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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

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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'
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) \[(a+a)+(-a)=a+(a+(-a))=a+0=a=a+(-a)=0\]
@jarmvel but the subtraction has not been defined

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Other answers:

Oh! sorry. I'm not using subtraction. I'm denoting the inverse of \[a\] by \[-a\]
additive inverse, of course. :)
ooo alright
is there special notation of additive inverse?

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