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Timeline for Prove that a Number is Algebraic

Current License: CC BY-SA 4.0

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Jun 11, 2024 at 16:59 history edited mbomb007 CC BY-SA 4.0
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Jun 17, 2020 at 9:04 history edited Community Bot
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Oct 28, 2016 at 14:03 vote accept mbomb007
Jun 22, 2016 at 20:04 comment added mbomb007 @AndersKaseorg Multiplying by negatives is fine. Not the others.
Jun 22, 2016 at 19:15 comment added Anders Kaseorg @mbomb007 Do you consider +-3, ++3, --3, -+3, *-3, *+3 to be valid output terms (adding/subtracting/multiplying the integer −3 or +3)?
Jun 22, 2016 at 18:49 comment added Anders Kaseorg @mbomb007 Division is not required. You can always move divisions to the end (/a +b+(a*b) /a, /a -b-(a*b) /a, /a *b*b /a, /a *x*x /a) and then delete them.
Jun 22, 2016 at 15:29 answer added LLlAMnYP timeline score: 1
Jun 22, 2016 at 14:31 history edited mbomb007 CC BY-SA 3.0
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Jun 21, 2016 at 22:02 history tweeted twitter.com/StackCodeGolf/status/745376381073227777
Jun 21, 2016 at 21:42 answer added Anders Kaseorg timeline score: 4
Jun 21, 2016 at 21:09 history edited mbomb007 CC BY-SA 3.0
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Jun 21, 2016 at 20:57 comment added mbomb007 It's a solution to the equation x^4-10*x^2+1. See WolframAlpha
Jun 21, 2016 at 20:52 comment added Mario Ishac @mbomb007 Whoops, my apologies, didn't catch that in the OP.
Jun 21, 2016 at 20:48 comment added Mario Ishac How is √2+√3 algebraic? If you multiply the number by itself you get 5 + 2√6... unless I am missing something you can never force out the radical.
Jun 21, 2016 at 20:47 comment added mbomb007 @TimmyD The answer needs to be exact, such that I could perform the operations and get zero. View the examples provided. There is no floating point arithmetic.
Jun 21, 2016 at 20:44 comment added AdmBorkBork How close to 0 do the results need to be? Given rounding errors and float precision, I could easily see problematic situations ...
Jun 21, 2016 at 20:25 history asked mbomb007 CC BY-SA 3.0
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