## Mathematicist Genesis

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I'll be in quarantine for a couple of weeks soon. I shall try and get the field of real numbers with its order defined in that time.

Oh are you or a loved one I'll? Sorry to hear. Be safe!
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Oh are you or a loved one I'll? Sorry to hear. Be safe!

Nah. Couldn't see my partner for a while, travelled to see her, quarantine restrictions got reimposed while I was here!
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I shall try and get the field of real numbers with its order defined in that time.

Look forward to this! Next, well-order an uncountable set of reals. :cool: :up:
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Nah. Couldn't see my partner for a while, travelled to see her, quarantine restrictions got reimposed while I was here!

Ah okay. That actually sounds quite nice.
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This is a very ambitious thread. But probably no more than the first 900 pages of Penrose's The Road to Reality. I may not live long enough to see its completion, but you guys are younger, so there is hope. :worry:
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This is a very ambitious thread.

Mostly because fdrake majorly upped the ante on the level of rigor I expected. I had hoped for a dozen or two posts about the size (and level of detail) of my OP, but instead he started giving us so much more.
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This is a very ambitious thread. But probably no more than the first 900 pages of Penrose's The Road to Reality. I may not live long enough to see its completion, but you guys are younger, so there is hope. :worry:

That's one of those books which is in dire need of a Miracle operator.

(vague description of mathematics)
A miracle occurs.
Now you try!
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What properties then characterise the ordering of naturals and fractions, and distinguish them from the ordering of province and country or the ordering of classifications of biological kinds?

On the first day of ordering, the Lord said let $\mathbb{N}$ be the set of all natural numbers, then for any pair of natural numbers x and y, y is the greatest and cometh after x if and only if x is contained within y and, conversely, y is the lesser of x and cometh earlier if and only if x containeth y:

$\forall x,y \in \mathbb{N}$
$y \gt x \leftrightarrow x \in y$
and
$y \lt x \leftrightarrow y \in x$

And thus the natural numbers were ordered and it was good.

On the second day of ordering, the Lord said let $\mathbb{Z}$ be the set of integers, then for any pair of integers x=(a,b) and y=(c,d), y be greater than x if and only if c+b > a+d:

$\forall x=(a,b) \in [x], y=(c,d) \in [y], [x],[y] \in \mathbb{Z}$
$[y] \gt [x] \leftrightarrow c+b \gt a+d$

and [y] be equivalent of [x] if and only if c+b = a+d. And thus the integers were ordered and it was good.

On the third day of ordering, the Lord said let $\mathbb{Q}$ be the set of all rationals, then for any pair of rationals x=(a,b) and y=(c,d), y be greater than x if and only if c*b > a*d:

$\forall x=(a,b) \in [x],y=(c,d) \in [y], [x],[y] \in \mathbb{Q}$
$[y] \gt [x] \leftrightarrow c*b \gt a*d$

and [y] be equivalent of [x] if and only if c*b = a*d. And lo the rationals were ordered and it was good.

On the fourth day of ordering, the Lord realised that Dedekind was late producing the reals and verily he flipped his fucking lid. He spake unto Dedekind:

Dedekind, though the set of rationals be infinite in number and uncountable in density, they be still discontinuous in separation. Go to the cutting the place and bring me forth a continuous set of numbers that I might truly count the horrors in store for Man. — The Lord

And Dedekind heard the Lord and said it would be done. He took the set of rationals and a saw, and he ordered the rationals as the Lord had commanded and he took his saw and he cut them in half and he spake to the Lord:

This sort of thing? — Dedekind

and the Lord looked upon the two sets, which he named A and B, and saw that A was infinite in number and that for any rational x in A, there was always a higher rational, for between any pair of rationals there exist an infinity of intermediate rationals as spaketh by @fdrake. But no matter which x he took, he could find no element y in B that was the equal or the lesser of it. The Lord sought for the smallest rational in B but found that, for any he looked upon, he could find a lesser. And the Lord spake unto Dedekind:

What number didst thou cut at, and upon which side did that number fall? — The Lord

and Dedekind confessed to the Lord that he did not know because he did not look. And the Lord spake to him that had Dedekind cut exactly at a rational x, then x would be either the greatest rational in A with none greater, or the least rational in B with none lesser, and that since neither A had a greatest rational nor B a lesser, the number at which Dedekind had cut could not be a rational. And he spake further unto Dedekind:

Be warned, for shit just got real. — The Lord

And he bade Dedekind make his cuts upon the set of rationals an infinite number of times and also bade him to be finished before midnight lest they run into a fifth day of ordering, and Dedekind did saw impossibly and created the set of real numbers, $\mathbb{R}$.

The Lord looked upon the set and said, Let A be the set of lesser rationals and B the set of greater, and (A,B) be a real number greater than the greatest of A and less than or the equal of the least of B. And let (C, D) be another such real. Then if A is a subset of C, then (A,B) is the lesser of (C,D):

$\forall x=(A,B),y=(C,D) \in \mathbb{R}$
$y \gt x \leftrightarrow A \in C$

and if A and C be brethren, then the two reals are as one. And he looked upon the ordering of the reals and it was good. Actually better than the ordering of the rationals.

*I'm figuring we covered union as part of generators.
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Christ, I think it'll take me longer to debug the mathjax than it did to write the comment.
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:clap: :clap: :clap: :rofl: :party: :fire: :100: :sparkle:

That's the kind of post I was hoping this whole thread would be. I didn't want to complain since fdrake was producing such good informative content, but it was the combination of that content with Biblical style that I was really hoping for in this thread, and you nailed it!
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Cool. I think there's a little more to do on multiplication before starting on the complex numbers, after which it seems sensible to head into group theory. Let us know if we're skipping over anything important.

I haven't really taken care to enumerate the axioms. For instance, I've dealt with selecting numbers from infinite sets without stating an axiom of choice. This is mostly because I haven't really unpicked the axioms covered so far. Might make a good summary post.

I'm also thinking that category theory has been handwaved a bit, but that can be put on a more formal footing once morphisms are covered in group theory. Technically that goes rather against the bottom-up approach you wanted but there's always something isn't there.
• 983
∀x=(a,b),y=(c,d)∈Z
y>x↔c+b>a+d

$\left( c,d \right)\in Z'=N\times N$
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Do you mean x is in Z and y in Z'? I thought the above was okay but I'm not used to it tbh.
• 983
x=(a,b) and y=(c,d) are ordered pairs of integers, not integers themselves. So x and y belong to the Cartesian product of the integers. You're defining an inequality of ordered pairs in terms of an inequality of integers.

Calling fishfry! :cool:

Christ, I think it'll take me longer to debug the mathjax than it did to write the comment.

MathType works well - it's WYSIWYG - using the Wikipedia cut and paste option. You only have to replace [itex] with [....]. Just a thought.
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I thought he was defining the integers x and y as (equivalence classes of) ordered pairs of naturals (with the equivalence class part implied by saying that when the ordered pairs of naturals return the same value under subtraction then the “two” integers thus defined are the same integer).
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I thought he was defining the integers x and y as (equivalence classes of) ordered pairs of naturals (with the equivalence class part implied by saying that when the ordered pairs of naturals return the same value under subtraction then the “two” integers thus defined are the same integer).

Yes, that's right.
• 983
A matter of notation and mathematical clarity. See definition 7.1

https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf

Not a big deal. You are to be applauded for wading into this.
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A matter of notation and mathematical clarity. See definition 7.1

https://www.math.uchicago.edu/~may/VIGRE/VIGRE2011/REUPapers/Lian.pdf

Not a big deal. You are to be applauded for wading into this.

It's worth getting right. You're quite right. The set was Z not Z', but I was representing the elements of Z as ordered pairs which is invalid, since they are sets, specifically equivalence classes. Further, rightly, I should have said y is equivalent to x, not equal to, when defining the equivalence relation.

I have edited the above. Can you check this? I'm not that practised at this.

$\forall x=(a,b) \in [x], y=(c,d) \in [y], [x],[y] \in \mathbb{Z}$
$[y] \gt [x] \leftrightarrow c+b \gt a+d$

is supposed to read, for each ordered pair x = (a,b) (from the set Z') in the integer [x] and y = (c,d) in the integer [y], where [x] and [y] are in Z, [y] is greater than [x] if and only if c+b > a+d.
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The Book of Imaginary Things

The Lord looked upon the naturals once more, reminding Himself that:

$a \cdot 0 = 0$
$a \cdot S(b) = a + (a \cdot b)$

and He said, let S(b) = a, and thus

$a \cdot a = a + (a \cdot (a-1)) = a + (a + a\cdot (a-2)) = ...$
$= \sum_{i=1}^a a$

And he said, Let this be denoted as $a^2$, and Me denote $a^3 = a \cdot a \cdot a$ and $a^n = \prod_{i=1}^n a$.

Then he asked, What then is $a^{m+n}$? And seeing that he could write $\prod_{i=1}^{m+n} a$ as

$a^{m+n} = \prod_{i=1}^m a \cdot \prod_{i=1}^m a$

he saw that $a^{m+n} = a^m a^n$.

And thus on that day the Lord created $a^m a^n$. And he said, Let n=d, and thus He named him $a^d a^m$.

He then asked, what is the exponent k such that $a^{m+n+k} = a^n$? And He remembered he had fought this devil before, for the answer lay not in the set of naturals, but in the integers.

Thus, he said, let k, m, n be integers and not naturals such that $m+n+k = n \rightarrow k=-m$, and therefore:

$a^{-m} \cdot (a^m a^n) = a^n$

such that $a^{-m}$ be the inverse of $a^m$. And further that since division be the inverse of multiplication, $a^{-m} = \frac{1}{a^m}$. And thus $a^0 = a^n a^{-n} = 1$.

Then He asked, what is $(a^n)^m$? From the above, He knew that $(a^n)^0 = 1 = a^0$ and $(a^n)^1 = a^n$, and He conjectured that $(a^n)^m = a^{m*n}$ and saw that it was true, since:

$(a^n)^m = (a^n)^{m-1} \cdot a^n = a^{n(m-1)} \cdot a^n = a^{nm - n + n} = a^{nm}$

(proceeding by induction).

Then He asked, what is m such that $(a^n)^m = a$ and saw that the answer lay not in the integers but in the rationals, since it must from the above be that $m=\frac{1}{n}$ such that $(a^n)^m = a^{mn} = a^{n/n} = a$.

Thus He turned his eye (for he had but one) to the rationals and he named these inverses of exponents roots.

And finally He turned his eye to the reals and asked, What is x such that $x^2 = -1$? And He looked but did not see among the reals an answer to this question, for $x^2 \ge 0 (\forall x \in \mathbb{R})$.

And knowing that for closure under subtraction He had to put aside the naturals for the integers, and for closure under division He had to put aside the integers for the rationals, He knew once again that he had been failed by his creations, and must create anew for His work to be good.

And he spake:

Let i be such that $i^2 = -1$, then $i^3 = -i$, $i^4 = 1$, and finally $i^5 = i$.
And let $(0 \cdot i)^2 = 0$ show that the real 0 and this spirit i lie on the same scale, and that any multiplier x of i yield $x \cdot i$ and move us along that scale, just as $x \cdot 1$ move us along the ordered real line.
And let it be seen that, whereas for any sum of two reals $a + b$, there exists a number c such that $a \cdot c = a + b$, there is no pair of reals x & c such that $x + i = c \cdot i$, since $c = \frac{c+i}{i}$ cannot be real if x is real and vice versa, and thus the addition of i to x doth not move us along the real line, and the addition of x to i doth not move us alone the line to which i belongs.
And let us name this line the imaginary number line, and let i be known as the imaginary number, for only I the Lord can imagine it.
And let the sum of real and imaginary numbers be called complex, and be written $a + b \cdot i$, where a and b are real numbers that scale 1 and i respectively.
And thus for each pair of real numbers a and b, there exists a complex number $a + b \cdot i$, and this defines the set of complex numbers $\mathbb{C}$.
And since the real and imaginary parts are so orthogonal, let not this set be ordered, but only subsets of fixed a or b be ordered.
— The Lord
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And thus on that day the Lord created aman
a
m
a
n
. And he said, Let n=d, and thus He named him adam
a
d
a
m
.

(quotes butcher mathjax but you know what I mean)

:rofl: :clap: :up: :fire:
• 983
And thus on that day the Lord created aman

Whoa, not so fast! Where is the link between complex numbers and the existence of man?

Inquiring minds want to know. :chin:
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And on that day, the Lord created $a^m a^n$.

Which fulfills @Pfhorrest 's original goal. Thread complete.
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You should keep it and publish, this is great stuff.
• 1.9k
Haha thanks!

Morphisms: Book I

And the Lord spake unto Morph and said:

Take the set of natural numbers $\mathbb{N}$, and count the ways that ye can fiddle with them, and note the result of your fiddling such that you can tellest me what thou did and what result it did have. — The Lord

And Morph did take the set of natural numbers, and the first thing he did was to take two elements from it and form of them an ordered pair as he had seen the Lord do:
$f(x, y \in \mathbb{N}) = (x, y)$

The second thing he did was to remove 0 and to put it back, then repeat for 1 and 2 and so on:
$f: \mathbf{N} \rightarrow \mathbf{N}$

The third thing he did was to swap 0 and 1:
$f: \mathbf{N} \rightarrow \mathbf{N} (f(0)=1, f(1)=0, f(n\gt1) = n)$

The fourth thing he did was to take an element at random and add it to all of the other elements and itself, and he saw that he had kept all the elements greater than or equal to the random element:
$f_{y \in \mathbf{N}}: \mathbf{N} \rightarrow \mathbf{N} + y = \mathbf{N'} \subset \mathbf{N}$

The fifth thing he did was to take an element > 0 at random and multiply with it all of the other elements and itself, and he saw that the sum of two such fiddles was the equal of the fiddle of the sum:
$f_{y \in \mathbf{N}}: \mathbf{N} \rightarrow y \cdot \mathbf{N} = \mathbf{N'} \subset \mathbf{N}$

The sixth thing he did was to take the element 0 = {} and multiply with it all of the other elements and itself, and he saw that he was left with unity:
$f_{0}: \mathbf{N} \rightarrow 0 \cdot \mathbf{N} = 1 \subset \mathbf{N}$

And thus Morph spent his days, fiddling with the natural numbers and noting the achievements of his fiddling, and he spake unto the Lord:

Lord, I have fiddled all of the members of the set of natural numbers, and I have seen the following:

First, for some fiddles I performed, I ended up with an object unlike the set of natural numbers, such as to select two and make an ordered pair.

Second, for some fiddles I performed, I ended up with a subset of the original set, such as to add numbers > 0 or to multiply by numbers different to 1.

Third, for some fiddles I performed, I ended up with the same set, though the elements had been adulterated or moved.

Fourth, for some fiddles I performed, not even the elements were affected.
— Morph

And the Lord heard Morph and said:

I have listened to your Morphisms and I am pleased. For you have discovered some Morphisms that leave each element in the set unaltered and thus the set as a whole unaltered, and we shall call this the identity morphism.

And you have discovered that some Morphisms alter the elements but leave the set as a whole unaltered, and we shall call these automorphisms.

And you have discovered that some Morphisms, such as the additions, alter the set but these alterings may be reversed by inverses called subtractions, and we shall call these isomorphisms.

And you have discovered that some Morphisms such as the multiplications act on sums of elements by acting on each element and summing, and we shall call these homomorphisms. And you have discovered further that some homomorphisms are not isomorphisms, such as multiplication by 0 which, in yielding unity, yields both an element and a subset of the set, but from which we cannot return to the set by any morphism bar the union with that set.

And finally you have discovered that some Morphisms do not yield objects of the same type you began with, such as the ordered pair you obtained from the set.
— The Lord

And He instructed Morph to continue to explore this set, and the other sets, and he spake:

Let the set of objects that you fiddle with be the domain of the morphism, and let the set of objects you end up with belong to the codomain, such that, in the additions and the subtractions, the codomain and the domain be as one, but the object yielded not be all that lies in the codomain but only that yielded by that morphism acting on that domain, and we shall call this subset of the codomain the image of the morphism. — The Lord

And Morph heard this and said:

What? — Morph

And the Lord in rage did smite Morph upon his head, causing Morph such pain and anguish that the Lord did feel sorry for him and gave him two white pills and a glass of water. And Morph swallowed the pills with the water and his head did heal and he was glad. And Morph did ask of the Lord:

My Lord, what do you call this small parcel of cephalic improvement?

and the Lord said, We shall call it the cocodomain and the Lord was booed for it was not good.
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You crack me up.... Best nerd jokes I ever read. :rofl:
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:blush:
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You should publish this.
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bring me forth a continuous set of numbers that I might truly count the horrors in store for Man. — The Lord
• 487
As I understand it, we’re really saying “all objects with this structure have these properties”, but that’s technically true whether or not there “really” are any objects with that structure at all.

Such as Odd Perfect Numbers https://medium.com/cantors-paradise/eulers-odd-perfect-numbers-theorem-82a393baa883
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