well x * x^-1 = e by definition so x^3 * x^-1 = e by substitution so x^2 = e so x = x^-1 I think that's all you need; R is just {e}. (Assuming I remember any abstract algebra, mind you)
You make it sound so easy! Unfortunately the existence of x^(-1) is only guaranteed in a field. In fact field's are commutative rings by definition, and so the result I'm looking for would be trivial.
It turns out that no one else in class got the answer. At least that's what I hear, from the more knowledgable people.
Also my ol friend Noah Simon sent me a link with some ol proofs.
Hey! You are posting on Livejournal again! Neato. For some reason, the filter I normally read wasn't letting your posts through, but I'll see to that shortly.
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so x^3 * x^-1 = e by substitution
so x^2 = e
so x = x^-1
I think that's all you need; R is just {e}. (Assuming I remember any abstract algebra, mind you)
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It turns out that no one else in class got the answer. At least that's what I hear, from the more knowledgable people.
Also my ol friend Noah Simon sent me a link with some ol proofs.
Thanks anyway though!
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Maybe I'll even start reading other livejournals again! No promises though ;-)
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