March 18th, 2013, 12:27 PM  #1 
Newbie Joined: Feb 2013 Posts: 10 Thanks: 0  Ring, proofs
Prove that, if in ring R multiplies xy and yx are reversible, than also the elements x and y are reversible. Prove that, if all elements in ring satisfy the equation , then this ring is commutative. 
March 22nd, 2013, 01:38 PM  #2 
Senior Member Joined: Aug 2011 Posts: 149 Thanks: 0  Re: Ring, proofs
I found this proof, but could anyone explain this little more detailed? Is this because of cancellation law? if so then how come ba = ba? Proof: implies . Since and so by the cancellation law. Now as so and so each element in R is its own additive inverse. Hence and so . 
March 22nd, 2013, 08:30 PM  #3 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Ring, proofs
Nice! is because in a^2 + ab + ba + b^2 = a + b, a^2 = a and b^2 = b so a and b from both sides cancel outs, leaving ab + ba = 0, so ab = ba.

March 23rd, 2013, 03:43 AM  #4 
Senior Member Joined: Aug 2011 Posts: 149 Thanks: 0  Re: Ring, proofs
When I thought about what you said I got: (a + b) = a^2 +ab + ba + b^2 since (a + b)^2 = (a + b) => a + ab + ba + b for that to be true ab+ba = 0 => ab = ba. But how can ab = ba become ab = ba? I mean how/why does this proof get rid of minus sign? 
March 23rd, 2013, 03:46 AM  #5 
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Ring, proofs
It follows from the proof above that every element is it's own additive inverse. Haven't you wrote the proof you gave over there by yourself?

March 23rd, 2013, 05:21 AM  #6 
Senior Member Joined: Aug 2011 Posts: 149 Thanks: 0  Re: Ring, proofs
Nope, I got it from some wiki (not that first half of my sentence is "I found this proof,"). There were only theorem and proofs with no explanations. And that's why I brought it here to discuss.


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