# Thread: The Golden Ratio Award Nominations 2016

1. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by Neverfly
Anything by 7777777 is fair game...
listen to my warning:

don't play with fire

2. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by 7777777
listen to my warning:

don't play with fire

Largest Integer...

3. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by Neverfly

Largest Integer...
who said so

4. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by Neverfly

Largest Integer...
Perfect, no?

5. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by grapes
Perfect, no?
It is certainly logic that cannot be argued with...

6. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by Neverfly

Largest Integer...
who said so

7. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by 7777777
ℵ 0 (aleph-naught, also aleph zero or the German term Aleph-null) is the cardinality of the set of all natural numbers, and is an infinite cardinal. The set of all finite ordinals, called ω or ω0 (where ω is the lowercase Greek letter omega), has cardinality ℵ 0. A set has cardinality ℵ 0 if and only if it is countably infinite, that is, there is a bijection (one-to-one correspondence) between it and the natural numbers.

The easiest case is with the transfinite cardinals. I'll use N-0 to denote aleph-0.

Note that although we use a "+" and "=" signs, they don't have their usual meaning.

The set of the natural numbers, N = {1,2,3,...} has cardinality N-0. Tee union {0} U N = {0,1,2,...} has cardinality N-0+1 = N-0
So. N-0 + 1 = N-0. By considering the cardinalities of various sets we deduce that
N-0 - 1 = N-0
N-0 + N-0 = N-0 => n*N-0 = N-0 where n is a natural number.
N-0^n = N-0, where n is a natural number.
2^N-0 = N-0^N-0 = N-1

The position within a set is called ordinal and and in your case since the set is a bijection with the set of natural numbers the ordinals are 1- aleph0 , all except of the last being natural transfinite numbers.

(This follows from the fact that a countable union of countable sets is countable, one of the most common applications of the axiom of choice.) This fact is analogous to the situation in ℵ 0: every finite set of natural numbers has a maximum which is also a natural number, and finite unions of finite sets are finite.

This is harder than most explicit descriptions of "generation" in algebra (vector spaces, groups, etc.) because in those cases we only have to close with respect to finite operations—sums, products, and the like. The process involves defining, for each countable ordinal, via transfinite induction, a set by "throwing in" all possible countable unions and complements, and taking the union of all that over all of ω1 +1.

In the end you will find the "position" being infinite. And yes this is not . If an infinite set's cardinality lies between that of an infinite set S and that of the power set of S, then it either has the same cardinality as the set S or the same cardinality as the power set of S. That is, for any infinite cardinal there is no cardinal such that < < .This is a generalization of the continuum hypothesis since the continuum has the same cardinality as the power set of the integers.

First, define a function over the natural numbers (that is, over the finite ordinals) f : N → Power(X), so that for every natural number n, f(n) is the set of finite subsets of X of size n (i.e. that have a bijection with the finite ordinal n). f(n) is never empty, or otherwise X would be finite (as can be proven by induction on n).

The image of f is the countable set {f(n)|n ∈ N}, whose members are themselves infinite (and possibly uncountable) sets. By using the axiom of countable choice we may choose one member from each of these sets, and this member is itself a finite subset of X. More precisely, according to the axiom of countable choice, a (countable) set exists, G = {g(n)|n ∈ N}, so that for every natural number n, g(n) is a member of f(n) and is therefore a finite subset of X of size n.

Expressed in category-theoretical terms, a set A is Dedekind-finite if in the category of sets, every monomorphism f : A → A is an isomorphism. A von Neumann regular ring R has the analogous property in the category of (left or right) R-modules if and only if in R, xy = 1 implies yx = 1. More generally, a Dedekind-finite ring is any ring that satisfies the latter condition. Beware that a ring may be Dedekind-finite even if its underlying set is Dedekind-infinite, e.g. the integers.

Since homomorphisms are morphisms in an appropriate category, we may consider the analogous specific kinds of morphisms defined in any category. However, the definitions in category theory are somewhat different. For endomorphisms and automorphisms, the descriptions above coincide with the category theoretic definitions; the first three descriptions do not. In category theory, a morphism f : A → B is called:
monomorphism if f ∘ g1 = f ∘ g2 implies g1 = g2 for all morphisms g1, g2: X → A, where "∘" denotes function composition corresponding to e.g. (f ∘ g1)(x) = f(g1(x)) in abstract algebra.
isomorphism if there exists a morphism g: B → A such that f ∘ g = 1B and g ∘ f = 1A, where "1X" denotes the identity morphism on the object X
epimorphism if g1 ∘ f = g2 ∘ f implies g1 = g2 for all morphisms g1, g2: B → X. (A sufficient condition for this is f having a right inverse)

For instance, the inclusion ring homomorphism of Z as a (unitary) subring of Q is not surjective (i.e. not epi in the set-theoretic sense), but an epimorphic in the sense of category theory. This inclusion thus also is an example of a ring homomorphism which is (in the sense of category theory) both mono and epi, but not iso.

Any homomorphism f : X → Y defines an equivalence relation ~ on X by a ~ b if and only if f(a) = f(b). The relation ~ is called the kernel of f. It is a congruence relation on X. The quotient set X / ~ can then be given an object-structure in a natural way, i.e. [x] ∗ [y] = [x ∗ y]. In that case the image of X in Y under the homomorphism f is necessarily isomorphic to X / ~; this fact is one of the isomorphism theorems. Note in some cases (e.g. groups or rings), a single equivalence class K suffices to specify the structure of the quotient, in which case we can write it X/K. (X/K is usually read as "X mod K".) Also in these cases, it is K, rather than ~, that is called the kernel of f (cf. normal subgroup).

We can start with making the natural numbers . Let us assume there is a set with the following properties.
1. Two unique elements denoted
2. We have a function
3. is injective
4. We have
5. We have no such that
6. Induction is applicable on

We then define this as the set . You may notice I did not include 0 while in the article they do, while it does not introduce an additional axiom I prefer it as it makes it possible to make the naturals into a proper topological semiring. Already here can we deduce a few of it's properties, one is that for any we have that . Some detractors may start yelling I am not using a natural number n already and hence presuming their existence, but I am. For that to be valid I must be using their existence as an assumption in my attempt to prove them. Not using our ordinary concept of natural numbers as merely a shorthand notation to ease reading and reasoning. Another we see is from this definition. This is indeed an equivalence relation. Refexivity is just the fact that
the sequences are Cauchy. Symmetry is obvious.The algebraic structure to be preserved may include more than one operation, and a homomorphism is required to preserve each operation. For example, a ring has both addition and multiplication, and a homomorphism from the ring (R, +, ∗, 0, 1) to the ring (R′, +′, ∗′, 0′, 1′) is a function such that f(r + s) = f(r) +′ f(s), f(r ∗ s) = f(r) ∗′ f(s) and f(1) = 1′ for any elements r and s of the domain ring. If rings are not required to be unital, the last condition is omitted. In addition, if defining structures of (e.g. 0 and additive inverses in the case of a ring) were not necessarily preserved by the above, preserving these would be added requirements.

Definition of infinite sets
A set is infinite if there exists an injection from to a proper subset

To see this we let be such that and , to see it we define as following, let , then , we clearly see its a member of our set T, and that it is injective is equally evident hence our set natural numbers is definitionally infinite.

From this definition we can show that it has all the properties of natural number addition. In the document provided they define , here we don't have to as we can just prove it with this

Quaint isn't it? For multiplication we do similarly.

In the document they don't prove multiplication by 0 is 0, it's fairly easy, we have

at which we have it must be 0. From here on it's just a mere formality verifying all the properties which is done to large degree in the document, read it or feel free to do it here! I am not. We will however impose an ordering on a set of naturals. We say that if for some k, if then we have , which clearly gives us that always.

The next step is to go from our full semiring to a fullfledge topological ring , in a previous post I showed a special case. In general for a semiring we can turn it into a ring by defining the operations of additoin and multiplication on as

and then taking the quotient of this by the ideal
and we get a full fledge ring out of a semiring. What's wonderful here is if we have an ordering on then we can extend it to by doing this, we say that if and only if there exist a such that and . We have the canonical function such that , which is injective. This can of course be done to natural numbers and we get

With the normal ordering and all evidently true. A natural question is if it's still infinite. We can simply construct a function if , otherwise and see it's injective, hence it's cardinality is the same as naturals and hence infinite. By our extension of ordering we can also see that

Cauchy sequences are not easily defined, they fullfill all criterias of a semiring, that null sequences forms an ideal demonstrates easily that all all operations are well-defined in the true sense of the word. It is also easy to see that completion renders the cardinality of the set greater than the original set if it's not already complete. Let be such that , it is evidently surjectively injective as . So we have that , however the ring is complete only if this function is an isomorphism, so for others it is not an isomorphism and hence not surjective, and as such we get for non-complete rings that .Check that this is an equivalence relation on
S. Let Z be the set of equivalence classes under this relation.But, now, alas we have lost any semblance of the minimal element property which was so important and desirable. In other words, there are now subsets of Q
which are bounded below with no minimal element in sight. (Not just in the set, but not even in Q).

For real numbers we have the underlying set being rationals and the norm just being the standard norm of rational numbers. For real numbers we can easily see that we can develop our standard notation of infinite decimals by having the decimal point alone being of the form by having the corresponding cauchy sequence being , using the normal algoritms for many we can see this is true for rationals, irrationals and much else.
I'm not sure where all this comes from, but the last part of the second-to-last paragraph appears to be directly lifted (and out of context) from this paper (Section 5, page 17):
https://www.scribd.com/doc/317032966...-de-numere-pdf
Let Z be the set of equivalence classes under this relation. But, now, alas we have lost any semblance of the minimal element property which was so important and desirable. In other words, there are now subsets of Q which are bounded below with no minimal element in sight. (Not just in the set, but not even in Q).
That paper, Construction of Number Systems, by N. Mohan Kumar, is posted to a few other Scribd sites, some of which are behind pay walls.

8. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by grapes
I'm not sure where all this comes from, but the last part of the second-to-last paragraph appears to be directly lifted (and out of context) from this paper (Section 5, page 17):
https://www.scribd.com/doc/317032966...-de-numere-pdf

That paper, Construction of Number Systems, by N. Mohan Kumar, is posted to a few other Scribd sites, some of which are behind pay walls.
Originally Posted by Neverfly
It feels just fine as a cardinal number. An infinite string with a value of one added to it will still remain an infinite string. Mathnerd already pointed out that would be different as a hyperreal or transfinite ordinal number.
But... I notice you dodged and avoided answering emperorzelos' direct question... Let me help you out be Re-Quoting it for you here:

See?
When confronted, you dodge.
He also made a thread, which he linked to in that same post where he points out:

Originally Posted by emperorzelos

From this definition we can show that it has all the properties of natural number addition. In the document provided they define , here we don't have to as we can just prove it with this

Quaint isn't it? For multiplication we do similarly.

In the document they don't prove multiplication by 0 is 0, it's fairly easy, we have

at which we have it must be 0. From here on it's just a mere formality verifying all the properties which is done to large degree in the document, read it or feel free to do it here! I am not. We will however impose an ordering on a set of naturals. We say that if for some k, if then we have , which clearly gives us that always.

The next step is to go from our semiring to a fullfledge ring , in a previous post I showed a special case. In general for a semiring we can turn it into a ring by defining the operations of additoin and multiplication on as

and then taking the quotient of this by the ideal
and we get a full fledge ring out of a semiring. What's wonderful here is if we have an ordering on then we can extend it to by doing this, we say that if and only if there exist a such that and . We have the canonical function such that , which is injective. This can of course be done to natural numbers and we get

With the normal ordering and all evidently true. A natural question is if it's still infinite. We can simply construct a function if , otherwise and see it's injective, hence it's cardinality is the same as naturals and hence infinite. By our extension of ordering we can also see that

which gives us that for and clearly and equally clearly we have that from natural number ordering, hence we have . As such by simple manner of construction we've seen that 7777777's claim is definitionally wrong. It's not only stupid at face value it is simply definitionally wrong.

Why have you not addressed any of that, either?

Look here

9. ## Re: The Golden Ratio Award Nominations 2016

Originally Posted by 7777777
Look here
Yikes. So, some of your post is from other users' posts, and some from other websites? Is that it?

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