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May 8th, 2017, 06:37 PM   #1
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If you remove the decimal from [ 0 - 1 ) what does that set match?

Quote:
Originally Posted by AplanisTophet View Post
$000 \in \sum^3$ whereas $0 \in \sum^1$.

That's ok for now. We're just establishing our alphabet. Currently, we have that $\mathbb{N} \subset \sum^*$. Agreed?



I agree, but that doesn't prove you wrong overall and I agreed to be objective.

I suggest that we focus only on the infinite decimal expansions on the interval [0, 1). These are numbers like 1/3 (non-terminating decimals), but not numbers like 1/2 = 0.5 (terminating decimals). The reason I suggest this is because we already know that the terminating decimals are countable. We need only try to enumerate the non-terminating decimals to accomplish your goal.

You want to 'flip' 0.333... across a decimal point and get ...333, yes? Would it be ok for us to just drop the "0." from 0.333... so as to get 333... please? We can call this function $f$ and then have, as examples:

$f(1/3) = 333... \in \sum^{\omega}$
$f(\pi - 3) = 14159... \in \sum^{\omega}$


More specifically:

$\{ f( \,x) \, : x \in [ \,0, 1) \, \land x \text{ is a non-terminating decimal} \} \subset \sum^{\omega}$

Agreed so far?
That's interesting. I didn't know you could just make a function that flips digits. Computationally speaking, yes. I understand it. I just didn't know that was allowed.

I have to be honest. I don't know what a lot of your symbols mean. But I guess
that N and the U shape opened to the right means that N is a subset or is in the set of Sigma*. Then yes. I agree. // oh look at that, if I actually read what I copied and pasted I'd know it said "subset"

$\{ f( \,x) \, : x \in [ \,0, 1) \, \land x \text{ is a non-terminating decimal} \} \subset \sum^{\omega}$

This looks like it's applying the function to the set and that $\land$ thing must be an "if condition" of some sort filtering the set and defining what's left as $\sum^{\omega}$

So yeah. I agree
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May 8th, 2017, 07:01 PM   #2
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The caret is "and". Personally, I think it's better to write words than colons, carets and similar.
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May 8th, 2017, 08:40 PM   #3
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Quote:
Originally Posted by InkSprite View Post
That's interesting. I didn't know you could just make a function that flips digits. Computationally speaking, yes. I understand it. I just didn't know that was allowed.
As long as your function is formally defined, anything is allowed. You are only limited by your imagination.


Quote:
Originally Posted by InkSprite View Post
I have to be honest. I don't know what a lot of your symbols mean. But I guess that N and the U shape opened to the right means that N is a subset or is in the set of Sigma*. Then yes. I agree. // oh look at that, if I actually read what I copied and pasted I'd know it said "subset"

$\{ f( \,x) \, : x \in [ \,0, 1) \, \land x \text{ is a non-terminating decimal} \} \subset \sum^{\omega}$

This looks like it's applying the function to the set and that $\land$ thing must be an "if condition" of some sort filtering the set and defining what's left as $\sum^{\omega}$

So yeah. I agree
Yes, $\subset$ means “subset.”

In English, $\{ f(x) : x \in [ \,0,1) \, \land x \text{ is a non-terminating decimal} \} \subset \sum^{\omega}$ means “(the set of all $f(x)$ such that x is in [0, 1) and x is a non-terminating decimal) is a subset of (the set of all infinite sequences over the alphabet $\sum$).” Since that is a mouthful, we can refer to this set as $A$ so long as we define it as such:

Let $A = \{ f(x) : x \in [ \,0,1) \, \land x \text{ is a non-terminating decimal} \}$ (now, going forward in our proof, we can simply refer to the set $A$).

The purpose of this was to allow you to formalize what you meant by the term “infinite integer.” Namely, the infinite integers are a precisely defined subset of $\sum^{\omega}$, so you can now refer to the elements of $A$ as infinite integers if you like, and people will understand precisely what you mean:

Let the set of all infinite integers $= A = \{ f(x) : x \in [ \,0,1) \, \land x \text{ is a non-terminating decimal} \}$.

So now, you have that $\mathbb{N} \subset \sum^*$ and $A \subset \sum^{\omega}$. It is your self-proclaimed goal to disprove Cantor’s Theorem, so formally, if you can find a surjective mapping from $\mathbb{N}$ onto $A$, then you would be successful. This is because you've already shown that the cardinality of $A$ is equal to the cardinality of [0,1) in the previous thread.
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May 8th, 2017, 09:34 PM   #4
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I guess they aren't countable because there's not a logical way to sequence them. That's interesting, because the real number set is uncountable for an entirely different reason. One's about reach, the other, description... Well, I guess you could get a description for them, but it would be continuous and rational so not like natural numbers.
I'm not sure it would be classified a number if it's counted with a description.

You seem to really know what you're talking about. I recently got hit with an idea that I just want to get a taste of. I'm clearly not educated in mathematics. I scrap things together for some web development. I'm really interested in some things. That's why I'm here. Can I PM you some exotic questions?

Last edited by InkSprite; May 8th, 2017 at 09:57 PM.
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May 8th, 2017, 11:00 PM   #5
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You seem to enjoy thinking, which means your questions are interesting.

We imagine the real numbers in the interval [0, 1). Numbers do not have decimal points. What do have decimal points are numerals in the decimal system of numeric representation when representing numbers that are not integers. I can in fact represent numbers in that interval without using a decimal point. Examples

$\dfrac{1}{2}$, or $\sqrt{\dfrac{3}{4}}$ or $\pi - 3.$

Now let us think of infinite strings in the form

$0.x_1x_2... \text { where } x_i \in \{0,\ 1,\ 2,\ 3,\ 4,\ 5,\ 6,\ 7,\ 8,\ 9\}.$

Then think of the infinite strings in the form $x_1x_2...$

Is it not easily demonstrable that those two sets can be placed in 1-to-1 correspondence and so have the same number of elements?

Zylo's whole thing about removing the decimal point is (bad joke coming) pointless.
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May 9th, 2017, 01:34 AM   #6
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What is all the fuss about Cantor's diagonal argument exactly? Not only here do I see plenty of attempts to refute it, but elsewhere too. I have not studied the ideas formally or informally (hence the ignorance in what follows), but I have to ask...

Is it simply the nature of the topic that allows for very subtle changes in usage of language that produce misconstrued definitions and results? Thus, poor interpretations by those lacking rigor.

Because that's what it always seems like, a back and forth of "what do you mean by this". Or is there actually some funny business going on... i.e., some unfinished business left by Cantor.

Quote:
Originally Posted by InkSprite View Post
Can I PM you some exotic questions?
Exotic eh.

Quote:
Originally Posted by JeffM1 View Post
... about removing the decimal point is (bad joke coming) pointless.
That's hilarious.

Last edited by skipjack; May 9th, 2017 at 02:33 AM.
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May 9th, 2017, 03:40 AM   #7
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I think I can prove $A$ is just as large as real numbers.
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May 9th, 2017, 06:57 AM   #8
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Quote:
Originally Posted by InkSprite View Post
I think I can prove $A$ is just as large as real numbers.
I can assure you that the cardinality of $A$ is equal to the cardinality of $\mathbb{R}$. It isn't a bad idea for you to prove that yourself though.
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May 9th, 2017, 07:10 AM   #9
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Quote:
Originally Posted by InkSprite View Post
You seem to really know what you're talking about. I recently got hit with an idea that I just want to get a taste of. I'm clearly not educated in mathematics. I scrap things together for some web development. I'm really interested in some things. That's why I'm here. Can I PM you some exotic questions?
Thank you, though I assure you there are some here that know far more than I do. They sometimes act like animals though...

I don't mind you asking questions I suppose, but I fear that asking only me would preclude you from obtaining the benefit of the otherwise bountiful amount of knowledge here. I think you'll find that people are very open to any type of mathematical questions here, no matter how "exotic" they might seem to you.
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May 9th, 2017, 07:27 AM   #10
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Quote:
Originally Posted by AplanisTophet View Post
I can assure you that the cardinality of $A$ is equal to the cardinality of $\mathbb{R}$. It isn't a bad idea for you to prove that yourself though.
I did so with a space filling infinite-dimensional fractal curve. I was quite proud of that. I guess winning an award in mathematics and pocketing some come-up money isn't going to be that easy.

I'm really glad this place exists.
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