Is 1 = 0.999... ? Really?

[Meno2 (Meno_)]

Very tru w. Further incursions about population’s right to exist, en mass, reflected by Oppeheir’s observance and subsequent exclamation, ‘Now, I am become death, the destroyer of worlds’- the relevance of which, fits perfectly , the relevance of a pre-verbiage logic superceeding those of analysis- is so fitting a reflection to the lack of transperancy to the public of what possible roads we are lead towardupon.

Again , interjections do have and can have mathematical underpinnings, likewise, which is not to say, they would be subject to the same kind of scrutiny.

Awareness of languages of transi finite values imply just that, and any comment reversely suggests the same insignification of signs within which the philosophies of language, logic and mathematics operate. That is in no way a diversion to place any one of them into a special sense of approbation.

Rather, it is an effort to balance a seemingly favored language, to show a break, instead of an interlace of derivatives, elevating priesthood from a sensible function into an esoteric device, as has occurred to repeat with the so called death of god.

Hoping this will not exhaust your patience, as has mine, since we have reached a stage in urgency, where interpretation must consider math above and beyond merely utilitarian aspects.

I refer the pre quantified sources of earliest mathematical principles, as those underlying logic, and the accession to logical sense, before those of math. The principle of the exclusion of the middle, comes directly from a psychological flight or fight, there was no holding or suspending at that time.

The priesthood of any holding came later, and lest these are forgotten, we err on standing on very thin ice.

For the forum , the only defense to the use of a function thereof, is to show the multiform reason for its analysis in the first place, and notto topple the priesthood in the second.

This is the last mathelogical comment I shall post, again no intent other than relevance here, neither any other kind of show.

[Carleas]

@James S Saint

No, because “A” is not one of the natural numbers being paired. You only pair the natural numbers, not the one left over.

But (f(x)) does pair A. (f(A)=1)

You’ve yet to point out a flaw in my proof. You pointed to a conclusion that you don’t like (which you helpfully highlighted in red), but you didn’t explain why it doesn’t follow from the proof. (f(x)) maps A. (f(x)) maps all the natural numbers. (f(x)) maps all of P onto all of N. You’ve said nothing to suggest otherwise, you’ve just pointed to the fact that the result is counter-intuitive because P clearly contains all the elements in N and one additional element.

You’re making an argument from consequences. (f(x)) is a bijection between P and N, even though (P = N \cup {A}). I don’t disagree that it’s counter-intuitive. Infinity is counter-intuitive. But (f(x)) is a bijective function between P and N, and my proof shows it.

If you disagree, point to a premise that I’ve used that is false, or explain why the conclusion doesn’t follow, or point to a specific element that doesn’t get paired uniquely by (f(x)).

[Meno2 (Meno_)]

the only reply here, which I am able to muster up is, Your premise, that non infinity is intuitional, and infinity is not, is arguable,and has been argued, but mathematically such arguments stand up only with a premise which disassumes that such intuition is even possible.

It is possibility of intuitive infinity can not be absolutely ruled out, without the a functional equivalent absolute proof. And non such exists in math, only probabilistic ones.

[James_S_Saint]

@Carleas

@James S Saint

No, because “A” is not one of the natural numbers being paired. You only pair the natural numbers, not the one left over.

But (f(x)) does pair A. (f(A)=1)

Then it doesn’t pair ALL of the rest of P.

@Carleas

You’ve yet to point out a flaw in my proof.

Yes I have, again. You’re just making excuses, ignoring the truth that even you recognize. The flaw is that you presume that merely because the two sets are infinite, the “n+1” ploy covers all members. It does not. And I figured out a way to make that more clear:

Let me restate the issue by reordering P.

P has two subsets within. It contains a subset of all of the natural numbers, as does N, but it also has the subset of the letter “A”. And just to make it even more clear, let me make that A be the entire alphabet, “a-z”. And also all sets are ordered, so we have to keep the order straight:

[list]A = [a,b,c … z]
N = [1,2,3…]
P = [A, N][/list:u]

Now what is your bijection function between N and P?

And even better;

[list]A = [a,1,2,3…]
N = [1,2,3…]
P = [N, A][/list:u]

What is your bijection function between N and P?

[Carleas]

@James S Saint

And even better;

A = [a,1,2,3…]
N = [1,2,3…]
P = [N, A]

What is your bijection function between N and P?

Your notation is a little unclear to me, so I’m going to make some assumptions, correct me if I’m wrong.

I assume you mean the element 1 in set A to be different from the element 1 in set N.
I assume you mean P to be the union of the two sets.
I’ll use (x_A) to mean x from set A and (x_N) to mean x from set N.

so P = ({a,1_A,1_N,2_A,2_N,3_A,3_N,…}), and (f(x) , P \to N):

(f(x)=\begin{cases}1&x=a\2x& x \in A \land x \neq a\2x+1& x \in N\end{cases})

This function uniquely maps every element in P to a corresponding element in N, and fully covers N.

[James_S_Saint]

@Carleas

@James S Saint

And even better;

A = [a,1,2,3…]
N = [1,2,3…]
P = [N, A]

What is your bijection function between N and P?

Your notation is a little unclear to me, so I’m going to make some assumptions, correct me if I’m wrong.

I assume you mean the element 1 in set A to be different from the element 1 in set N.
I assume you mean P to be the union of the two sets.
I’ll use (x_A) to mean x from set A and (x_N) to mean x from set N.

so P = ({a,1_A,1_N,2_A,2_N,3_A,3_N,…}), and (f(x) , P \to N):

(f(x)=\begin{cases}1&x=a\2x& x \in A \land x \neq a\2x+1& x \in N\end{cases})

This function uniquely maps every element in P to a corresponding element in N, and fully covers N.

No.

P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

[Carleas]

@James S Saint

No.

P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

That is the same set, just reordered. Just as both ({…,-3,-2,-1,0,-1,-2,-3,…}) and ({0,1,-1,2,-2,3,-3,…}) are the set of integers.

(f(x)) is a bijection, no matter how you order P.

[James_S_Saint]

@Carleas

@James S Saint

No.

P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

That is the same set, just reordered. Just as both ({…,-3,-2,-1,0,-1,-2,-3,…}) and ({0,1,-1,2,-2,3,-3,…}) are the set of integers.

(f(x)) is a bijection, no matter how you order P.

It is not the “same set”. It is an ordered set. You cannot reorder it.

N = ({1_N, 2_N, 3_N …})
P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

What is your bijection function?

[Carleas]

@James S Saint

What is your bijection function?

(f(x)=\begin{cases}1&x=a\2x& x \in A \land x \neq a\2x+1& x \in N\end{cases})

That (f(x)) is a bijection does not depend on the order.

[wtf1 (wtf)]

1   2   3   4   5   6   7   ...
^   ^   ^   ^   ^   ^   ^
|   |   |   |   |   |   | 
|   |   |   |   |   |   |   
v   v   v   v   v   v   v   
A   1   2   3   4   5   6   ...   

The above is a visualization of the bijection between the sets (\mathbb N) and (\mathbb N \cup {A}).

Imagine that these are wooden blocks like a child plays with. Alphabet blocks. On each wooden cube we write one of the top row numbers on the block. So we have a set of wooden blocks labelled (1, 2, 3, \dots), one natural number for each block.

Now we come along and repaint the label on each block with the corresponding label from the bottom row. The block labeled (1) we repaint (A). The block labelled (2) we repaint as (1), and so forth.

Are not adding or subtracting any blocks or changing them in any way other than to label them differently.

There is a line of houses on one side of a street with addresses 101, 102, 103, 104, 105. One day the city tells them that their street is being converted to even numbers only, with the odd numbers across the street, and they’ll have to change their address.

Each resident goes down to the hardware store, buys some new decorative metal numerals, and nails the numerals to the front of their house. Now they read 202, 204, 206, 208, 210.

But the houses don’t change. Each house is identical before and after the readdressing, with the sole exception of the decorative numerals representing its address.

A bijection is nothing more than a relabeling or renaming of underlying objects that do not change.

So of course there’s a 1-1 correspondence between a given set before and after repainting. It’s the exact same set, completely unchanged except for a different label.

[James_S_Saint]

@Carleas

@James S Saint

What is your bijection function?

(f(x)=\begin{cases}1&x=a\2x& x \in A \land x \neq a\2x+1& x \in N\end{cases})

That (f(x)) is a bijection does not depend on the order.

That function pairs only half of the set P with N (as if there was any other choice).

You have half of set A paired (with N).
And you have half of N paired (with N).

N = ({1_N, 2_N, 3_N…})
P = ({1_N, 2_N, 3_N…, a, 1_A, 2_A, 3_A…})

…and it’s obvious.

You have a choice to pair the first half of sets A and N to N or to pair all of N to almost all of A.
But you cannot pair all of both A and N to N.

You are claiming that the same set N can be paired to itself as well as another infinite set. It is more than obvious that a set once paired with itself cannot also pair with more than itself.

(N = {1_N, 2_N, 3_N…})
(A = {a, N})

(P = {N, A})

Your function;
(f(x) = {N, a, N} \to N)

To pair P with N means to pair N with itself twice plus one more pairing for “a”.

Nothing can be paired with more than itself.

[Carleas]

@James S Saint

That function pairs only half of the set P with N (as if there was any other choice).

If that’s true, it should be trivial to identify a single, specific element in P that isn’t mapped to N. Can you?

[James_S_Saint]

@Carleas

@James S Saint

That function pairs only half of the set P with N (as if there was any other choice).

If that’s true, it should be trivial to identify a single, specific element in P that isn’t mapped to N. Can you?

That would be an erroneous hypothesis of yours. When dealing with infinite strings, one cannot point to what is happening “at the end” or beyond. That is what you use to hide the failure of your function to pair all members. And that is the failure of bijection as a principle.

Bijection cannot be used with infinite sets for that very reason. The word “infinite” is not well defined and thus cannot be precisely used in mathematics. Two sets can be infinite and yet of very different size or degrees of infinite (such as Aleph0 and Aleph1). Although once an infinite set is well defined, as in the hyperreals, normal reasoning applies again because math is merely logic applied to quantities.

You cannot know that you have paired an infinite set to anything other than itself. But you can know that once paired with itself, there is nothing left of it to pair with anything else. Do you disagree with that?

In Cantor’s day and prior, they didn’t have the hyperreals understanding.

[Carleas]

@James S Saint

When dealing with infinite strings, one cannot point to what is happening “at the end” or beyond.

There is no ‘at the end’; by their construction the natural numbers have no upper bound. The bijection I provided relies only on properties of the natural numbers, namely that every natural number has a successor (from which it’s straightforward to show that doubling a natural number also produces a natural number).

@James S Saint

Two sets can be infinite and yet of very different size or degrees of infinite (such as Aleph0 and Aleph1).

I don’t disagree. But the union of a set with cardinality (\aleph_0) and a set of cardinality 1 does not produce a set of cardinality (\aleph_1) (it produces a set with cardinality (\aleph_0)).

@James S Saint

But you can know that once paired with itself, there is nothing left of it to pair with anything else. Do you disagree with that?

Yes, I disagree. A countably infinite set can be paired with a countably infinite subset of itself, so that e.g. the set of natural numbers can be paired with the set of even natural numbers (the bijection being (g(x) = 2x)).

Even using your ordering of P, all elements are mapped to elements of N. So
(f(1_N) = 2(1)+1 = 3 \
f(2_N) = 2(2)+1 = 5\
…\
f(a) = 1 \
f(1_A) = 2(1) = 2 \
f(2_A) = 2(2) = 4 \
…)
James, your objections here lack rigor. You’ve been presented with a proof that shows that the set of natural numbers has a bijection to the set of natural numbers plus (a), and you don’t like that conclusion. And you keep saying, in various ways, that you just really don’t like that conclusions. But that’s not a rigorous argument. The proof works, you’ve made no rigorous attempt to show otherwise, and you can’t identify an element that isn’t mapped by the bijection.

If you want to reject the conclusion, just point to a premise that you don’t accept (e.g., the construction of the natural numbers), or to a step in the proof that you don’t like (e.g. (x) is a natural number (\therefore 2x ) is a natural number), or any element for which it fails (e.g. (x=?) (can’t suggest one because one doesn’t exist)). Those would be rigorous responses.

Continuing to assert your preferred conclusion is not.

[wtf1 (wtf)]

@James S Saint

The word “infinite” is not well defined and thus cannot be precisely used in mathematics.

The other day I showed you the definition. A set is called infinite if it can be put into bijective correspondence with a proper subset of itself. Since I’m interested in math history I mentioned that this definition is due to Richard Dedekind. You responded by not only insulting me, which is par for the course; but also by throwing out some insults at Dedekind. That really makes no sense. It seemed a little off frankly.

But you can’t say infinite sets aren’t well defined in math because they clearly are.

@James S Saint

Two sets can be infinite and yet of very different size or degrees of infinite (such as Aleph0 and Aleph1).

Yes, infinite sets can have various cardinalities. Which does not alter the fact that infinite sets are perfectly well defined.

@James S Saint

Although once an infinite set is well defined, as in the hyperreals, normal reasoning applies again because math is merely logic applied to quantities.

You’ve said that many times. “Logic applied to quantities.” But you have your own private definition of logic, and don’t even appear to accept the existential and universal quantifiers of predicate logic. But “logic applied to quantities?” It’s like “Better living through chemistry.” Sounds like it means something but it’s just a slogan.

@James S Saint

In Cantor’s day and prior, they didn’t have the hyperreals understanding.

We’ve established beyond all doubt at this point that the hyperreals don’t help any aspect of your argument.

And the hyperreals are a technical construction in modern set theory. You can’t reject modern set theory and predicate logic yet claim to believe in the hyperreals.

[James_S_Saint]

@Carleas

James, your objections here lack rigor. You’ve been presented with a proof that shows that the set of natural numbers has a bijection to the set of natural numbers plus (a), and you don’t like that conclusion.

Pure BS and you know it.

Although I did figured out a way to “point to” an element that your function doesn’t pair with anything.

You use x as your pointer, a natural number, I presume. And you reference 2x and 2x+1. What happens when x goes to infinity? You have 2 * infinity.

So for 2x, x from 1 to infinity, your pointer is at;
2,4,6,8,10,… 2*infinity and those are paired with A

Then you have 2x+1, x from 1 to infinity;
3,5,7,9,11,… 2*infinity+1 and those paired with N

[b]What kind of number is “2x+1” such that it can go to 2*infinity+1?

And exactly what is “infinity+1”?[/b]

@Carleas

And you keep saying, in various ways, that you just really don’t like that conclusions.

What is that a motto of your’s or something. Why do you join those who always accuse of their own guilt? You guys formed a club or something?

[Carleas]

@James S Saint

What happens when x goes to infinity? You have 2 * infinity.

Unless you’re changing the definition of your sets, infinity is not a member of P, so we never have 2*infinity. Every element of P is finite, even though P is infinite.

[James_S_Saint]

@Carleas

@James S Saint

What happens when x goes to infinity? You have 2 * infinity.

Unless you’re changing the definition of your sets, infinity is not a member of P, so we never have 2*infinity. Every element of P is finite, even though P is infinite.

Not so. For you to match EACH element in A while skipping every other number, your 2x must go from 2 to 2* infinity;
2x ( \to ) 2,4,6,8,10,… 2*infinity and those are paired with A (an infinite subset of P).

And again, what is “infinity+1”?

[Carleas]

Infinity isn’t a natural number. (x \notin P: x = \infty). So, granted, (\infty) isn’t mapped by the bijection, but (\infty) doesn’t have to be mapped for it to be a bijection between P and N.

Are you trying to take the limit of the function to show that it’s not bijective?

[James_S_Saint]

@Carleas

Infinity isn’t a natural number. (x \notin P: x = \infty). So, granted, (\infty) isn’t mapped by the bijection, but (\infty) doesn’t have to be mapped for it to be a bijection between P and N.

Are you trying to take the limit of the function to show that it’s not bijective?

I didn’t say that it has to “map infinity”. I said it has to go 2*infinity in order to map A, and also to map the subset N, both within P.