My Math Forum A simple exercise in the number field

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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 $xy+x(-y)=0$. Since $-xy$ is by definition the additive inverse of $xy$ – i.e. $xy+(-xy)=0$ – and additive inverses are unique, this would imply that $-xy=x(-y)$. Then you can use the result Rudin has already shown, namely $(-x)y=-xy$. To show $xy+x(-y)=0$, use the distributive law to factor out the $x$.

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