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.
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$\begingroup$Recall the two axioms about addition:
- There exists a unique element $0$ such that for every $x$, $x+0=x$.
- 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 $1+1=0$.
$\endgroup$ 1 $\begingroup$As a Finite Field, the distribute law must hold: The distributive law holds: (a+b) · c = a · c+b · c for all a,b, c ∈ F. You can chose 1 or 0.
Now, consider Boolean Logic. Using the Or operator, then 1 is the correct answer. 0,1,1,1 If Exclusive Or, then 0,1,1,0 (as you have in your table)
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