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March 18th, 2013, 12:27 PM   #1
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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.
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March 22nd, 2013, 01:38 PM   #2
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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 .
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March 22nd, 2013, 08:30 PM   #3
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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.
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March 23rd, 2013, 03:43 AM   #4
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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?
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March 23rd, 2013, 03:46 AM   #5
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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?
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March 23rd, 2013, 05:21 AM   #6
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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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