- 01 Jan '13 13:31 / 2 editsAnyone:

Correct me if I am wrong but; two of the implications of the proven Godel's theorem is that:

1, there must be some mathematical theorems that can be made that are unprovable and yet are nevertheless true.

2, we cannot ever know exactly which mathematical theorems are the ones that are both unprovable and yet are nevertheless true.

My question is this;

is the number of possible mathematical theorems that are both unprovable and yet are nevertheless true finite or infinite?

Regardless of the answer to this question, I find it interesting to note that one of the implications of 1, and 2, above is that research into general mathematics can logically never end with an ultimate climax of the gaining of complete knowledge and understanding of mathematics because that would have to include a reliable way of determine the truth or falsity of any possible mathematical theorem without exception which would be impossible. In other words, there would always be something to research in mathematics for, at the very least, there will always be a mathematical theorem ( actually an infinite number of them? -that's my question ) that may be both true and provable but have yet to be mathematically proven true and so yet to be also proven provable. This means general research in mathematics will be research without end. - 01 Jan '13 19:15 / 7 edits

sorry, edit error:*Originally posted by humy***Anyone:**

Correct me if I am wrong but; two of the implications of the proven Godel's theorem is that:

1, there must be some mathematical theorems that can be made that are unprovable and yet are nevertheless true.

2, we cannot ever know exactly which mathematical theorems are the ones that are both unprovable and yet are nevertheless true.

My quest also proven provable. This means general research in mathematics will be research without end.

disregard the " ( actually an infinite number of them? -that's my question ) " bit because that was not*exactly*my question.

Actually, I didn't ask my question quite right anyway for what I really want to know is:

Is there is an infinite number of possible mathematical theorems that are both true**AND****UN**provable**AND**which cannot be*proven*to be**UN**provable? - 03 Jan '13 13:13 / 1 edit

Also, are there an infinite number of mathematical theorems at all?*Originally posted by humy***sorry, edit error:**

disregard the " ( actually an infinite number of them? -that's my question ) " bit because that was not*exactly*my question.

Actually, I didn't ask my question quite right anyway for what I really want to know is:

Is there is an infinite number of possible mathematical theorems that are both true [b]AND**UN**provable**AND**which cannot be*proven*to be**UN**provable?[/b]

For instance, you could have a theorem that specifies a 'log' function in it, then and entirely different theorem where everything is the same except the 'log' becomes 'log' squared or some such.

Where is Erdos when we need him?

Speaking of which, what is your Erdos number?:

http://www.oakland.edu/enp/compute/ - 03 Jan '13 15:11

actually I haven't published in math journals, I suspect I have an (astronomiccal) Erdos number anyway, but the link will be difficult to establish.*Originally posted by sonhouse***Also, are there an infinite number of mathematical theorems at all?**

For instance, you could have a theorem that specifies a 'log' function in it, then and entirely different theorem where everything is the same except the 'log' becomes 'log' squared or some such.

Where is Erdos when we need him?

Speaking of which, what is your Erdos number?:

http://www.oakland.edu/enp/compute/ - 04 Jan '13 02:22

Theorem: 1 is greater than 0*Originally posted by sonhouse***Also, are there an infinite number of mathematical theorems at all?**

For instance, you could have a theorem that specifies a 'log' function in it, then and entirely different theorem where everything is the same except the 'log' becomes 'log' squared or some such.

Where is Erdos when we need him?

Speaking of which, what is your Erdos number?:

http://www.oakland.edu/enp/compute/

Theorem: 2 is greater than 1

Theorem: 3 is greater than 2

.

.

.

.

. - 04 Jan '13 10:21 / 3 edits
*Originally posted by sonhouse***Also, are there an infinite number of mathematical theorems at all?**

For instance, you could have a theorem that specifies a 'log' function in it, then and entirely different theorem where everything is the same except the 'log' becomes 'log' squared or some such.

Where is Erdos when we need him?

Speaking of which, what is your Erdos number?:

http://www.oakland.edu/enp/compute/Also, are there an infinite number of mathematical theorems at all?

according to what is implied by a mathematician I have just spoken to on another forum, the answer appears to be 'yes' although he was specifically referring to those theorems that can neither be proved nor disproved:

http://mathhelpforum.com/new-users/210631-some-questions-godels-therum.html

“....

there must be an infinite number of statements that can neither be proven nor disproven. That's the case because, if there were only a finite number of such statements, we could add those statements themselves to the axioms, getting a new system of axioms in which all statements can be either proven or disproven, contradicting the original statement

….”

( he was referring to mathematical theorems when he spoke of “statements” above )

I think I understand his deduction. - 04 Jan '13 10:41 / 12 editsAfter getting a reply from a mathematician about my questions but not getting all the answers I really wanted, I realize that my original questions are badly worded and flawed for, as he implied, my original questions imply that we can KNOW a statement is true even if we cannot prove it!

OK, in that case, forget about my OP. Instead, this is what I really want to know:

One of the implications of Godel's theorem is that:

There is an INFINITE number of possible mathematical theorems that can neither be proved nor disproved (I think at least that much has now been safely established here)

Now, for what what I really to know which is close to the above but a bit more subtle:

Is there an INFINITE number of possible mathematical theorems that can neither be proved nor disproved that ALSO have the characteristic that there is NO possible proof that can be given that it can NOT be either proved or disproved (so we can never KNOW that the theorem cannot ever be proved or disproved)?

-THAT is what I really want to know.

For clarity, expressing the same question in a different way:

Is there an infinite number of mathematical theorems that are not provable and not disprovable

and cannot be proved to be either not provable or not disprovable?

in other words (and I think this is the clearest expression of my question yet):

Is there an infinite number of mathematical theorems that have ALL four of the following properties:

1, it cannot be proved

2, it cannot be disproved

3, it cannot be proved to be impossible to prove

4, it cannot be proved to be impossible to disprove

?

Anyone? - 04 Jan '13 12:37 / 5 edits

oh wait! If a theorem is wrong then it MUST be possible to disprove it because it wouldn't be in any sense 'wrong' if there was no possible application of the theorem that would expose it as being wrong!?*Originally posted by humy***After getting a reply from a mathematician about my questions but not getting all the answers I really wanted, I realize that my original questions are badly worded and flawed for, as he implied, my original questions imply that we can KNOW a statement is true even if we cannot prove it!**

OK, in that case, forget about my OP. Instead, this is what I really wa to prove

4, it cannot be proved to be impossible to disprove

?

[/quote]

Anyone?

Is my thinking right here? Because, if so, then there cannot be a theorem that is both wrong and impossible to disprove and, therefore, only theorems that are right cannot be disproved so, in the last expression of my question, property 2, i.e. “ 2, it cannot be disproved “ is redundant because it should be replaced with simply “ it is correct”? Is that right?

If so, shouldn’t my question really be:

Is there an infinite number of mathematical theorems that have ALL four of the following properties without exception:

1, it is correct

2, it cannot be proved

3, it cannot be proved to be impossible to prove

4, it cannot be proved to be impossible to disprove

?

-BUT*without*the above question implying we may be able to determine that a PARTICULAR theorem is correct even though we cannot prove it! (because that would obviously be impossible) ?

Does that make sense? Is my thinking correct here? - 05 Jan '13 05:05 / 1 edit

You can set up a branch of mathematics by laying down one or more axioms, which are statements taken to be true without proof. Euclidean geometry, for instance, is based on five postulates, no one of which can be proven using the other four. It follows that an infinite number of branches of mathematics can be constructed, since there exist an infinite number of sets of mathematical statements that can serve as axioms. So, if any one branch of mathematics has even a single theorem for which your four properties are all true, we can be sure there must be an infinite number of such theorems across all the possible branches of mathematics.*Originally posted by humy*Is there an infinite number of mathematical theorems that have ALL four of the following properties:

1, it cannot be proved

2, it cannot be disproved

3, it cannot be proved to be impossible to prove

4, it cannot be proved to be impossible to disprove

Even if we constrain ourselves to a single branch of mathematics of sufficient complexity (i.e. a branch founded on more than one nontrivial axiom), I believe the answer to your question is yes. Henceforth I will assume that we are operating within a single mathematical system with a fixed set A of axioms.

Before going forward, I have to point out that the truth value (truth or falseness) of your four properties are not independent. It'll be easier for us to consolidate them thus:

There exist an infinite number of mathematical conjectures C that have the following two properties:

(P1) C is impossible to prove or disprove.

(P2) It is impossible to prove or disprove that C is impossible to prove or disprove.

Certainly P1 can be true and P2 false, just considering the Continuum Hypothesis. However, if P2 is true for a given conjecture C, then P1 is necessarily also true for C! For if P1 were false, that would mean that C can be either proven or disproven, and the proof itself would stand as proof that C is*not*impossible to prove or disprove -- making P2 false and contradicting our initial assumption that P2 is true. Simply put, conjectures that satisfy P2 are a subset of the set of conjectures that satisfy P1. Because of this, we can dispense with P1 altogether and ask whether or not the following is true:

There exist an infinite number of mathematical conjectures C that have the following property:

(P2) It is impossible to prove or disprove that C is impossible to prove or disprove.

The set S1 of conjectures satisfying P1*is*infinite (not to be established here) and the set S2 of conjectures satisfying P2 is, again, a subset of S1. Is S2 infinite? For a conjecture C to be placed in S1 (i.e. satisfy P1), one just has to prove that your given set A of axioms cannot prove C, and also prove that your given set A of axioms cannot prove ~C (the negation of C). The best way to do this, usually, is to show that neither C nor ~C contradict any of the axioms in A.

What does it take for a conjecture C to be placed in S2 (i.e. satisfy P2)? Really not much has changed, for the statement "C is impossible to prove or disprove" is*itself*a conjecture which we could denote by C'. If C' satisfies P1 (i.e. the statement "C' is impossible to prove or disprove" is true), then C satisfies P2 and therefore is placed in set S2. So, we must prove that neither C' nor ~C' contradict the relevant set of axioms A' (not the same as A). We could say that C' is a "meta-conjecture" and A' a set of "meta-axioms," but the game is the same, and thereby we can make the case that S2 must be an infinite set by the same reasoning that establishes S1 as an infinite set.

That is, the answer to your question is yes. Admittedly I rushed it toward the end, but my time was running out. :/ - 05 Jan '13 10:06 / 1 edit

Thanks for that!*Originally posted by Soothfast***You can set up a branch of mathematics by laying down one or more axioms, which are statements taken to be true without proof. Euclidean geometry, for instance, is based on five postulates, no one of which can be proven using the other four. It follows that an infinite number of branches of mathematics can be constructed, since there exist an infinite nu is yes. Admittedly I rushed it toward the end, but my time was running out. :/**

That is brilliant!

I will definitely be mulling over this. - 05 Jan '13 10:33 / 1 editI have to make one correction. I said this above:

For if P1 were false, that would mean that C can be either proven or disproven, and the proof itself would stand as proof that C is not impossible to prove or disprove -- making P2 false and contradicting our initial assumption that P2 is true.

I should have said this:

For if P1 were false, that would mean that C can be either proven or disproven. For the sake of argument suppose C has been proven. That is, we have found a proof that C is true. Then we know that C is impossible to disprove. Now, the statement P2 is really a conjunction of four simple statements:

(1) It is impossible to prove that C is impossible to prove

&

(2) It is impossible to disprove that C is impossible to prove

&

(3) It is impossible to prove that C is impossible to disprove

&

(4) It is impossible to disprove that C is impossible to disprove

Since we now know that C is impossible to disprove (by virtue of having furnished a proof for C), it follows that (3) is false, and therefore P2 is false -- and that contradicts our initial assumption that P2 is true. A similar contradiction arises if we suppose C has been disproven.