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I am self-studying the wonderful book, Elementary Geometry from an Advanced Standpoint.

In chapter 1, problem 19 it says: Suppose that the elements of R were 0 and 1, with addition and multiplication defined by these tables. Which of the postulates are true?

+  |  0  1       
0  |  0  1
1  |  1  0

.  |  0  1       
0  |  0  0
1  |  0  1

One of the postulates is this:

A-4 For every a in R there is exactly one number -a in R, called the negative of a, such that

a + (-a) = (-a) + a = 0

Obviously there are no negative numbers in the above tables. Thus I might argue that A-4 does not hold.

On the other hand, from some responses to other questions that I have read, in mathematics it is not the particular symbols that matter, but rather that the symbols (whatever they may be) exhibit the desired properties such as commutativity and associativity. Do I understand correctly? This is a real epiphany for me.

Thus, for each number in the above tables there is indeed a number that when added to the number yields 0:

0 + 0 = 0
1 + 1 = 0

Thus I will now argue that A-4 does hold.

Which is correct, A-4 holds or A-4 does not hold?

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There's no axiom saying that $-1 \ne 1$! – Zhen Lin Jul 25 '12 at 9:46
@Zhen Indeed, I showed Robert the field of $2$ elements in his prior question, to show that it is not a consequence of the field axioms that $\:1+1 \ne 0,\:$ or, equivalently, $\rm\:-1 \ne 1.\:$ Perhaps this is what sparked the current question. – Bill Dubuque Jul 25 '12 at 14:57
up vote 3 down vote accepted

As you noted, axiom $\rm\,A\!\!-\!\!4\,$ holds true. This axiom asserts that every element $\rm\,a\,$ of a field has an additive inverse $\rm\: -a,\:$ called negative $\rm a\:$ or minus $\rm a\:$ (just as reciprocal denotes multiplicative inverses). One easily checks that such inverses are unique, so this yields an operation $\rm\:a\to -a\:$ called "negation". This terminology, which denotes an inverse element or operation, should not be confused with the (related) use of "negative" to denote the property of being $< 0$ in a ordered field.

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You answered your own question--A-4 holds because 0+0=0 and 1+1=0. Your R is the field with 2 elements.

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