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October 4th, 2018, 07:39 PM   #71
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Originally Posted by JeffM1 View Post
Thus your definition of "neighborhood of zero" appears to contain every number that is included in the traditional understanding of real numbers. Is that correct?
Absolutely yes. I have only put the hat on the additive identity to mimic the style of notation I use for numbers in the other neighborhood.


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why do you claim that it is a strawman for me to say that your proposition 2 says that numbers in the neighborhood of R_0 result in real numbers when subtracted from $\hat \infty$
If you agree that real numbers can have the form "inf minus b" then I agree that
(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.
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October 4th, 2018, 11:15 PM   #72
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So how do you prove that a real number can have the form infinity - b ?
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October 5th, 2018, 02:25 AM   #73
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If you read what I wrote, you will see that I have divided one set into two sets, but if you divide one of them into two sets again then there will be three.
The boundary is a big problem for your definitions as it is not possible to divide R into two disjoint open sets.

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.
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October 5th, 2018, 02:44 AM   #74
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Originally Posted by studiot View Post
The boundary is a big problem for your definitions as it is not possible to divide R into two disjoint open sets.

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.
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.
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October 5th, 2018, 04:16 AM   #75
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Originally Posted by Micrm@ss View Post
So how do you prove that a real number can have the form infinity - b ?
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.
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October 5th, 2018, 04:21 AM   #76
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I don't get it.

Any real number has a decimal form, right?
So what is the "standard form" of infinity - 1?
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October 5th, 2018, 04:38 AM   #77
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Originally Posted by Micrm@ss View Post
I don't get it.

Any real number has a decimal form, right?
So what is the "standard form" of infinity - 1?
$-\hat{1}.\hat{0}\hat{0}\hat{0}\dots$
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October 5th, 2018, 05:00 AM   #78
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Originally Posted by Micrm@ss View Post
Any real number has a decimal form, right?
A real number is a cut in the real number line. All other definitions/requirements pertain to concepts other than whether or not something is a real number.
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October 5th, 2018, 05:18 AM   #79
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Originally Posted by sevensixtwo View Post
Absolutely yes. I have only put the hat on the additive identity to mimic the style of notation I use for numbers in the other neighborhood.




If you agree that real numbers can have the form "inf minus b" then I agree that
(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.
OK. That was polite. Thank you.

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 hyper-reals. 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 R-hat. For example, you will have to prove that

$x \in \mathbb R \text { and } y \in \mathbb R-hat \implies \dfrac{x}{y} = 0.$

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 non-Euclidian geometry does not make it false in Euclidian geometry.
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October 5th, 2018, 08:44 AM   #80
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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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