Question

field proof A field consists of a set F, distinguished elements 0 ∈ F, 1 ∈ F, and functions + : F ×F → F , and · : F × F → F . We will write a + b for +(a, b) and ab for ·(a, b). These data are subject to the axioms:
i) ∀a,b,c∈F, (a+b)+c=a+(b+c) ii) ∀a∈F, 0+a=a
iii) ∀a∈F, ∃a ̃∈F, a ̃+a=0 iv) ∀a,b∈F, a+b=b+a
v) ∀a,b,c ∈ F, (ab)c = a(bc) vi) ∀a∈F, 1a=a
vii) ∀a∈F, ∼(a=0)⇒∃a′ ∈F,a′a=1 viii) ∀a, b ∈ F, ab = ba
ix) ∀a,b,c∈F, (a+b)c=ac+bc

Answer 1

Question 2
Prove the following results about fields; use only the axioms or a fact that you have already proved. Do not jump to conclusions based on what happens with the real numbers; note that division and subtraction have not been defined, and that you cannot divide by 1+1 since it could be zero. Each line of your argument should specify which axioms or previous results are used.
(1) If a+a=a then a=0.
(2) ∀a∈F, 0a=0
(3) If 0 = 1, then ∀a ∈ F,a = 0. Hence we usually assume that 0 ̸= 1; (this is sometimes
rephrased as “F has at least two elements”).
(4) Ifx∈F is an element such that∀a∈F,x+a=a then x=0
(5) Ifa∈F andx∈F is an element such that x+a=0,then x=a'
(6) ∀a∈A,a''=a
(7) ∀a∈1'a=a'

Answer 2

Is nor very complicated. For example for the first
(1) a+a=a then a=0
proof: \[a \in F \Rightarrow \exists -a \in F\]
then we have that, using i), ii) and iii)
\[(a+a)+(-a)=a+(a+(-a))=a+0=a=a+(-a)=0\]

Answer 3

@jarmvel but the subtraction has not been defined

Answer 4

Oh! sorry. I'm not using subtraction. I'm denoting the inverse of \[a\] by \[-a\]

Answer 5

additive inverse, of course. :)

Answer 6

ooo alright

Answer 7

is there special notation of additive inverse?