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October 6th, 2018, 12:01 PM   #81
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Quote:
 Originally Posted by sevensixtwo 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.
Ok, so if $x$ is a real radius that simply means $x$ is a real number, correct?:

$$x \in (0,\infty) \implies \exists y \in \mathbb{R} \text{ such that } y > x$$

Quote:
 Originally Posted by sevensixtwo Cuts in the part beyond the radius are real numbers in the neighborhood of infinity.
Ok, so:

$$y > x \implies y \in \text{ 'the neighborhood of infinity'}$$
$$x = 1 \implies (1,\infty) = \text{ 'the neighborhood of infinity'}$$

Quote:
 Originally Posted by sevensixtwo I have defined these types of real numbers to have the form "infinity minus b."
This seems nonstandard. I believe the standard would be that $\infty - b = \infty \neq y \in (1,\infty)$.

Quote:
 Originally Posted by sevensixtwo Since it is known that one can make definitions, I have proven that a real number can have this form.
The real numbers are already defined. One can define a number in any fashion they see fit, but that doesn't make it a real number. Have you proven that $\infty - b \neq \infty$ and instead that $\infty - b \in \mathbb{R}$ or something? What am I missing...?

 January 11th, 2019, 07:16 PM #82 Senior Member     Joined: Jul 2012 From: DFW Area Posts: 642 Thanks: 99 Math Focus: Electrical Engineering Applications Unfortunately, Dr. Atiyah has passed away. NY Times article
January 12th, 2019, 10:18 AM   #83
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 Originally Posted by jks Unfortunately, Dr. Atiyah has passed away. NY Times article
That is so sad! I hope I can still be doing math until I'm 90!

 January 12th, 2019, 12:19 PM #84 Banned Camp   Joined: Jun 2010 Posts: 17 Thanks: 0 It should be born in mind that this proof is irrelevant as )Mathematics/science end in contradiction an integer= a non-integer. When mathematics/science end in contradiction it is proven in logic that you can prove anything you want in mathematics ie you can prove Fermat's last theorem and you can disprove Fermat's last theorem http://gamahucherpress.yellowgum.com...e-possible.pdf
January 12th, 2019, 02:06 PM   #85
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Quote:
 Originally Posted by bas It should be born in mind that this proof is irrelevant as )Mathematics/science end in contradiction an integer= a non-integer. When mathematics/science end in contradiction it is proven in logic that you can prove anything you want in mathematics ie you can prove Fermat's last theorem and you can disprove Fermat's last theorem http://gamahucherpress.yellowgum.com...e-possible.pdf
Can you tell me something? You claim to be Australia's leading erotic poet. Yet in the ten or so years I've been reading your mathematical expositions online, I've never seen you post any erotic poetry. Why is that?

January 12th, 2019, 09:21 PM   #86
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Quote:
 Originally Posted by bas . . . it is proven in logic that you can prove anything you want in mathematics
That would include proving that any of your mathematical assertions are mistaken.

January 13th, 2019, 09:39 AM   #87
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Quote:
 Originally Posted by bas It should be born in mind that this proof is irrelevant as )Mathematics/science end in contradiction an integer= a non-integer. When mathematics/science end in contradiction it is proven in logic that you can prove anything you want in mathematics ie you can prove Fermat's last theorem and you can disprove Fermat's last theorem http://gamahucherpress.yellowgum.com...e-possible.pdf
I'm guess I'm not surprised but haven't you found out that the decimal representation of a real number is not necessarily unique? There's nothing hard or abstract about this whole thing. That's one reason why Mathematicians tend to choose to use abstract symbols rather than decimals.

-Dan

January 13th, 2019, 10:03 AM   #88
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Quote:
 Originally Posted by topsquark I'm guess I'm not surprised but haven't you found out that the decimal representation of a real number is not necessarily unique?
Yes, any rational number not equal to $0$ that may be expressed as $\frac{n}{10^m}$, where $n$ is an integer and $m$ is a natural number, will have precisely two decimal representations. The rest of the real numbers will have only a single decimal representation.

The same goes for any base, as base 10 is just one of the infinite number of bases we could use when representing real numbers. E.g., if $B$ is the base, then rationals that may be expressed as $\frac{n}{B^m}$ will have precisely two representations in base $B$. Every other real number will have only a single representation in base $B$.

So now the question is, how does this ruin all of mathematics again? It seems like a very consistent and trivial thing to me. On the other hand, confusing notation with the notion of consistency to the point where "math is broken" is a ridiculously inconsistent position to take.

January 13th, 2019, 10:06 AM   #89
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Quote:
 Originally Posted by topsquark I'm guess I'm not surprised but haven't you found out that the decimal representation of a real number is not necessarily unique? There's nothing hard or abstract about this whole thing. That's one reason why Mathematicians tend to choose to use abstract symbols rather than decimals. -Dan
Dan

He has no clue that there is a difference between the symbol and the thing symbolized.

January 13th, 2019, 04:02 PM   #90
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Quote:
 Originally Posted by Maschke Can you tell me something? You claim to be Australia's leading erotic poet. Yet in the ten or so years I've been reading your mathematical expositions online, I've never seen you post any erotic poetry. Why is that?
Australia's leading erotic poet? There is such a thing? Is it an annual or lifetime award? Are the judges exclusively placental (I ask because I am not sure I trust the taste of monotremes in this area)? Can a prize be meaningful if logic does not exist. Many questions! More details are needed urgently.

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