Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
 2 years ago
 2 years 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
 2 years ago
 2 years ago

This Question is Closed

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'
 2 years 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) \[(a+a)+(a)=a+(a+(a))=a+0=a=a+(a)=0\]
 2 years ago

math_proof Group TitleBest ResponseYou've already chosen the best response.0
@jarmvel but the subtraction has not been defined
 2 years 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 \[a\]
 2 years ago

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

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

math_proof Group TitleBest ResponseYou've already chosen the best response.0
is there special notation of additive inverse?
 2 years ago
See more questions >>>
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.