RE: The Mathematical Proof Thread
September 10, 2017 at 10:38 pm
(This post was last modified: September 10, 2017 at 10:46 pm by Kernel Sohcahtoa.)
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The Mathematical Proof Thread
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RE: The Mathematical Proof Thread
September 21, 2017 at 10:59 am
(This post was last modified: September 21, 2017 at 12:24 pm by Kernel Sohcahtoa.)
(September 21, 2017 at 5:28 am)pocaracas Wrote: Kernel, I think you're ready for this link: https://artofproblemsolving.com/communityYou have shown it to me before, and it was in relation to math proofs. The site looks pretty cool: I've certainly got a lot more to learn and could always improve. Out of curiosity, is the site being recommended due to poor/incorrect mathematical reasoning/proof-writing in my previous post? Thanks. P.S. In general, to all AF members, if I've made an error in a proof/mathematical work that I've posted here, then if it's okay with you, could you please let me know, so that I can post a corrected version? Thanks, and I appreciate everyone's participation in this thread. Live long and prosper.
Kernel's humility and diplomacy is breathtaking and astounding.
All I've got going for me is my ability to constantly speak my mind. Or perhaps rather: my inability to fail to constantly speak my mind
(September 21, 2017 at 1:52 pm)Kernel Sohcahtoa Wrote: In your post, you said that you have a physics background. Out of curiosity, are you more interested in applied mathematics or pure mathematics (or do you like both)? Thanks again for your response. I'd say I like both, but I prefer the applied variety. Give me 3 dimensions and I can still visualize the thing.... go higher and things become tricky.
When you analyse an algebraic division Ring, a set E with with 2 operations called '+' and '*' with the axioms of the algebraic goup (E.+) previously proven with commutivity a+b = b+a, the exitence of one element we called 0 where for all x in E: x+0 = x, with the set not being empty or singular;
Given the properties of the operation '*': 1: there exists a member we call u where for all set members a, a*u = a 2: for all members of E, a*b = b*a 3: for all a,b,c in E, a*(b+c) = a*b+a*c With these in mind prove that a*0=0 and while you're at it prove that 0 is not equal this 'u' mentioned on the axioms.
Also, this 'u' is commonly used in arithmetic as the number 1. The neutral set member in multiplication.
Our numeric system with the common + - * / operations forms a division ring. I prefer an analytical approach to math. |
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