1 = 2$ modulo $3$ gives $1 + 1 = x$ as desired.
A: Suppose that we have a field $\{0,1,x\}$ in which $1 +1 = 0$.
Now, what is $1 + x$?  It can't be $1$ or $x$ (unique additive identity), so we must have $1 + x = 0$.
However, $1 + 1 = 0$.  So, $1 = x$.
