October 4th, 2018, 07:39 PM  #71  
Newbie Joined: Aug 2013 Posts: 13 Thanks: 0  Quote:
Quote:
(infinity  b)  (infinity  b') = b'  b , but otherwise there is no number X in neighborhood of the origin such that (infinity  b)  b' = X . Indeed, (infinity  b)  b' = infinity  (b + b') conforms exactly to the definition of a real number in the neighborhood of infinity. I claim that you used a straw man because I believe you incorrectly paraphrased me and then made a counterargument to your own paraphrasing rather than to what I wrote. However, if you do concede that real numbers can have the form "inf minus b" then I will be happy to have misunderstood your post. Last edited by skipjack; October 5th, 2018 at 12:15 AM.  
October 4th, 2018, 11:15 PM  #72 
Senior Member Joined: Oct 2009 Posts: 783 Thanks: 280 
So how do you prove that a real number can have the form infinity  b ?

October 5th, 2018, 02:25 AM  #73  
Senior Member Joined: Jun 2015 From: England Posts: 915 Thanks: 271  Quote:
And a neighborhood is an open set. You can do it with a closed set and an open set, but then the closed set is not a neighborhood. Last edited by skipjack; October 5th, 2018 at 03:24 AM.  
October 5th, 2018, 02:44 AM  #74 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  There are different conventions here. The one I see most often is: an open neighbourhood of x is an open set containing x, while a neighbourhood of x is any set containing an open neighbourhood of x.
Last edited by skipjack; October 5th, 2018 at 03:25 AM. 
October 5th, 2018, 04:16 AM  #75 
Newbie Joined: Aug 2013 Posts: 13 Thanks: 0  Consider the definition of a real number: it is a cut in the real number line. For any real radius around the origin, the real number line leaves that radius. Cuts in the part beyond the radius are real numbers in the neighborhood of infinity. I have defined these types of real numbers to have the form "infinity minus b." Since it is known that one can make definitions, I have proven that a real number can have this form.

October 5th, 2018, 04:21 AM  #76 
Senior Member Joined: Oct 2009 Posts: 783 Thanks: 280 
I don't get it. Any real number has a decimal form, right? So what is the "standard form" of infinity  1? 
October 5th, 2018, 04:38 AM  #77 
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 311 Thanks: 109 Math Focus: Number Theory, Algebraic Geometry  
October 5th, 2018, 05:00 AM  #78 
Newbie Joined: Aug 2013 Posts: 13 Thanks: 0  
October 5th, 2018, 05:18 AM  #79  
Senior Member Joined: May 2016 From: USA Posts: 1,310 Thanks: 551  Quote:
I am now going to make some meta comments. I do not have the knowledge to say when a different set of axioms about the real numbers results in a consistent system. Most people believe that Robinson created an alternative, but logically consistent, system, namely that of the hyperreals. I myself have not tried to confirm that assessment. So, be my guest: develop the consequences derivable from a new axiomatic scheme and demonstrate that it is logically consistent system. But it is disingenuous to use the language of traditional mathematics about a different system. Your "real" numbers are not what others mean by real numbers. In that system, no real number subtracted from infinity results in a real number. You are free to develop your own mathematics, but you are not free to imply that others must accept your axioms or that your vocabulary supplants that of the broader community. Words like "neighborhood" and "real number" have generally accepted meanings. When you use them with a different meaning, people will think that you are confused or deceptive or both. So I now retract my comment about the hats. Wherever you are dealing with concepts that differ from standard definitions, use hats or entirely different symbols. Moreover, you cannot use any of the results from traditional mathematics about real numbers unless you prove that they obtain under your revised axioms. Obviously, if $\mathbb R_0$ is the same as what is generally called $\mathbb R$, you can use all the traditional results with respect to numbers in that set. But you cannot just assume that those results apply to any number in Rhat. For example, you will have to prove that $x \in \mathbb R \text { and } y \in \mathbb Rhat \implies \dfrac{x}{y} = 0.$ Finally, if you disprove the Riemann Hypothesis in your system, that entails nothing about the hypothesis in the traditional system, which is what most mathematicians are interested in. The fact that something is false in nonEuclidian geometry does not make it false in Euclidian geometry.  
October 5th, 2018, 08:44 AM  #80 
Newbie Joined: Aug 2013 Posts: 13 Thanks: 0 
I proved the thing you said I need to prove on page five of this thread. I didn't disprove RH in a new arena, I disproved it in a neglected region of the normal arena: C.


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