In a field $F=\{0,1,x\}$,  $x + x = 1$ and $x\cdot x = 1$ Looking for some pointers on how to approach this problem:

Let $F$ be a field consisting of exactly three elements $0$, $1$, $x$. Prove that 
  $x + x = 1$ and that  $x  x = 1$.

 A: Hint 1: We know that $1x=x$ and $0x=0$. But $x$ must have a multiplicative inverse, since $x\neq 0$. So the multiplicative inverse has to be <fill in the blank>
Hint 2. Note that $1+x$ must be either $0$, $1$, or $x$. Can it be $1$? Then we would have $1+x=1$, which implies <fill in the blank>. Can it be $x$? What does that leave for $x+x$?
A: Suppose that $xx=x$. Then because $x\neq 0$, we can multiply both sides by $x^{-1}$ (whatever it may be) to get $x=1$. But this contradicts $x\neq 1$, so we cannot have $xx=x$.
Suppose that $xx=0$. Then because $F$ is a field, it is in particular an integral domain, so for any $a,b\in F$ we have $ab=0\implies a=0$ or $b=0$. Thus, $xx=0$ implies $x=0$, which contradicts $x\neq 0$. 
Thus, the only remaining possibility is that $xx=1$.
Similar reasoning can be used for $x+x$.
A: Write down the addition and multiplication tables. Much of them are known immediately from the properties of 0 and 1, and there's only one way to fill in the rest so that addition and multiplication by nonzero are both invertible.
