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Oct 28, 2010 at 20:55 answer added Aaron Meyerowitz timeline score: 0
Oct 28, 2010 at 16:13 vote accept mmm
Oct 28, 2010 at 1:59 answer added Gerry Myerson timeline score: 6
Oct 28, 2010 at 0:08 answer added Gerhard Paseman timeline score: 0
Oct 27, 2010 at 23:54 answer added Tony Huynh timeline score: 1
Oct 27, 2010 at 23:49 answer added Nick S timeline score: 0
Oct 27, 2010 at 23:47 answer added Bugs Bunny timeline score: 0
Oct 27, 2010 at 21:43 history edited mmm CC BY-SA 2.5
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Oct 27, 2010 at 21:43 comment added mmm @Fiktor: true, thank you!...I'd delete my answers. The question still remains, though.
Oct 27, 2010 at 21:21 comment added Fiktor Your claim about "it's an arithmetical progression in the case A=B" is wrong. You can just write the terms of minimal $S$ containing 1 as polynomials in A, and note that for any $d$ there is only finite number of polynomials in $S$, whose degree is less than $d$. And in the case $A=2$, $B=1$ you get $S$ equal to the set of odd positive numbers, i.e. just one arithmetical progression.
Oct 27, 2010 at 20:49 history edited mmm CC BY-SA 2.5
added 142 characters in body; added 6 characters in body
Oct 27, 2010 at 20:47 comment added mmm @Yuan: yes, A and B are fixed. @Robin: yes, true. (Shall add this in the main post). Sorry about being unclear.
Oct 27, 2010 at 20:26 comment added Robin Chapman I can't figure out what you mean: $Ax+By$ whatever it is, is not a "binary operation". So does "$A=2$, $B=1$" mean that if $x$ and $y$ are elements of your set, then $2x+y$ is?
Oct 27, 2010 at 20:21 answer added ohai timeline score: 0
Oct 27, 2010 at 20:17 comment added Qiaochu Yuan Are A and B fixed?
Oct 27, 2010 at 20:15 history asked mmm CC BY-SA 2.5