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math_proof
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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
 one year ago
 one year ago
math_proof Group Title
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
 one year ago
 one year ago

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math_proof Group TitleBest ResponseYou've already chosen the best response.0
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'
 one year ago

jarmvel Group TitleBest ResponseYou've already chosen the best response.0
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) \[(a+a)+(a)=a+(a+(a))=a+0=a=a+(a)=0\]
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
@jarmvel but the subtraction has not been defined
 one year ago

jarmvel Group TitleBest ResponseYou've already chosen the best response.0
Oh! sorry. I'm not using subtraction. I'm denoting the inverse of \[a\] by \[a\]
 one year ago

jarmvel Group TitleBest ResponseYou've already chosen the best response.0
additive inverse, of course. :)
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
ooo alright
 one year ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
is there special notation of additive inverse?
 one year ago
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