```Date: Jan 10, 2013 5:48 AM
Author: William Elliot
Subject: Re: Division without the axiom of choice

On Thu, 10 Jan 2013, pepstein5@gmail.com wrote:> Let A and B be sets.  Assume ZF without assuming choice.  Then, for all > non negative integers n, I believe (correct me if I'm wrong) that, if n > x A is equipotent to n x B, then A is equipotent to B.It's false for n = 0.If A is infinite, the so is B andA equipotent n x A equipotent n x B equipotent BIf A finite, then so is B and for n > 0, if n|A| = n|B| in N.then |A| = |B|.> There's a famous Conway/Doyle paper which proves this for n = 2 and n = > 3.What do you mean this for n = 2 and n = 3, when this is about all n >= 0?> However, it doesn't seem rigorous or clear and I have trouble > understanding it.I don't uunderstand what you're talking about.> Does anyone know a more axiomatic treatment?  (I don't have access to a > university, and I'm not in the market for maths purchases, so only free > references would be helpful.)> How about an explicit statement of "this for n = 2 and n = 3"?
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