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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    $\begingroup$ You can't prove it, it's by definition. At best you can ask for the motivation behind this definition. $\endgroup$ – Git Gud Oct 31 '14 at 20:27
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    $\begingroup$ 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 $1$ . $\endgroup$ – MJD Oct 31 '14 at 20:27
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    $\begingroup$ 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 $0$. $\endgroup$ – Dilip Sarwate Oct 31 '14 at 20:29
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    $\begingroup$ 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. $\endgroup$ – Mirko Oct 31 '14 at 20:34
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    $\begingroup$ @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 $1+1=1$. $\endgroup$ – hmakholm left over Monica Oct 31 '14 at 22:13

Recall the two axioms about addition:

  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 $1+1=0$.

  • $\begingroup$ Ah yes. I should have re-looked-up these two axioms our prof gave us. Thank you! $\endgroup$ – user1511417 Oct 31 '14 at 20:47

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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