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December 9th, 2013, 09:41 AM   #1
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A simple exercise in the number field

Hello, I have been working to understand mathematics on a rigorous level. The book that I have been using is Principles of Mathematical Analysis 3rd edition by Walter Rudin. I am currently working on an example from the first chapter dealing with field axioms. I am trying to prove that (-x)y=x(-y) only using the field axioms. Modeling off the proof that Rudin gave for (-x)y=-(xy) I started like this:

Proof that (- x)y=x(-y):
By the additive inverse axiom, for every element x in the field there exists an element -x in the field such that -x+x=0. Application yields:
(-x)y+(-(-x)y)=0. Now, by the law of distribution (-(-x)y)=x(-y)? This is where my problem is. How can (-x)y=x(-y) and simultaneously x(-y) is the additive inverse of (-x)y?

I hope that by shedding light on this I can complete the proof.
Thanks!
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December 12th, 2013, 08:57 AM   #2
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Re: A simple exercise in the number field

First show that . Since is by definition the additive inverse of i.e. and additive inverses are unique, this would imply that . Then you can use the result Rudin has already shown, namely .

To show , use the distributive law to factor out the .
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