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December 9th, 2013, 09:41 AM  #1 
Newbie Joined: Dec 2013 Posts: 1 Thanks: 0  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! 
December 12th, 2013, 08:57 AM  #2 
Member Joined: Mar 2013 Posts: 90 Thanks: 0  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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