# Why is 1+1=0 in a finite field F={0,1}?

This table: $$\begin{array}{|c|cc|} \hline +& 0& 1\\ \hline 0& 0& 1\\ 1& 1& 0\\ \hline \end{array}$$ "feels" right, but how can you prove that $1+1=0$? What is the reason? I assume that due to $F \times F \rightarrow F$, the result of $1+1$ must be within the field F after all.

I'm looking for a logical explanation.

• You can't prove it, it's by definition. At best you can ask for the motivation behind this definition. – Git Gud Oct 31 '14 at 20:27
• For $\{0,1\}$ to be a group, $1$ must have an additive inverse $x$ for which $1+x = 0$. $x$ can't be $0$ because $1+0=1$, so $x$ must be $1$ . – MJD Oct 31 '14 at 20:27
• If $1+1=1$ (the only choice since you refuse to consider the possibility that $1+1=0$), then, on subtracting $1$ from both sides, we arrive at $1=0$ which is not permitted by those who insist that in a field the multiplicative identity $1$ must be different from the additive identity $0$. – Dilip Sarwate Oct 31 '14 at 20:29
• previous comments are right, but also to say that $0$ represents all even integers, and $1$ represents all odd ones (in the usual way where $F=\Bbb Z_2$ comes from). So, $1+1=2$, but $2=0$ in this model. – Mirko Oct 31 '14 at 20:34
• @Dilip: Even if you allow the null ring to be a field, you can't have $1=0$ once you have decided on $1+1\ne 0$ yet $1+1=1$. – Henning Makholm Oct 31 '14 at 22:13

1. There exists a unique element $0$ such that for every $x$, $x+0=x$.
2. For each element $x$ there exists a unique element $y$ such that $x+y=0$.
If $1+1=1$ it has to be the case that $1+0=0$. But in that case you just reversed the roles of $0$ and $1$, as dictated by the axioms. That's fine, and it still defines a group structure, but it's easier to just relabel them to the traditional roles, where $0$ is the additive unit.
So if $1+1\neq 1$, it has to be the case that $1+1=0$.