Basic field theory proof

elic307

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Oct 31, 2021
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Hi, can someone provide a proof to:
For any a,b in a field F, prove using field axioms only that
-ab = a(-b) = (-a)b
 
Hi, can someone provide a proof to:
For any a,b in a field F, prove using field axioms only that
-ab = a(-b) = (-a)b
Please show us what you have tried and exactly where you are stuck.

Please follow the rules of posting in this forum, as enunciated at:


Please share your work/thoughts about this problem.
 
Whether it is an automatic reply or not is not important. What seems important to me is that your post does not satisfy this rule from the referenced page:

Show your beginning work, or ask a specific question about the exercise, or explain why you're stuck.

If you don't even know where to start then say so, and someone might provide a hint.

Good luck!
 
Hi, can someone provide a proof to:
For any a,b in a field F, prove using field axioms only that
-ab = a(-b) = (-a)b
In the attached, look at the last theorem.
 

Attachments

  • Field.pdf
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In the attached, look at the last theorem.
Thanks a lot for your reply. Much appreciated.
I went over the document (first path :)) and it looks very clear.

Regarding theorem 2. Shouldn't it exclude the 0? That is,
theorem 2: For each a, not including 0, in the field, the -a is unique.
 
Thanks a lot for your reply. Much appreciated.
I went over the document (first path :)) and it looks very clear.

Regarding theorem 2. Shouldn't it exclude the 0? That is,
theorem 2: For each a, not including 0, in the field, the -a is unique.
What is not unique about 0 ?
 
What is not unique about 0 ?
I take it back.
At first I did not understand the meaning of unique.
I believe that unique in this context means that for each "a" there is only one item (-a) that fulfills the condition a + (-a) = 0
 
I take it back.
At first I did not understand the meaning of unique.
I believe that unique in this context means that for each "a" there is only one item (-a) that fulfills the condition a + (-a) = 0
Your definition sounds right to me. And it still applies to 0 as well as to any other a, doesn't it ?
 
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