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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 $x x = 1$.

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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

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 . Can it be $x$? What does that leave for $x+x$?

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    @BrandonK. You don't have to do that much work: again, if you look at the addition table, every row and every column must contain each element exactly once. In other words, the set $\{x+a\mid a\in\{0,1,x\}\}$ has to be $\{0,1,x\}$. Since you already know that $x+0=x$ and that $x+1=0$, that means that $x+x$ has to be equal to $1$, because that's the only thing left.2012-07-09
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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 $x+x$.

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    The reasoning for $\,x + x\,$ is not so trivial, and not what I would call "similar" - try it.2012-07-08
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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.