Timeline for Classifying countable sets of weighted dots on a real line
Current License: CC BY-SA 3.0
20 events
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| Apr 13, 2017 at 12:19 | history | edited | CommunityBot |
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| Apr 14, 2016 at 11:11 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 23:03 | comment | added | Anixx | @Vidit Nanda I am thinking about a numerical system that would extend the real numbers, each class would correspond to an extended number. There would be rules on arithmetical operations, for instance, multiplication goes as follows: put two real axes perpendicular to each other with sets of the classes you intend to multiply and draw the lines parallel to the axes over all the dots. Find the intersection dots on the plane and put them on a new real axis in order according to the greatest coordinate, multiplying weights in process. The class of this new set will be the product. | |
| Apr 11, 2016 at 16:11 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 16:00 | history | edited | Andrés E. Caicedo |
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| Apr 11, 2016 at 15:20 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 15:18 | comment | added | Anixx | @Vidit Nanda you can move all the negative dots to the symmetric positive positions (where they would coincide with existing dots, you can sum up the weights), you can move the dot from zero (and other finite amount of dots ) freely wherever you want it. | |
| Apr 11, 2016 at 14:29 | comment | added | Vidit Nanda | Also, just to see if I get the rules: if we have "all positive integers, zero and all halved negative integers", how would you find an equivalent representative with only positive dots? Let's say all weights are 1. | |
| Apr 11, 2016 at 14:25 | comment | added | Vidit Nanda | How intriguing. While we think about it, could you please explain briefly why this gadget is of interest to you? | |
| Apr 11, 2016 at 13:47 | comment | added | Anixx | @Yaakov Baruch I have added a rule for accumulation points, second from the end. | |
| Apr 11, 2016 at 13:46 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 12:23 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 12:12 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 12:10 | history | undeleted | Anixx | ||
| Apr 11, 2016 at 12:01 | history | deleted | Anixx | via Vote | |
| Apr 11, 2016 at 11:46 | comment | added | Anixx | @Yaakov Baruch what about cases without accumulation point? | |
| Apr 11, 2016 at 11:39 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 11:33 | history | edited | Anixx | CC BY-SA 3.0 |
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| Apr 11, 2016 at 10:55 | comment | added | Yaakov Baruch | If the set has an accumulation point, it seems that applying the rules a finite number of times, doesn't allow undoing that; so that the set will not be equivalent to one based on on the integers. | |
| Apr 11, 2016 at 10:20 | history | asked | Anixx | CC BY-SA 3.0 |