I have two questions, and I address them in two seperate paragraphs below.
I read that if we do not accept the Completeness Axiom for the reals and fields but accept all the other axioms for fields and the reals, then given any two-element field, whose elements are denoted by $0$ and $1$, we can show that the field does not have a supremum. I'm not sure how to prove this, but I believe it may be a consequence of the fact this two-element field cannot be totally ordered but I don't know how to then show that this field does not have a supremum; for, it seems to me that the supremum is simply $1$ since it is the smallest element such that all the other elements ($0$) in the field are smaller than it.
Also, I have another question, in my textbook, where it tries to verify that this set of reals will be field it assumes that $0+0=1+1=0$. Now, I can't simply understand how $1+1=0$ but I do believe that the author is trying to show, somehow, the inverse of $1$, but surely that would be $-1$? Now, if it were group theory, this might have made sense since $1+1=0$ for the set of integers in modulo $2$. After some speculation, I noticed that the author said, for fields, we define two operations to have the properties of inverses, identities, associative law Et Cetera. So, does it mean that for this field $1+1=0$ is defined this way?
(Note: By the other "axioms" I mean the distributive, associative, and commutative laws, the identity property and the inverse property. Also, the reason I am asking this question is to clarify a statement the author makes to show that without the completeness axioms some fields may not have a supremum: "we will see in Example 1.1.2 (the field wih elements $0$ and $1$) that properties A-H does not gurantee that every nonempty set that is bounded above has a supremum". And, the text I am quoting from is http://ramanujan.math.trinity.edu/wtrench/texts/TRENCH_REAL_ANALYSIS.PDF)