the product of infinitely many natural numbers
Yuwu Xiong
i tried to construct an injection from interval (0,1) to N, to
test the cantor's concept of cardinal, alephs, actual infinity,
etc, which i do not really understand...
the injection is based on euclid's second theorem (the number of primes
is infinite) and the fundamental theorem of arithmetic (unique
factorization of natural numbers).
put it shortly, by Schroeder-Bernstein Theorem, if we can prove
that there is an _injection_ from R into N, then these two sets
are equipollent. further, since R is equipollent to interval
I=(0, 1), we only need to find an inject from I to N.
the injection could be established like this:
for any real number in I, it could be denoted as
r=0.abcd...
where digits a,b,c.. is <=9 & >=0.
we denote P_i as the ith prime number, where i ranges from 1 to
infinity. so P_1=2, P_2=3, P_3=5, and so on...
the injection G from I to N is:
G(r)=(P_1^a)(P_2^b)(P_3^c)...
i think G is an injection into N because:
1. G(r) belongs to N
2. if r1<>r2, then G(r1)<>G(r2)
3. since a/b/c/... is not greater then 9, G(r) can not
cover all N, e.g, 2^10 is not in ran G.
i asked dr. math on whether or not this is a valid injection.
the reply was negative, because as dr. math told me that when r
is infinite, G(r) is not in N. so my next question is, if G(r)
is not in N, i.e., G(r) is not a natural number, then what is G(r)?
or is there any books or links related to this topic?
it seems the key point here is whether or not to accept the concept
of actual infinity. if we do not accept this concept, it's logicall
to say that G(r) is a natural number...
thanks,
/bruin
Denis Feldmann
> hi, all,
>
> i tried to construct an injection from interval (0,1) to N, to
> test the cantor's concept of cardinal, alephs, actual infinity,
> etc, which i do not really understand...
Looks like it
>
> the injection is based on euclid's second theorem (the number of primes
> is infinite) and the fundamental theorem of arithmetic (unique
> factorization of natural numbers).
Bad start, but lets read on
>
> put it shortly, by Schroeder-Bernstein Theorem, if we can prove
> that there is an _injection_ from R into N, then these two sets
> are equipollent.
Right, and by Cantor, we deduce math is inconsistent
further, since R is equipollent to interval
> I=(0, 1), we only need to find an inject from I to N.
>
> the injection could be established like this:
>
> for any real number in I, it could be denoted as
> r=0.abcd...
> where digits a,b,c.. is <=9 & >=0.
>
> we denote P_i as the ith prime number, where i ranges from 1 to
> infinity. so P_1=2, P_2=3, P_3=5, and so on...
>
> the injection G from I to N is:
>
> G(r)=(P_1^a)(P_2^b)(P_3^c)...
>
So let's calculate G(1/9)=G(0.11111....). We find G(1/9)=2*3*5*7*11*13*....
1) This doesn't look very much like an integer to me (especially because
of Euclid theorem on the infinitusde of primes)
2) Even if you allow for infinite integers (a classical proof of
complete misunderstanding of basic mathematics), you will still have to
explain how the theorem of unique factorisation can be proved for them...
> i think G is an injection into N because:
>
> 1. G(r) belongs to N
> 2. if r1<>r2, then G(r1)<>G(r2)
Why? Because of unique factorisation ?
> 3. since a/b/c/... is not greater then 9, G(r) can not
> cover all N, e.g, 2^10 is not in ran G.
This is useless : this argument proves G is not surjective, which doen't
add anything to yourcresult
>
> i asked dr. math on whether or not this is a valid injection.
> the reply was negative, because as dr. math told me that when r
> is infinite, G(r) is not in N. so my next question is, if G(r)
> is not in N, i.e., G(r) is not a natural number, then what is G(r)?
Oh, and you were not convinced... Crank behaviour, that. Ok, I
understand your subject line, now. But do you really expect someone will
somehow save your argument by giving you some kind of infinite
integers, with the user's manual, allowing you to prove Cantor was
mistaken all akong?
> or is there any books or links related to this topic?
>
> it seems the key point here is whether or not to accept the concept
> of actual infinity. if we do not accept this concept, it's logicall
You don't even know how to spell that word. As for its meaning...
Dave Seaman
> hi, all,
> i tried to construct an injection from interval (0,1) to N, to
> test the cantor's concept of cardinal, alephs, actual infinity,
> etc, which i do not really understand...
> the injection is based on euclid's second theorem (the number of primes
> is infinite) and the fundamental theorem of arithmetic (unique
> factorization of natural numbers).
> put it shortly, by Schroeder-Bernstein Theorem, if we can prove
> that there is an _injection_ from R into N, then these two sets
> are equipollent. further, since R is equipollent to interval
> I=(0, 1), we only need to find an inject from I to N.
> the injection could be established like this:
> for any real number in I, it could be denoted as
> r=0.abcd...
> where digits a,b,c.. is <=9 & >=0.
> we denote P_i as the ith prime number, where i ranges from 1 to
> infinity. so P_1=2, P_2=3, P_3=5, and so on...
> the injection G from I to N is:
> G(r)=(P_1^a)(P_2^b)(P_3^c)...
> i think G is an injection into N because:
> 1. G(r) belongs to N
This is where the your argument fails. What is G(1/3), for example?
> 2. if r1<>r2, then G(r1)<>G(r2)
> 3. since a/b/c/... is not greater then 9, G(r) can not
> cover all N, e.g, 2^10 is not in ran G.
> i asked dr. math on whether or not this is a valid injection.
> the reply was negative, because as dr. math told me that when r
> is infinite, G(r) is not in N. so my next question is, if G(r)
> is not in N, i.e., G(r) is not a natural number, then what is G(r)?
Since you are the one who is attempting to define a function G, it is for
you to decide what G(1/3) is. As it stands, your attempted definition
makes no sense. You certainly have not defined an injection G : I -> N,
because you have not defined G(r) for every r in I.
> or is there any books or links related to this topic?
Related to what topic? Understanding why G(1/3) is not in N? It's an
infinite product, and it diverges to oo.
> it seems the key point here is whether or not to accept the concept
> of actual infinity. if we do not accept this concept, it's logicall
> to say that G(r) is a natural number...
Really? Which natural number is G(1/3)?
> thanks,
> /bruin
--
Dave Seaman
U.S. Court of Appeals to review three issues
concerning case of Mumia Abu-Jamal.
<http://www.mumia2000.org/>
Dave L. Renfro
[snip]
> I=(0, 1), we only need to find an inject from I to N.
[snip]
> we denote P_i as the ith prime number, where i ranges
> from 1 to infinity. so P_1=2, P_2=3, P_3=5, and so on...
>
> the injection G from I to N is:
>
> G(r)=(P_1^a)(P_2^b)(P_3^c)...
>
> i think G is an injection into N because:
[snip]
> i asked dr. math on whether or not this is a valid injection.
> the reply was negative, because as dr. math told me that when r
> is infinite, G(r) is not in N. so my next question is, if G(r)
> is not in N, i.e., G(r) is not a natural number, then what
> is G(r)? or is there any books or links related to this topic?
You don't need books or links. Just pick a number r whose
decimal expansion contains infinitely many nonzero digits,
such as 1/3, pi, sqrt(2), etc. Using any of these numbers
for r, your expression for G(r) clearly diverges to infinity,
and hence it will not be a real number. In particular, G(r)
will not be a natural number. Therefore, the domain of G is
not all of I.
As for what G(r) is, why would you want it to be anything?
To me, this seems completely unrelated to your original
concerns. If you insist on trying to make it be anything,
then why not start with something easier. What is the sum
1 + 2 + 4 + 8 + ... + 2^n + ...? Suppose we call it S. Then
2S = 2(1 + 2 + 4 + 8 + ...) = 2 + 4 + 8 + 16 + ... = S - 1.
But, from 2S = S - 1, we get S = -1. Does this mean that the
original sum is -1? Not necessarily. It could mean that some
of the ordinary rules of arithmetic break down when we try to
make S a real number. Perhaps you could try to develop some
consistent (with some, or maybe all, of the rules of arithmetic)
way of defining infinite sums of natural numbers before trying
to define infinite products of natural numbers.
Dave L. Renfro
Yuwu Xiong
so the problem (or at least one of the problems) of my injection, in your eyes, is that G(r) is not defined (for e.g., 1/3 or pi), or G(r) diverge...
but my understanding is that what i needed is not to tell what G(r) equals to, but to just to prove that G(r) belongs to N.
yes, G(r) will diverge to +oo in most cases, but does it implies G(r) is not in N, or G(r) is not a number at all?
put it in another way, we know <N;*> is a groupoid (or magma), this means N is closed under the operation *.
G(r) is just an combined infinite operations on N, if u say G(r) is not in N, that looks likes to say N is not closed under *.
let me know how do u think about it.
thanks again,
/bruin
imagin...@despammed.com
> thanks for the feedback, Denis Feldmann, Dave Seaman, and Dave L. Renfro.
>
> so the problem (or at least one of the problems) of my injection, in your eyes, is that G(r) is not defined (for e.g., 1/3 or pi), or G(r) diverge...
>
> but my understanding is that what i needed is not to tell what G(r) equals to, but to just to prove that G(r) belongs to N.
I think that would do.
>
> yes, G(r) will diverge to +oo in most cases, but does it implies G(r) is not in N, or G(r) is not a number at all?
Yes, precisely: "+00" is not a natural number, so anything that
"diverges" to it is not in N. For r=1/3 (for example), G(r) is not in
N, and with usual terminology G(r) is not a number at all.
> put it in another way, we know <N;*> is a groupoid (or magma), this means N is closed under the operation *.
Yes. Closed under an operation simply means that for any two(!)
elements a and b of N, the product a*b is also a member of N.
The rationals are closed under addition; but the limit of an infinite
sequence of rationals may be irrational.
If you don't like this, just join one of the other crank threads - you
will find plenty of people who share your dislike...
Brian Chandler
http://imaginatorium.org
Yuwu Xiong
thanks for the feedback.
> > yes, G(r) will diverge to +oo in most cases, but
> does it implies G(r) is not in N, or G(r) is not a
> number at all?
>
> Yes, precisely: "+00" is not a natural number, so
> anything that
> "diverges" to it is not in N. For r=1/3 (for
> example), G(r) is not in
> N, and with usual terminology G(r) is not a number at
> all.
this is the subtle point to me. i agree that +00 is not a natural number, but how can u get the statement that anything diverges to +00 is also not a natural number? that's also the reason why i think the difference roots in the acceptance of actual infinity or not.
>
>
> > put it in another way, we know <N;*> is a groupoid
> (or magma), this means N is closed under the
> operation *.
>
> Yes. Closed under an operation simply means that for
> any two(!)
> elements a and b of N, the product a*b is also a
> member of N.
yes, the definition for closed is for one operation. i just applied the induction principle here:
1. one operation is closed.
2. if n operation is closed, we can prove n+1 operation is also closed.
3. thus N is closed for any number of operations...
>
> The rationals are closed under addition; but the
> limit of an infinite
> sequence of rationals may be irrational.
>
> If you don't like this, just join one of the other
> crank threads - you
> will find plenty of people who share your dislike...
one of the best point of math, as i understand, is that whether or not u like it, there is not ambiguity for a specific problem, yes or no, no alternatives. do not know how u define crank, but i am willing to change/accept any result if it convinces me somehow. though i am not doing any research, i think the following quote from this forum is good for me:
"Doing any sort of research in mathematics with a fixed idea of what
you will allow yourself to find is not really a good idea."
i am very clear that yet another possibility also exist: it's not that the theory can not convinces me, but that i am too dull to get the point. even for this case, i still want to ask if there is a way to get me through with helps on the forum (e.g., from u)...
thanks again.
/bruin
Kees
It's wise to use induction correctly! 1 and 2 above are correct but 3 should be:
Thus N is closed under any *finite* number of operations.
I hope you now understand that N being closed under product doesn't give any meaning to an inifinite product.
A general remark: Whenever you use some mathematical symbol or operations you have to say what you mean by that. In your question you use an infinite product of natural numbers. My question to you is what is the definition of an infinite product of naturals. Notice that I'm not asking what is a specific infinite prodcut equal to, that would come only after you tell me what at all is an infinite product of naturals. Before you do that there is no point for discussion since your terms are not well defined.
All the best,
Kees
>
Yuwu Xiong
> operation.
> > i just applied the induction principle here:
> > 1. one operation is closed.
> > 2. if n operation is closed, we can prove n+1
> > operation is also closed.
> > 3. thus N is closed for any number of
> operations...
>
>
> It's wise to use induction correctly! 1 and 2 above
> are correct but 3 should be:
>
> Thus N is closed under any *finite* number of
i would say like this: N is closed under any number (n) of operations as long as n is a natural number.
>
> I hope you now understand that N being closed under
> product doesn't give any meaning to an inifinite
> product.
>
> A general remark: Whenever you use some mathematical
> symbol or operations you have to say what you mean by
> that. In your question you use an infinite product of
> natural numbers. My question to you is what is the
> definition of an infinite product of naturals. Notice
> that I'm not asking what is a specific infinite
> prodcut equal to, that would come only after you tell
> me what at all is an infinite product of naturals.
> Before you do that there is no point for discussion
> since your terms are not well defined.
in the definition of injection, i only used the product operation. +00 is just a symbol for infinity which u can never reach (according to potential infinity view point), +00 is not a term in the operation, nor the number of operations. i think "diverges to" dose not mean "equals to". by "an infinite product of naturals", it actually means the product of a bunch of naturals, where the number of naturals, as well as the value of those numbers, can go towards +00, but never reach it.
by the way, if we say G(0.111....) is not a number, how it can be used in an arithmatic expression? for example, in calculating zeta(2) by euler, it used as a denominator. e.g., a link on this forum: <http://mathforum.org/library/drmath/view/55801.html>
or divergeness is also serves as the criteria for judging if the value of an expression a number or not?
thanks for the feedback.
/bruin
>
> All the best,
>
> Kees
>
> >
imagin...@despammed.com
> hi, brian,
>
> thanks for the feedback.
>
> > > yes, G(r) will diverge to +oo in most cases, but
> > does it implies G(r) is not in N, or G(r) is not a
> > number at all?
> >
> > Yes, precisely: "+00" is not a natural number, so
> > anything that
> > "diverges" to it is not in N. For r=1/3 (for
> > example), G(r) is not in
> > N, and with usual terminology G(r) is not a number at
> > all.
> this is the subtle point to me. i agree that +00 is not a natural number, but how can u get the statement that anything diverges to +00 is also not a natural number?
Very simply. Natural numbers do not "diverge". An infinite sequence may
diverge (e.g. 1+2+3+4...) or an infinite sequence may converge to a
natural (e.g. 1+1/2+1/4+1/8...), but neither of these infinite
sequences _is_ a natural number.
Since natural numbers do not "diverge", anything that does diverge
cannot be a natural number.
> that's also the reason why i think the difference roots in the acceptance of actual infinity or not.
(Hint: the expression "acceptance of actual infinity" is a diagnostic
for crankhood.)
> > > put it in another way, we know <N;*> is a groupoid
> > (or magma), this means N is closed under the
> > operation *.
> >
> > Yes. Closed under an operation simply means that for
> > any two(!)
> > elements a and b of N, the product a*b is also a
> > member of N.
> yes, the definition for closed is for one operation. i just applied the induction principle here:
> 1. one operation is closed.
> 2. if n operation is closed, we can prove n+1 operation is also closed.
> 3. thus N is closed for any number of operations...
This is invalid. The point of the following example:
> > The rationals are closed under addition; but the
> > limit of an infinite
> > sequence of rationals may be irrational.
... is that if you use some new word "sclosed" to mean closed and also
that for an infinite sequence of elements, the "infinite operation" (a1
* a2 * a3 * ...) is necessarily defined, and also in the relevant set,
then "sclosed" does not apply to the rationals under addition, for
example. Actually, I suspect you lose most if not all infinite groups
for example.)
Brian Chandler
http://imaginatorium.org
Dave L. Renfro
> in the definition of injection, i only used the product operation.
> +00 is just a symbol for infinity which u can never reach
> (according to potential infinity view point), +00 is not a term
> in the operation, nor the number of operations. i think "diverges
> to" dose not mean "equals to". by "an infinite product of naturals",
> it actually means the product of a bunch of naturals, where the
> number of naturals, as well as the value of those numbers, can
> go towards +00, but never reach it.
The product operation inputs TWO (2) numbers and outputs ONE (1)
number. If we want to use it *twice*, there are two possibilities:
a(bc) and (ab)c
It just so happens that each of these ways of applying the product
operation *twice* results in the same number (associative law).
Thus, we simply write abc when one or the other of these is
intended. (Sometimes we might want to keep the parentheses
because, in some situations, this might be useful in order
to clue a reader in as to what manipulations we did to come
up with the product abc.)
In the case of *three* applications of the product operation
(to 4 numbers a, b, c, and d), there are 5 possibilities
(you might want to see if you can find all 5 of them).
Using the ordinary associtivity law, all 5 of these can be
shown to give the same number. The verifications that these
are all equal are somewhat like the derivation you can
find in this post: <http://tinyurl.com/px8zk>.
In general, there are a certain number of ways that a binary
operation can be applied n times (google "catalan number"),
and it can be proved from ordinary associativity that all of
the different ways give rise to the same output (when the
inputs aren't changed). This can be proved by induction,
and I can post a proof and a number of related references
if you'd like. (I thought I'd already posted an essay on
this topic, but apparently not. However, I have the write-up
in a LaTeX document, which I can re-write as a usenet post
if anyone is interested.)
However, none of this is of any immediate direct relevance
to what an infinite product (or an infinite sum) is. The
usual way to define these is to take the limit of the
first n partial products (or sums, in the case of infinite
sums). Sometimes this limit exists and sometimes it doesn't.
By the way, there are also other ways one can take the limit
of the partial products. (Note: We always have to work with
partial products, and the term means a finite number of the
inputs being used by the way, because that's the only thing
we can obtain by applying the binary operation over and
over again.) The usual way is what you'll find in elementary
calculus texts, where sequences and series are discussed
(and you should go back and review this, at least the
beginning part of the sequences and series chapter), and
this is what almost everyone here will be talking about
(the limit, as n --> oo, of the n-1 product operations
applied to the first n inputs).
It is important to note also that the use of infinity in
n --> oo is very different from the use of infinity in the
sense of infinite cardinal numbers. In the case of n --> oo,
this is simply a short hand for a limit, and it is not
necessary to bring infinity onto the scene (certainly not
infinity in the cardinal number sense) to define and work
with limits. (Although, in more advanced math, it is sometimes
convenient to do so -- using the extended real line, for
example. However, even in these situations, the notion of
infinity is not the same as infinite cardinal numbers.)
I'm not going to get into the details about what limits
are here -- this is something you should review in an
elementary high school or college calculus text. In fact,
even many precalculus texts include some discussion of
limits, although not usually the epsilon-delta formulation,
which is what you want to look at to be convinced that
limits have nothing to do with "completed infinities"
(at least, not of the type you're dealing with).
> by the way, if we say G(0.111....) is not a number, how it
> can be used in an arithmatic expression? for example, in
> calculating zeta(2) by euler, it used as a denominator.
> e.g., a link on this forum:
> <http://mathforum.org/library/drmath/view/55801.html>
None of the denominators I saw at this web page had an
infinite product. They each had various finite products
(whose lengths had no upper bound, however ... but each
of them was still a product involving a finite number of
inputs).
Dave L. Renfro
Dave Seaman
> hi, brian,
> thanks for the feedback.
>> > yes, G(r) will diverge to +oo in most cases, but
>> does it implies G(r) is not in N, or G(r) is not a
>> number at all?
>> Yes, precisely: "+00" is not a natural number, so
>> anything that
>> "diverges" to it is not in N. For r=1/3 (for
>> example), G(r) is not in
>> N, and with usual terminology G(r) is not a number at
>> all.
> this is the subtle point to me. i agree that +00 is not a natural number, but how can u get the statement that anything diverges to +00 is also not a natural number? that's also the reason why i think the difference roots in the acceptance of actual infinity or not.
No, that's not the problem. I think everyone agrees that your definition
of G(1/3) leads to an infinite product, and that the number of terms in
the product is an actual infinity. Everyone except you is saying that
the product does not evaluate to a natural number.
>> > put it in another way, we know <N;*> is a groupoid
>> (or magma), this means N is closed under the
>> operation *.
>> Yes. Closed under an operation simply means that for
>> any two(!)
>> elements a and b of N, the product a*b is also a
>> member of N.
> yes, the definition for closed is for one operation. i just applied the induction principle here:
> 1. one operation is closed.
> 2. if n operation is closed, we can prove n+1 operation is also closed.
> 3. thus N is closed for any number of operations...
That is not the correct conclusion. When you apply mathematical
induction, the conclusion is that N is closed for any *finite* number of
operations.
>> The rationals are closed under addition; but the
>> limit of an infinite
>> sequence of rationals may be irrational.
>> If you don't like this, just join one of the other
>> crank threads - you
>> will find plenty of people who share your dislike...
> one of the best point of math, as i understand, is that whether or not u like it, there is not ambiguity for a specific problem, yes or no, no alternatives. do not know how u define crank, but i am willing to change/accept any result if it convinces me somehow. though i am not doing any research, i think the following quote from this forum is good for me:
> "Doing any sort of research in mathematics with a fixed idea of what
> you will allow yourself to find is not really a good idea."
> i am very clear that yet another possibility also exist: it's not that the theory can not convinces me, but that i am too dull to get the point. even for this case, i still want to ask if there is a way to get me through with helps on the forum (e.g., from u)...
> thanks again.
> /bruin
G. A. Edgar
If K = 2*3*5*7* ...
where for each prime we have one factor, then what is K+1 ?
Hmm.. not divisible by ANY prime ??
--
G. A. Edgar http://www.math.ohio-state.edu/~edgar/
Kees
> > operation.
> > > i just applied the induction principle here:
> > > 1. one operation is closed.
> > > 2. if n operation is closed, we can prove n+1
> > > operation is also closed.
> > > 3. thus N is closed for any number of
> > operations...
> >
> >
> > It's wise to use induction correctly! 1 and 2
> above
> > are correct but 3 should be:
> >
> > Thus N is closed under any *finite* number of
> > r of operations.
> i would say like this: N is closed under any number
> (n) of operations as long as n is a natural number.
>
Which is exactly what I said. In this case you correctly interpreted the meaning of "finite".
I asked you a simple question: what do you mean by a product of infinitely many naturals. Before you give me a definition of that thing we can't discuss this further.
Yuwu Xiong
> If K = 2*3*5*7* ...
> where for each prime we have one factor, then what is
> K+1 ?
> Hmm.. not divisible by ANY prime ??
thanks, edgar.
this is a good question, and i have no answer for it at the moment. it also puzzled me a lot, just like cantor's ordinal number omega, omega+1, and omega+omega etc puzzled me before...
thanks again.
/bruin
Kees
> numbers?
> > If K = 2*3*5*7* ...
> > where for each prime we have one factor, then what
> is
> > K+1 ?
> > Hmm.. not divisible by ANY prime ??
>
> thanks, edgar.
>
> this is a good question, and i have no answer for it
> at the moment. it also puzzled me a lot, just like
> cantor's ordinal number omega, omega+1, and
> omega+omega etc puzzled me before...
The difference between Cantor's well defined ordinar numbers and your vague notion of a product of infinatly many naturals is that Cantor's oridnal numbers are defined and your notion is not. It is customary in mathematics to give precise definitions before making any claims about vague things. While Cantor's ordinal numbers can be confusing at first, once can *prove* properties about them because they are well defined. If you will give a definition of your notion of infinite product of naturals it would be possible to start looking for properties of these. I suggest that you first think carefully what do you mean by an infinite product of naturals and then, if you wish to discuss this with others, convey your notion by a precise definition.
Kees,