Deeper than primes - Continuation

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The failure here is that this is not an answer to the question.

Let me try, against all experience, to explain the question and then explain why this answer is wrong.

Jsfisher said:
If there is at least 1 N-to-N mapping, you can not show that there are *no* N-to-N mappings.

Your 'answer' shows 1 N-to-N mapping, 1 N-to-nil mapping and 3 N-to-M mappings.

So therefore you failed the assignment.

Wrong, my argument shows two different cases with OR condition between them, about N to N mapping, where N is some case of an infinite set.

Here are the two cases and the OR condition between them:

(A case of (1-to-1 and onto from N to N):
1 ↔ 1
2 ↔ 2
3 ↔ 3
4 ↔ 4
5 ↔ 5
... )

OR

(A case of no (1-to-1 and onto from N to N):
1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

Edit:

exactly as used by Hilbert's Hotel case, which is prepared to include more visitors.

It is clear that no N member is missing in both sides, yet there is no 1-to-1 and onto from N to N in the given case above.

)

No one of the cases prevents the other case, as jsfisher tries to address by his "If there is at least 1 N-to-N mapping, you can not show that there are *no* N-to-N mappings." argument, simply because the two cases have OR logical condition between them (it means that no case prevents the other if we deal with an infinite set like N).

Edit: In order to enter to the wonderful world of infinities, one can't omit any of the two possible cases.
 
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Wrong, my argument shows two different cases with OR condition between them, about N to N mapping, where N is some case of an infinite set.

Here are the two cases and the OR condition between them:

(A case of (1-to-1 and onto from N to N):
1 ↔ 1
2 ↔ 2
3 ↔ 3
4 ↔ 4
5 ↔ 5
... )

OR

(A case of no (1-to-1 and onto from N to N):
1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
... )

No one of the cases prevents the other case, as jsfisher tries to address by his "If there is at least 1 N-to-N mapping, you can not show that there are *no* N-to-N mappings." argument, simply because the two cases have OR logical condition between them (it means that no case prevents the other if we deal with an infinite set like N).

Oh, for crying out loud, doron. OR? Really? Do you know what OR means? This is beyond ridiculous. You desperately try to argue that it's either the first or the second, but not both (of course this is idiocy as well), but fail to even express that by using OR?

Here's a laughing dog for you:
:dl:
 
Wrong, my argument shows two different cases with OR condition between them, about N to N mapping, where N is some case of an infinite set.

Here are the two cases and the OR condition between them:

(A case of (1-to-1 and onto from N to N):
1 <-> 1
2 <-> 2
3 <-> 3
4 <-> 4
5 <-> 5
... )

OR

(A case of no (1-to-1 and onto from N to N):
1 <-> 1
2 <->
3 <-> 2
4 <-> 3
5 <-> 4
...

Edit:

exactly as used by Hilbert's Hotel case, which is prepared to include more visitors.

It is clear that no N member is missing in both sides, yet there is no 1-to-1 and onto from N to N in the given case above.

)

No one of the cases prevents the other case, as jsfisher tries to address by his "If there is at least 1 N-to-N mapping, you can not show that there are *no* N-to-N mappings." argument, simply because the two cases have OR logical condition between them (it means that no case prevents the other if we deal with an infinite set like N).

Edit: In order to enter to the wonderful world of infinities, one can't omit any of the two possible cases.

Utterly wrong! The assignment *explicitly * states *AND*

Don't pretend you missed it.
 
It's not. Who said it was? It just doesn't mean what you <a class="ktg6us78hf8vdu7" href="javascript:void(0)">think</a> it means, which only goes to further emphasize your inability to follow the line of conversation.
Please explain us in details what


1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

means.
 
Please explain us in details what


1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

means.

Us? Is there anyone else besides you who doesn't know? Seriously, you can't even ask the right question. It's not about that mapping, it's about what you think that the existence of both mappings means. And no, I won't explain that either for you. You're going to have to do it yourself.
 
realpaladin said:
Utterly wrong! The assignment *explicitly * states *AND*

Don't pretend you missed it.
Wrong, the two cases are independent of each other if one deals with an infinite set like N, and this is exactly why the world of infinity is so wonderful.
 
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Oh, for crying out loud, doron. OR? Really? Do you know what OR means? This is beyond ridiculous. You <a class="ktg6us78hf8vdu7" href="javascript:void(0)">desperately</a> try to argue that it's either the first or the second, but not both (of course this is idiocy as well), but fail to even express that by using OR?

It is very simple.

A = (no (1-to-1 and onto from N to N))

B = (1-to-1 and onto from N to N)

Code:
A OR B
F    F --> F
F    T --> T
T    F --> T
T    T --> T

So you see, T T --> T by OR.
 
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Wrong, the two cases are independent of each other if one deals with an infinite set like N, and this is exactly why the world of infinity is so wonderful.

Wut? Really? You are really claiming that because of infinity you can have conflicting properties on one and the same set?

You can not prove that.
 
It is very simple.

A = (no (1-to-1 and onto from N to N))

Bzzzt! Wrong, but thanks for playing. Here's your laughing dog:
:dl:

Just because I'm feeling generous:
A = (there exists (no (1-to-1 and onto from N to N)))

What you wrote is either false or incorrectly written. Take your pick.
 
Bzzzt! Wrong, but thanks for playing. Here's your laughing dog:
:dl:

Just because I'm feeling generous:
A = (there exists (no (1-to-1 and onto from N to N)))

What you wrote is either false or incorrectly written. Take your pick.

So you can't deal with http://www.internationalskeptics.com/forums/showpost.php?p=9706767&postcount=2808.

In that case let us deal only with the case of no (1-to-1 and onto from N to N))

For example:

1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

Is this mapping is valid, and if not, why not?
 

Sure I can, it's you who doesn't even know what he's talking about.

In that case let us deal only with the case of no (1-to-1 and onto from N to N))

What does that mean? "no (1-to-1 and onto from N to N)"? Can you write it concisely?

For example:

1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

Is this mapping is valid, and if not, why not?

That is a mapping, yes. For 10 extra points and a cookie, write it like a man.
 
What do you think that means?
It means that also

1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

is a valid mapping from N to N.

My claim is that this valid case is ignored by Traditional Mathematics, as a legitimate case for detailed mathematical work.
 
It means that also

1 ↔ 1
2 ↔
3 ↔ 2
4 ↔ 3
5 ↔ 4
...

is a valid mapping from N to N.

OK, although I have to say it's a really weird way of saying that. Why don't you use mathematical notation?

My claim is that this valid case is ignored by Traditional Mathematics, as a legitimate case for detailed mathematical work.

Do you have anything to substantiate your claim? How is it "ignored"? Who said it wasn't valid?
 
laca said:
That is a mapping, yes.
You did not answer my question.

EDIT: Is this mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is a valid mapping from N to N?

Please answer only by Yes or No.
 
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You did not answer my question.

LOL. Says the king of not answering questions. Here's your laughing dog again:
:dl:

EDIT: Is this mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is a valid mapping from N to N?

Please answer only by Yes or No.

It's not a valid mapping from N because 2 is a member of N and it doesn't seem to have a mapping. It's a valid mapping from N \ {2} though.

Anything else I can help you with?
 
It's not a valid mapping from N because 2 is a member of N and it doesn't seem to have a mapping.
What you actually argue is that since mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is not mapping

1 → 1
2 → 2
3 → 3
4 → 4
5 → 5
...

then mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is invalid.

In other words, you have failed to understand the OR condition between these mappings, which enables their independence of each other.

Moreover, you do understand Hilbert's Hotel argument, which explicitly provides the validity of a mapping like

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

It is about time that you read and carefully learn how Hilbert's Hotel is prepared to include more visitors, if you really wish to get the validity of

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

mapping.
 
What you actually argue is that since mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is not mapping

1 → 1
2 → 2
3 → 3
4 → 4
5 → 5
...

then mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

is invalid.

No, that is not what I'm arguing. Can't you READ?

In other words, you have failed to understand the OR condition between these mappings, which enables their independence of each other.

In other words, you have failed to understand what I'm saying. Try again.

Moreover, you do understand Hilbert's Hotel argument, which explicitly provides the validity of a mapping like

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

The mapping is valid, but it's not from N. Why is that so hard to understand?

It is about time that you read and carefully learn how Hilbert's Hotel is prepared to include more visitors, if you really wish to get the validity of

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

mapping.

For the umpteenth time, it is valid, but not from N. It is about time that you read and carefully learn.
 
The mapping is valid, but it's not from N.

The mapping is indeed valid.

Moreover, it is from N to N because the right and left sides of it include any possible N member, such that no N member is left out.

Yet, as clearly shown by

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

valid mapping, there is also the case of no 1-to-1 and onto from N to N, even if no N member is missing in both sides.

This is another property of infinite sets, in addition to the property of the 1-to-1 correspondence between some given infinite set to its proper subset.

Traditional Mathematics works with mapping of an infinite set to its proper subset, but it does not work with (it ignores) the valid case of mapping of a given set to itself (such that no member of a given set is missing in both left and rights sides of the mapping), which is not 1-to-1 and onto.

Edit:
It's not a valid mapping from N because 2 is a member of N and it doesn't seem to have a mapping.
Since no N member is missing in both left and right sides of the mapping, this mapping is from N to N, even if it is not 1-to-1 and onto.
 
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Pray tell doron, if no member is missing from the left side, what member of N does 2 map to?
 
Pray tell doron, if no member is missing from the left side, what member of N does 2 map to?
To no member at all.

But that does not prevent the fact that the mapping is from N to N, since no N member is missing in both left and right sides of the mapping.

In other words, by dealing with infinite sets, it is shown that there is also no (1-to-1 and onto) from N to N, even if no N member is missing in both left and right sides of the mapping.

EDIT: If this property of infinite sets is not taken in account by Traditional Mathematics, we have here a novel thing for further mathematical work.
 
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My example shows that (no 1-to-1 (and onto) from N to N) is also a property of N, in addition to the property of (1-to-1 (and onto) from N to N).

No, your example does not. Your language skills are the equal to your math skills.

Your example is a mapping that is not one-to-one. It says nothing about there being no one-to-one mapping, or as you choose to say "no 1-to-1 from N to N". And, oh gee, we already know that there are mappings that are one-to-one.

Traditional Mathematics deals only with the case of (1-to-1 (and onto) from N to N) and does not deal with the case of (no 1-to-1 (and onto) from N to N).

Utter rubbish. There are infinitely many mappings from N to N. Mathematics does not limit which ones it "deals with".

Moreover, you continue to use this bogus expression: "case of (no 1-to-1...from N to N)". Your case is mapping that is not one-to-one. It is definitely not a case of no one-to-one mapping.

"Not" and "no" are not synonyms.
 
To no member at all.

But that does not prevent the fact that the mapping is from N to N, since no N member is missing in both left and right sides of the mapping.

Actually, it does. Since you had other, far more basic lapses in comprehension and reason, it seemed best to pass over this boo-boo, but since you raised it....

A mapping from N to N is a function with domain N and range N. The number, 2, is clearly in N and therefore clearly in the domain of the mapping. The fact you excluded it as a mapped value disqualifies your example as a mapping from N to N.

It is, however, a mapping from N-{2} to N, and it is, curiously enough, one-to-one.
 
The fact you excluded it as a mapped value disqualifies your example as a mapping from N to N.
You are wrong jsfisher.

Since no N member is missing in both left and right sides of the mapping, this mapping must be from N to N.

Yet this mapping is not onto since the left member 2 is not mapped to any member at all.

All what you did is to omit (EDIT:exclude, yes, you are the one that actually exclude 2 from the domain (what I call "the left side")) this member, and by doing this you are missing the fact that there is also the case that there is mapping from N to N (without missing any N member in both left and right sides of the mapping) which is not onto.

EDIT:

Here it is:

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

in spite of your limited treatment with mappings between infinite sets (you can't deal with the case of the mapping of room number 2 (of Hilbert's Hotel) to its empty state).
 
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To no member at all.

:dl:

But that does not prevent the fact that the mapping is from N to N, since no N member is missing in both left and right sides of the mapping.

LOL. What do you think a mapping is? It needs both sides. Seriously, doron, you're only making a fool of yourself.

In other words, by dealing with infinite sets, it is shown that there is also no (1-to-1 and onto) from N to N, even if no N member is missing in both left and right sides of the mapping.

EDIT: If this property of infinite sets is not taken in account by Traditional Mathematics, we have here a novel thing for further mathematical work.

Good stuff, man. Just one question. What in the blue hell are you on?
 
You are wrong jsfisher.

Since no N member is missing in both left and right sides of the mapping, this mapping must be from N to N.

Keep repeating it. 2 more and you get a cupcake! Won't be any less wrong, though...

Yet this mapping is not onto since the left member 2 is not mapped to any member at all.

Mapped and not mapped at the same time. Is this really your big breakthrough?

All what you did is to omit (EDIT:exclude, yes, you are the one that actually excludes 2) this member, and by doing this you are missing the fact that there is also the case that there is mapping from N to N (without missing any N member in both left and right sides of the mapping) which is not onto.

Look, doron. If your definition of mapping allows that one member of the "left" is not mapped (i.e. no mapping is mapping), then by the same definition it must be "onto" as well. Unless you come up with an inconsistent and/or idiotic definition for onto as well. Just keep the failings of your own retarded "system" within that "system" and don't try to project it onto (:)) Useful Maths.
 
It needs both sides.
No side is missing, the right side is simply empty.

For example, by using Hilbert's Hotel, 2 is the number of a given room (the left side of the mapping) and no member at all is its empty state (the right side of the mapping).

Edit: A room does not disappear because it is empty, and so is the case about mapping 2 →
 
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You are wrong jsfisher.

Since no N member is missing in both left and right sides of the mapping, this mapping must be from N to N.

Yet this mapping is not onto since the left member 2 is not mapped to any member at all.

You might just want to review the meanings of map and domain.

...
in spite of your limited treatment with mappings between infinite sets (you can't deal with the case of the mapping of room number 2 (of Hilbert's Hotel) to its empty state).

"Empty state" is not an element of N. So, either 2 is unmapped in your example, in which case your domain is N-{2} or 2 is mapped to "empty state", in which case your range is N+{{}}. Which is it?

By the way, since you seem all wrapped up in "Hilbert's Hotel", it would be much more natural to express the mapping of an infinite number of people into an infinite number of rooms while leaving Room 2 unassigned would be:

Person 1 --> Room 1
Person 2 --> Room 3
Person 3 --> Room 4
Person 4 --> Room 5
Person 5 --> Room 6
. . .

Then again, you have a habit of doing most things mathematical exactly backwards. See laca for your cupcake.
 
"Empty state" is not an element of N.
EDIT: Exactly (actually it is no member at all).

But the mapping to an empty state is done by using N element.

Again, by using Hilbert's Hotel, 2 is the number of a given room (the left side of the mapping) and no member at all is its empty state (the right side of the mapping).

A room does not disappear because it is empty, and so is the case about mapping 2 →

Edit: It is well known that |R| > |N| because there are R members in the mapping, which are not mapped to anything (because they are beyond the range of N members in the mapping between R and N members).

If we omit these R members we can't concluded that |R| > |N|, so your argument against mapping to nothing does not hold water even in terms of Traditional Mathematics.
 
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EDIT: Exactly (actually it is no member at all).

But the mapping to an empty state is done by using N element.

Again, by using Hilbert's Hotel, 2 is the number of a given room (the right side of the mapping) and no member at all is its empty state (the left side of the mapping).

A room does not disappear because it is empty, and so is the case about mapping 2 →

Before he edits it again after seeing Hilbert's Hotel explained.
 
EDIT: Exactly (actually it is no member at all).

But the mapping to an empty state is done by using N element.

So, Doronetics has its own non-standard definition for mapping, domain, and range. Would you please post those unique definitions so that we might better understand your posts?
 
So, Doronetics has its own non-standard definition for mapping, domain, and range. Would you please post those unique definitions so that we might better understand your posts?
jsfisher,

Again you wish to get things only in terms of verbal_symbolic skills.

But in order to understand what is going on in Hilbert's Hotel argument, also visual_spatial skills have to be used.

By using verbal_symbolic AND visual_spatial skills on the following mapping

1 → 1
2 →
3 → 2
4 → 3
5 → 4
...

one enables to understand that there are more rooms than visitors even if the names of the rooms are identical to the names of the visitors (where emptiness is no-visitor (and therefore no-name), where no-visitor can't be considered as some form of visitor, exactly as no-member (for example in the case of the empty set) can't be considered as some form of member)).

By using verbal_symbolic AND visual_spatial skills at Hilbert's Hotel argument, one enables to conclude that the exact cardinality of an infinite set like N is not well-defined, because there is difference between the amount of the rooms and the amount of the visitors, even if there is 1-to-1 and onto between the names of the rooms and the names of the visitors.

More things about mapping, take for example the mapping between the members of N and the members of P(N), as done in |P(N)| > |N| Cantor's proof.

During this proof we check the identity between some N member and the members of some P(N) member, or in other words, the mapping from some N member to some P(N) member penetrates into the checked P(N) member in order to find an identity. The non-matched cases (where one of the cases is the mapping between some N member and the "content" of {}, where in this case the mapping is of the form n → ) enable us to define some P(N) member that is not in the range of any one of the N members, which enables us to conclude that |P(N)| > |N|.

So as you see, the mapping of the form x → is used also by Traditional Mathematics, so you are invited to provide its verbal_symbolic-only definition.

Edit: Here is the standard formal definition (http://en.wikipedia.org/wiki/Domain_of_a_function):
Given a function f:X→Y, the set X is the domain of f; the set Y is the codomain of f. In the expression f(x), x is the argument and f(x) is the value. One can think of an argument as an input to the function, and the value as the output.

The image (sometimes called the range) of f is the set of all values assumed by f for all possible x; this is the set {f(x) | x ∈ X}. The image of f can be the same set as the codomain or it can be a proper subset of it. It is in general smaller than the codomain; it is the whole codomain if and only if f is a surjective function.

A well-defined function must carry every element of its domain to an element of its codomain.

Please explain how this definition enables to conclude that there is no copy of n (which is a member of N) in {} (which is a member of P(N)), without using a mapping of the form n → , which actually enables to penetrate into {} in order to conclude that there is no copy of n in {}.

EDIT: A room does not disappear because it is empty, and so is the case about mapping 2 → , which is understood only if one uses also his\her visual_spatial skills.
 
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...By using verbal_symbolic AND visual_spatial skills on the following mapping...


No, by doing it wrong you get something that isn't a mapping but call it one anyway.

You alleged different skills don't change definitions, they just assist you in your dilutions. Carry on, though. Maybe after another decade of cranking without result you'll see just how much of your life was wasted on nonsense. I suspect not, though.
 
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