Re: Logarithm of transfinite numbers
- From: Virgil <ITSnetNOTcom#virgil@xxxxxxxxxxx>
- Date: Mon, 03 Apr 2006 15:42:04 -0600
In article <MPG.1e9b3ec6a13dfafc98abd5@xxxxxxxxxxxxxxxxxxxxxxxxx>,
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Virgil said:
In article <MPG.1e971f4a718dc8b198abba@xxxxxxxxxxxxxxxxxxxxxxxxx>,< snipposaurus rex>
Tony Orlow <aeo6@xxxxxxxxxxx> wrote:
Inverse Function Rule of Bigulosity Theory
If the nth element in an inductively defined quantitative set
Absent any definition of what are or are not "quantitative sets",
the definition fails at this point.
Sets of real numbers.
So what are inductively defined sets of real numbers in TOmatics? The
reals are not inductively defined in any system I know of.
is given by f(n), and f(g(n))=g(f(n))=n for all n in N,
According to the above, f and g must be bijections of the set N to
itself, i.e., permutations of N.
Uh, no. f maps naturals to some set of reals (which may all be
naturals) and g maps that set back to the set of naturals.
Not according to TO's reqirement that "f(g(n))=g(f(n))=n for all n in N".
That requires that f and g have naturals as both arguments and as
values, so that both must be functions from N to N.
then the number of elements e, such that x<e<y, is given by
floor(g(y)-g(x)+1).
Since neither f or g, at least as so far described, need preserve
any order relation on N, this makes no sense at all.
Actually that is an astute observation, so it ust have been a
mistake, ;) I thought about it this weekend, and reminded myself that
I forgot to include the stipulation that the sets be monotonically
increasing.
A set cannot be "monotonically increasing" in any mathematical sense.
So, you're correct. If the values fluctutate all over the
place you have a more complicated situation, but for formulas where x
increases as y increases, IFR applies.
Not til it has beeen debugged.
Since TO is so inept at definitions, he may never get things sorted out.
If you are measuring a set over the entire positive domain
What sort of sets are being "measured"? And how is "positive
domain" defined with respect to that set?
Sets of real numbers (not ALL of them), mapped from the naturals, in
quantitative order, over the entire domain of the reals, that is,
over the real number line.
Does TO mean the image of an order preserving mappping from the naturals
to the reals?
then we use x=1 and y=N,
As most standard set theories prohibit a set being a member of
itself, what sort of theory is TO using in which he can have y a
member of N and simultaneoulsy equal to N?
Remember, N is the length of the real line, the count of the
hypernaturals, and of the reals in the unit interval. Let me call it
BigOne from now on, to avoid "confusion".
Then is TO now distinguishing the set of naturals from his whatever it
is "bigone" beast?
Can you explain exactly how you think boundedness affects the set
size, directly? What's the formula for that? As an example, in
the naturals starting at 1, if n is a natural and p_n is its
position in the set, n=p_n.
Not in vN. And what is not true in EVERY model is not required by
the axiom system.
And, in any case, what is true for every BOUNDED set of naturals
need not be true for any UNBOUNDED set of naturals.
And no TO hand waving will change that.
Yeah, well, I'm asking how you think boundedness tells you
something about the set size. Like, what does it equal?
In "serially ordered" sets, where any non-extreme member has exactly one
predecessor ans exactly one successor, bounded sets are finite and
unbounded one are infinite, at least in mathematics.
Bounded subsets of the naturals have finite cardinality. Unbounded
ones don't.
Sorry, but you'll have to explain how I would know whether
"boundedness" (as sketched out above) is an "equality between
expressions..." or not. Could you give examples of "equality
between expressions other than simply "x=x", or is that the
only case it covers?
Sure. Let's have the set of squares of naturals, S, and go ahead
and apply the Inverse Function Rule.
TO's last attempt to define his "inverse function rule" was such a
complete disaster, that he must produce a workable definition
before any argument based on "inverse function rules" is
acceptable.
S is defined by f(n)=n^2,
What are the domain and range of f? If they are both some set of
naturals, N, then f does not have an inverse function at all.
What are you on? If f(n)=n^2, and f(g(n))=g(f(n))=n, then g(n) is
sqrt(n), at least in the positive domain.
What about g(2)? Or g(3), and so on. Accordint to TO's last definition,
f(g(n)) must be defined for each n in N, so what is g(2)?
so g(n)=sqrt(n),
This g is not inverse to f on N. Following garbage deleted.
So, for n>0, sqrt(n^2)<>n?
Every inequality is based on a difference, right?
Wrong! Order inequalities, like x < y or x < = y, are based on
ORDER RELATIONS on the set in which the objects occur.
If there were no difference between x and y, then x<y would be false
and x<=y would only be true because x=y.
TO has it backwards, if neither x < y nor y < x , THEN x = y. The order
comes first before there can be a direction of inequality. Absent order,
one can never say x < y or x > y, only x =/= y.
This same rule will still give finite values for the infinite
series having the same formula IF AND ONLY IF the number of
non-zero bit is mathematically finite.
If any string has a unlimited number of non-zero bits, then any
"sum" of this form will have an unlimited number of terms greater
than equal to 1, and will thus be a divergent series, and will NOT
have any finite value.
I know that's what you think.
Does TO claim that a sum of endlesly many positive naturals has a finite
sum?
what is the point here? I'm missing it.
That infinitely many non-zero naturals can "add up" to more tha any
finite value.
If there ARE infinitely many, which is at issue here.
There are endlessly many, which makes a sum of endlessly many positive
naturals. if TO claims this is a finite sum let him state its value.
Since it has been explained so clearly by so many, TO's intrasigent
position can only be a result of invincible ignorance or deliberate
trolling.
Or, clear vision of an alternative perspective.
Anything knowingly in violation of all the systems extant, which TO's
maunderings are, is eitherf invincible ignorance or deliberate trolling.
Most of my studies in
this area have been for my own edification
In that respect, they have been total failures.
but you've changed all
that Virgil. Because of you, and you alone, I must vanquish this
theory.
TO may try, but will fail as all such fools fail.
You have given new meaning to my life, to take meaning away
from yours.
It will provide considerable amusement to those who have not yet
killfiled all your threads to watch TO try.
But in the matter of mathematical creativity, TO is a eunuch, he would,
but he can't.
.
- References:
- Re: Logarithm of transfinite numbers
- From: Tony Orlow
- Re: Logarithm of transfinite numbers
- Prev by Date: Re: Prime numbers and finite sets
- Next by Date: Re: Logarithm of transfinite numbers
- Previous by thread: Re: Logarithm of transfinite numbers
- Next by thread: Re: Logarithm of transfinite numbers
- Index(es):
Relevant Pages
|
Loading
