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]]>Please note that the 'ijk' part above is meant to be in subscript but not sure how to do that.

Thanks for any help.

]]>I think it can. Does this work?

Let b be the right inverse of a.

I have left out some associativity steps to make the proof simpler.

(ba)(ba) = b(ab)a = bea = ba

Let c be the right inverse of ba

(ba)(ba)c = (ba)c

(ba)e = e

ba = e

So b is a left inverse of a.

Note: Every other property of the group is required for this.

Bob

]]>https://en.wikipedia.org/wiki/Hippasus]]>

I agree with thickhead's reasoning and that the frequency and score should be switched for the problem to make sense.

If those two are switched, then the solution is one of those in the multiple-choice list.

Monox D. I-Fly has made an arithmetical error that accounts for his answer not appearing in the list.

]]>Remember the Cipolla formula has a much wider range, it must handle from 1 to infinity as best it can. The polynomial is expected to be better in the small range we fit for.

]]>Closure should be achievable, too, right? Let's say A o B = C, then A o C = B -> Whatever you do, you still land in the group. It's late at night here, so please someone correct me if I'm wrong, but it sounds plausbile in my head.]]>

http://www.mathisfunforum.com/search.ph … 1038166535

And there is a formula in terms of the coefficients.

But as I have been saying to evene and corrupted, it is much easier to do numerically.

]]>Thanks for the link.

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