My Math Forum Ring, proofs

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 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 $x^2=x$ , 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? $-ba=ba$ Is this because of cancellation law? if so then how come -ba = ba? Proof: $(a+b)^2=a+b$ implies $a^2+ab+ba+b^2=a+b$. Since $a^2=a$ and $b^2=b$ so $ab=-ba$ by the cancellation law. Now as $a+a=(a+a)^2=a^2+2a^2+a^2=a+a+a+a$ so $a+a=0 \ \forall a\in R$ and so each element in R is its own additive inverse. Hence $-ba=ba$ and so $ab=ba$.
 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! $ab= -ba$ 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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