What do you call a function that extracts an algebraic number from a transcendental number?

[Kharakov]

I mean, other than a transcendental function.

So if I extract (-1) from pi with cosine, I'm using a transcendental function on a transcendental "number". But is there another more specific term, like "decomposition function", or something like that to describe getting a rational or algebraic from a transcendental?


Detranscendentalizing?



Other thing, I was thinking about how there is a 1 to one mapping of rationals to transcendentals with the following transcendental generating function (which presumably can generate all transcendentals >= .693....).

Is this standard? I remember there are uncountable infinites of reals, but countable infinites of algebraics. Would the following directly map an algebraic (x) to every transcendental value > log(2)?


There is a set of numbers defined by (with k= to the number of radicals + 1):

\( Q_{x,n,k} = \left( nx^{n-1} \right)^k \, \times \left[x - \sqrt[n]{x^n-x + \sqrt[n]{x^n-x + \sqrt[n]{x^n-x+...}}} \right]\)

with constants associated with the numbers defined by:

\(q_{x,n} = Q_{x,n,k \to \infty}\)

with

\(q_{2,2} = \pi^2\)


Because one can increase or decrease x by a rational amount, one can adjust what transcendental one generates by increasing or decreasing x by whatever rational one desires. This means transcendental values are directly mapped to rational values with this function.

Does this mean there are the same amount of transcendentals as rationals?

 

[Bomb#20]

No. It means your presumption that your generating formula can generate all transcendentals >= .693... is wrong. Why would you assume that increasing or decreasing x by a rational will always let you hit whatever transcendental someone picks? For instance, here's a famous transcendental number: L = 0.110001000000000000000001... (the digits are 1 in positions 1, 2, 6, 24, 120, ... n!, ...; otherwise 0). Sure, there's bound to be some x you can plug in to make (1+L) come out of your q function -- you just invert q_n -- but what reason do you have to think x = q_n^-1(1+L) is rational?

 

[steve_bank]

The answer to ne is simple. In applied math we truncate and round to the level needed to maintain accuracy and minimize error propagation in calculations.

The general rule is to carry more decimal places for constants than the number of decimal places in the data.

3.14159 is burned into my brain. It was what I generally used until pi was a button on a calculator, then limited to 8 or 9 places.

It doesn't answer your question from a theoretical view, it is how I view it. I started out with a slide rule. pi was a highlighted spot on a scale.

 

[Kharakov]

your presumption that your generating formula can generate all transcendentals >= .693... is wrong.

I didn't get back in time to mention my reservations- which were numbers such as pi+3, or whatever. I just thought (without thinking deeply enough), that there would be a perfect continuum, without thinking that in order to have a perfect continuum generated by the function, I might need the transcendentals as well (to generate the rationals... right?).

I assumed that because every rational would generate a slightly larger or smaller transcendental, that they would smoothly generate all of them. However, as I lay in my friend's van, thinking about it, I thought "what about pi^2+1? What ensures that it is generated by a rational number? "

I was envisioning a smooth continuum of transcendentals generated by varying x, without varying another component (1 to 1 mapping), without acknowledging that it doesn't seem likely that some higher transcendental would follow the exact same pattern as a lower transcendental... offhand.

Why would you assume that increasing or decreasing x by a rational will always let you hit whatever transcendental someone picks? For instance, here's a famous transcendental number: L = 0.110001000000000000000001... (the digits are 1 in positions 1, 2, 6, 24, 120, ... n!, ...; otherwise 0). Sure, there's bound to be some x you can plug in to make (1+L) come out of your q function -- you just invert q_n -- but what reason do you have to think x = q_n^-1(1+L) is rational?

I'd think 1/L would be generated by some x and n, but once again, why would I assume that it is generated by a rational or algebraic number?


If x and n are both algebraic, can the equation generate every transcendental? If x and n are both rational, can it?

 

[steve_bank]

pi+3 is going back to the debate on infinity. What is 1 + 0.333...

I'd say the answer is adding a finire number to an infinite quantity is undefined to begin with . There is no way to quantitatively evaluate it.

sin( 2 * (3.13159...) * x) is not computable.

 

[steve_bank]

I am jumped into this without much foundation.

Algebraic means addition, subtraction, multiplication, and division.

https://en.wikipedia.org/wiki/Transcendental_function

If f is an algebraic function and alpha is an algebraic number then f(alpha ) is also an algebraic number. The converse is not true: there are entire transcendental functions f such that f(alpha ) is an algebraic number for any algebraic For a given transcendental function the set of algebraic numbers giving algebraic results is called the exceptional set of that function

https://en.wikipedia.org/wiki/Elementary_function

https://en.wikipedia.org/wiki/Algebraic_function

In mathematics, an algebraic function is a function that can be defined as the root of a polynomial equation. Quite often algebraic functions are algebraic expressions using a finite number of terms, involving only the algebraic operations addition, subtraction, multiplication, division, and raising to a fractional power

Looks like you are talking about is determining exceptional sets?







So this is what mathematicians do....

 

[Juma]

Why would the function y = x - 1 be called transcendental?

 

[steve_bank]

https://en.wikipedia.org/wiki/Transcendental_number


Pi (π) is the best-known transcendental number
In mathematics, a transcendental number is a real or complex number that is not algebraic—that is, it is not a root of a nonzero polynomial equation with integer (or, equivalently, rational) coefficients. The best-known transcendental numbers are π and e. Though only a few classes of transcendental numbers are known (in part because it can be extremely difficult to show that a given number is transcendental), transcendental numbers are not rare. Indeed, almost all real and complex numbers are transcendental, since the algebraic numbers are countable while the sets of real and complex numbers are both uncountable. All real transcendental numbers are irrational, since all rational numbers are algebraic. The converse is not true: not all irrational numbers are transcendental; e.g....

The set of transcendental numbers is uncountably infinite. Since the polynomials with rational coefficients are countable, and since each such polynomial has a finite number of zeroes, the algebraic numbers must also be countable. However, Cantor's diagonal argument proves that the real numbers (and therefore also the complex numbers) are uncountable. Since the real numbers are the union of algebraic and transcendental numbers, they cannot both be countable. This makes the transcendental numbers uncountable.

No rational number is transcendental and all real transcendental numbers are irrational. The irrational numbers contain all the real transcendental numbers and a subset of the algebraic numbers, including the quadratic irrationals and other forms of algebraic irrationals.

 

[steve_bank]

Why would the function y = x - 1 be called transcendental?

Can it be expressed as a polynomial with integer exponents?

 

[Bomb#20]

Why would you assume that increasing or decreasing x by a rational will always let you hit whatever transcendental someone picks? For instance, here's a famous transcendental number: L = 0.110001000000000000000001... (the digits are 1 in positions 1, 2, 6, 24, 120, ... n!, ...; otherwise 0). Sure, there's bound to be some x you can plug in to make (1+L) come out of your q function -- you just invert q_n -- but what reason do you have to think x = q_n^-1(1+L) is rational?

I'd think 1/L would be generated by some x and n, but once again, why would I assume that it is generated by a rational or algebraic number?


If x and n are both algebraic, can the equation generate every transcendental? If x and n are both rational, can it?

No and no, by a cardinality argument. The set of (x,n) pairs where x and n are both algebraic is countably infinite; the set of transcendentals is uncountably infinite. All rationals are algebraic, so one argument settles both questions.

 

[Kharakov]

If x and n are both algebraic, can the equation generate every transcendental? If x and n are both rational, can it?

No and no, by a cardinality argument. The set of (x,n) pairs where x and n are both algebraic is countably infinite; the set of transcendentals is uncountably infinite.

Can you explain the difference (countable and uncountable), Cantor's proof, and why it is indisputably valid in all cases? Is there another proof?

Varying n AND x allows one to control the starting point and "flow" of the values created. So you can start with the first digits of pi^2, but by varying n and x (to keep the first digits), you can create other transcendentals that are arbitrarily close to pi^2, starting to diverge by however much you'd like determined by whatever rational values of x and n you use. So varying x by some small amount, and n, can give you various incremental changes in the evolution of the number generated. You can increase x, and decrease n, keeping the same starting digits (an arbitrary amount determined by the rationals you use), varying the digits after a point however you would like, by selecting whatever difference in x and n you want to determine the number you create.


You need to use 2 infinite sets to do it though.... set of all x>1, and set of all n>1.

 

[steve_bank]

Kharakov

I don't get what your question is that you sent me.

Assuming the links are correct algebraic functions are addition, multiplication, subtraction, division. For a function to be algebraic it must be expressible as a polynomial with a finite number roots.

g(z) = 3z

If f(g(x)) is algebraic then f(x) is algebraic. is algebraic f(g(x) is algebraic.
From the links algebraic and transcendental functions are definitions.

As to the OP, sounds like you are trying to go back to an algebraic function passed to a transcendental function.

Given g(z) is algebraic and f(x) is transcendental you want to get back g(z) from F(x)?

 

[Bomb#20]

If x and n are both algebraic, can the equation generate every transcendental? If x and n are both rational, can it?

No and no, by a cardinality argument. The set of (x,n) pairs where x and n are both algebraic is countably infinite; the set of transcendentals is uncountably infinite.

Can you explain the difference (countable and uncountable), Cantor's proof, and why it is indisputably valid in all cases? Is there another proof?

Varying n AND x allows one to control the starting point and "flow" of the values created. So you can start with the first digits of pi^2, but by varying n and x (to keep the first digits), you can create other transcendentals that are arbitrarily close to pi^2, starting to diverge by however much you'd like determined by whatever rational values of x and n you use. So varying x by some small amount, and n, can give you various incremental changes in the evolution of the number generated. You can increase x, and decrease n, keeping the same starting digits (an arbitrary amount determined by the rationals you use), varying the digits after a point however you would like, by selecting whatever difference in x and n you want to determine the number you create.


You need to use 2 infinite sets to do it though.... set of all x>1, and set of all n>1.

Well, really, beero1000 is your man for explaining stuff like this. He seems to be on extended hiatus, so I'll give it a go, but keep in mind I'm just filling in so I don't have to do a good job.

I don't know of another proof besides Cantor, but that doesn't mean there isn't one. Cantor's proof show's there are more reals than integers, by a reductio ad absurdum. Assume there's a 1-to-1 mapping M from the positive integers to all of the reals between 0 and 1 that have no nines in their decimal expansions. So whatever that mapping is, M[1] = 0.abcde..., M[2] = 0.fghij..., and so forth, for some set of digits a, b, c etc., each of them a single digit from 0 to 8. Picture this mapping as an infinitely large 2-dimensional table of digits, with the M[j]'s running from 1 to infinity down from the top, and the digits of each number starting at the left and running infinitely off to the right. Now construct D = 0.agm... as the number formed by reading along the diagonal, so the nth bit of D is just the nth bit of M[n]. That's a number with a digit that matches every M at a given position. Then we make a change: we construct a new number that's sort of like D + 0.1111..., except that for all the 8's in the decimal expansion of D, instead of changing that digit from 8 to 9, we change it from 8 to 0. So for instance if D = 0.35828403..., then C = 0.46030514... This can be done no matter what D actually is. So we've constructed a number C that's different from M[1] at digit 1 after the decimal point; and C is different from M[2] at digit 2, and it's different from M[3] at digit 3, and so forth. So it's different from every number in the table, and that means C isn't in the table. It isn't one of the M[n] numbers, for any n at all. But C is a real number between 0 and 1, and all its digits are 0 to 8, because we changed all the 8's in D into 0's instead of into 9's. So the existence of C contradicts our original assumption that M is a 1-to-1 mapping M from the positive integers to all of the reals between 0 and 1 that have no nines in their decimal expansions. So the assumption must have been wrong. You can't put the positive integers into a 1-1 mapping with that subset of the reals. That means there are more reals in that subset than there are positive integers. When you add in all the reals we skipped -- i.e., the ones that have 9's in their decimal expansions and all the reals greater than 1.0 -- that doesn't bring the number of reals down. So there are more positive reals than positive integers. Symmetrically, there are more negative reals than negative integers. Therefore there are more reals than integers.

Is that indisputably valid? It looks indisputably valid to me, though I'm sure there are people who would dispute it. There are people called "intuitionists" who insist reductio ad absurdum isn't a valid proof procedure; but I don't really grok how anyone can think that way. Anyway, that's Cantor's argument, for whatever you think it's worth.

So what does all this mean for transcendentals? Well, a transcendental number by definition is just a number that isn't algebraic. The algebraic numbers can be put into a 1-to-1 mapping with the integers -- it's possible to define a formula for taking in an arbitrary algebraic number and computing an integer n, that gives a different n for every different algebraic number. So you can make the giant infinite table and have it include every algebraic number. Of course you can then do Cantor's same diagonalization trick and construct a number C that isn't in the table, but that table was never assumed to have all the reals. It was only assumed to have all the algebraic numbers. The C you get by changing all the digits along the diagonal will not be algebraic, so there's no contradiction. So there are as many algebraics as integers.

But if there were also only as many transcendentals as integers, then that would mean there are only twice as many reals as integers, since the reals are just the algebraics plus the transcendentals. But twice as many as the integers is the same thing as the number of integers. (It's simple to put the even numbers into a 1-to-1 mapping with the integers: {...-2, -1, 0, 1, 2...} : {...-4, -2, 0, 2, 4...}. The same goes for the odds. So two integers-sized sets added together make an integers-sized set.) Therefore there have to be more transcendentals than algebraics. Q.E.D.

Why can you make a formula for taking in an arbitrary algebraic number and computing an integer n that's different for every algebraic number, when you can't do that for arbitrary reals? It's because it takes a finite amount of information to specify an arbitrary algebraic number, but an infinite amount of information to specify an arbitrary real. An algebraic number is the root of a polynomial with integer coefficients: (axⁿ + bx^n-1 + ... + yx + z = 0). So you can just pack all those integer coefficients together into a single integer according to any rule that lets you tell where one ends and the next begins. For instance, write all the coefficients in base 8, and run them together with 8's and 9's separating them, depending on whether the next coefficient is positive or negative, and tack on a number at the end to specify which root of the polynomial you're specifying. So (- sqrt(3)), the first root of (1x² +0x - 3 = 0) would be M[1809381]. That gets all the algebraic numbers into your table. But if you try to do that same trick to the reals you'll have to pack infinitely many integers into an integer, and they won't fit. An infinitely long string of digits is not an integer.

Of course, your original plan had two algebraic numbers to work with, your x and your n. But having two doesn't help you. Every combination of two algebraics can still go into the table, because putting together two finite amounts of information still gives you a finite amount. If you want to pack two algebraics together into one integer, you can do it using a similar technique to the one above. Just use base 7 instead of base 8 for the coefficients, so the resulting integers from each algebraic number won't have any 7 digits in them. Then you can write one integer after the other with a 7 in between to mark where the first ends and the second begins.

Anyway, I hope all that helps. Best I can do.

 

[Kharakov]

It looks more like Q² > R(t)¹. IOW, the complex plane of rationals is greater in size than the transcendental reals (although you'd have to have the whole R₂ to have all the reals.... to get rationals from the function I mentioned, you need transcendentals). You do have infinite Q* infinite Q.... nicely defined as bigger than R(trans).

It's more of a "what's a bigger infinity defined by" thing. I think more dimensions>more "uncountable".

Cantor's diagonalization "proof" is poofed by logic- Cantor presents a false argument by massaging infinites (assuming a starting depth from the left .000...0, .000...001, etc.) or reordering the set of naturals (0000, followed by 11111 infinity, starts in the middle somewhere, etc.). Or simply not using binary to simplify the argument.

The columns in the below? 1/11, 1/101, 1/10001..... although technically they should all be 0... because they should have an infinite amount of preceding 0s....:

.00...000
.00...001
.00...010
.00...011
.00...100
.00...101
.00...110
.00...111

I assure you, you will find every pattern in the above (simplified binary distribution). Cantor's argument is bullshit. It basically assumes an infinite amount of digits in some number, and everyone knows how easy it is to massage infinite series into different forms.

from other side:

.11111..=1 so we'll start with .01111...=.1 and compress trailing 1111s


.001
.010
.011
.100 *** <<-- you have to arrange them like the first series in order to get every possible
.101 <<--diagonal combination.... unless you're Cantor, and then you just bullshit people
.110 <<-- to fuck with them.... stars are first value.... you see how this could rapidly
.111 <<-- get hard to fucking explain... if you thought Cantor was valid because AUTHORITY

The problem I have, is that I'll forget that Cantor is bullshit in a couple months, and someone will represent it to me as fact, and I'll have to figure it out again, and again.

Bomb, did we have this conversation (about binary and proper ordering breaking Cantor) before?


Anyways: Cantor argument= bullshit. It's simply based on incorrectly arranging infinite series, or obfuscating the nature of infinite series with invalid series manipulation, and not simplifying to binary so one can see EXACTLY what's going on.


Since varying rational n and x result in different trailing ends of transcendentals generated (with same starting digits), why would one ever assume that the rational complex plane (x,n) is not greater to or equal in size to the real transcendentals? Does this not prove that more dimensions implies greater size when dealing with infinities (almost definitely not...)?

 

[Kharakov]

Kharakov

I don't get what your question is that you sent me.

Assuming the links are correct algebraic functions are addition, multiplication, subtraction, division. For a function to be algebraic it must be expressible as a polynomial with a finite number roots.

g(z) = 3z

If f(g(x)) is algebraic then f(x) is algebraic. is algebraic f(g(x) is algebraic.
From the links algebraic and transcendental functions are definitions.

As to the OP, sounds like you are trying to go back to an algebraic function passed to a transcendental function.

Given g(z) is algebraic and f(x) is transcendental you want to get back g(z) from F(x)?

My whole idea was that there was a one to one mapping of rationals to transcendentals (which there is, infinity=infinity), although it appears that using the equation I mentioned, it requires 2 rational values to map the transcendentals, and 3 would be easier, 4 even more so.

rational a,b,x,n (easiest):

all trans = a + b*q_x,n

 

[Kharakov]

Wait a second... I didn't realize Cantor's argument was that there are SOME infinite sets that cannot be put on a one to one correspondence with natural numbers. I thought it was that there were NO infinite sets (other than algebraics) that can be put on a one to one correspondence with natural numbers. After all, the natural numbers include every possible permutation of digits.


I don't see the pertinence of the diagonal argument to whether or not transcendent numbers outnumber rationals?

 

[steve_bank]

Kharakov

I don't get what your question is that you sent me.

Assuming the links are correct algebraic functions are addition, multiplication, subtraction, division. For a function to be algebraic it must be expressible as a polynomial with a finite number roots.

g(z) = 3z

If f(g(x)) is algebraic then f(x) is algebraic. is algebraic f(g(x) is algebraic.
From the links algebraic and transcendental functions are definitions.

As to the OP, sounds like you are trying to go back to an algebraic function passed to a transcendental function.

Given g(z) is algebraic and f(x) is transcendental you want to get back g(z) from F(x)?

My whole idea was that there was a one to one mapping of rationals to transcendentals (which there is, infinity=infinity), although it appears that using the equation I mentioned, it requires 2 rational values to map the transcendentals, and 3 would be easier, 4 even more so.

rational a,b,x,n (easiest):

all trans = a + b*q_x,n

It is all Greek to me.

 

[Kharakov]

Kharakov

I don't get what your question is that you sent me.

Assuming the links are correct algebraic functions are addition, multiplication, subtraction, division. For a function to be algebraic it must be expressible as a polynomial with a finite number roots.

g(z) = 3z

If f(g(x)) is algebraic then f(x) is algebraic. is algebraic f(g(x) is algebraic.
From the links algebraic and transcendental functions are definitions.

As to the OP, sounds like you are trying to go back to an algebraic function passed to a transcendental function.

Given g(z) is algebraic and f(x) is transcendental you want to get back g(z) from F(x)?

My whole idea was that there was a one to one mapping of rationals to transcendentals (which there is, infinity=infinity), although it appears that using the equation I mentioned, it requires 2 rational values to map the transcendentals, and 3 would be easier, 4 even more so.

rational a,b,x,n (easiest):

all trans = a + b*q_x,n

It is all Greek to me.

It's a statement that every transcendental can be generated by the q_x,n function * some rational value + some rational value.

So you could generate pi^2+2 pretty easily: q_2,2*1 + 2. However, this still isn't a one to one mapping of transcendentals to rationals. I can't even remember why I was on this tangent at this point.

Obsessed with pointing out flaws that don't exist in loosely stated logic, I suppose.

"but I said almost all, not all, and you only found \(2^\infty\) exceptions to the rule...."

 

[Bomb#20]

Wait a second... I didn't realize Cantor's argument was that there are SOME infinite sets that cannot be put on a one to one correspondence with natural numbers. I thought it was that there were NO infinite sets (other than algebraics) that can be put on a one to one correspondence with natural numbers.

Infinite sets of numbers come in nested categories, and people have come up with lots of categories that can be put on a one to one correspondence with natural numbers, and lots of other categories that can't be. Here's a nesting of categories, where each infinite set is a subset of all the later infinite sets:

...
Integral eighth powers
Integral fourth powers
Integral squares
Non-prime numbers
Natural numbers
Integers
Rational numbers
Algebraic numbers
Computable numbers
Describable numbers
Real numbers
Complex numbers
Quaternions
...

All of those sets up through the describable numbers can be put on a one to one correspondence with natural numbers; starting with the real numbers, they cannot be. In addition, you can get an infinite set by deleting one of these sets from one of its supersets. The prime numbers are the natural numbers with the non-primes deleted; the negative numbers are the integers with the natural numbers deleted; the irrational numbers are the real numbers with the rational numbers deleted; and so forth. In particular for our topic, the transcendental numbers are the real numbers with the algebraic numbers deleted. And in general, any time you have a category like that, defined by taking one of the basic categories and deleting one of its smaller subsets, what you have left will have the same number of elements as the larger category you deleted a subset from. So there are exactly as many prime numbers as natural numbers; there are exactly as many negative numbers as integers; there are exactly as many irrational numbers as real numbers; and there are exactly as many transcendental numbers as real numbers.

"Exactly as many" has a defined meaning when we're talking about infinite sets. When set Q has exactly as many elements as set S, it means there exist subsets P and R, where P is a subset of Q and R is a subset of S, such that P can be put on a one to one correspondence with S and Q can be put on a one to one correspondence with R. You don't have to actually put Q and S on a one to one correspondence with each other. That can be technically very difficult for a variety of boring reasons, so we use the subset method. The idea is that when P can be put on a one to one correspondence with S, it means the number of elements in P and S are equal, so since P is a subset of Q, the number of elements in Q must be greater or equal to the number in S. So if we can do that subset matching in both directions, we get |Q| >= |S| and |S| >= |Q|, and from the two inequalities we deduce |Q| = |S|.

Here's an example. Are there the same number of real numbers R with 0 <= R < 1 as there are infinite strings of decimal digits? On first glance, obviously yes -- they're the same set, right? .358 = .358; .4747... = .4747...; sqrt(1/2) = .70710678118...; and so forth. The problem comes in when we remember that .73999... and .74000... are the same real number, but they're different infinite strings of decimal digits. So if you match them up that way, with .4747... mapped to .4747... and so forth, then you don't have a proper one to one correspondence between the sets. Finding a way to match them up so there are no repetitions is going to be an incredible pain in the ass if it can be done at all. So how can we show the two sets have the same number of elements? The simplest solution is the subset method. For one direction it's trivial: we simply delete the set of strings of decimal digits that end in an infinite string of 9's. So we have a one to one correspondence between the reals from 0 to 1 and a subset of the infinite strings of decimal digits. For the other direction we need a one to one correspondence between infinite strings of decimal digits and a subset of the reals from 0 to 1. There are a lot of ways to do that; here's an easy one: We map each infinite string of decimal digits to the real number with the same expansion in base 11. So now "999..." will map to .999... in base 11 instead of in base 10, i.e., it maps to 9/11 + 9/121 + 9/1331 + ... This makes the ".73999... = .74000..." problem go away, since they aren't equal in base 11. Of course, base 11 has exactly the same problem with its "a" digit ("a" means ten over some power of eleven), .73aaa... = .74000...; but that won't hurt us since our set of infinite strings of decimal digits doesn't have any "a" digits in it. So we've successfully created a one to one correspondence between infinite strings of decimal digits and a subset of the reals from 0 to 1. Note that this isn't one-to-one between the two sets we care about -- we're missing a lot of reals in this direction. There's no string of decimal digits that maps to 0.0a = 10/121, for instance. But that's okay. There are as many strings of decimal digits as a subset of the reals, and there are also as many reals as a subset of the strings of decimal digits, so Bob's your uncle.

After all, the natural numbers include every possible permutation of digits.

They include every finite permutation of digits. So all the numbers that can be specified with a finite amount of information can be put in a 1-to-1 correspondence with the natural numbers. That's not just the algebraics. The algebraic numbers are the numbers that can be specified with a finite amount of information in the language of polynomials. But there are other languages that are more expressive than polynomials. All the familiar transcendental numbers like pi and e can also be specified with a finite amount of information, using notations with summation symbols or integrals or trigonometric functions or whatever. So to put them in a 1-to-1 correspondence with the natural numbers you merely have to encode those operators as digits, the same way we encode polynomial coefficients as digits when we map algebraics to natural numbers. The transcendental numbers you can't put in a 1-to-1 correspondence with the natural numbers are precisely the ones that can't be specified with a finite amount of information in any language whatsoever -- it's the ones whose digit sequences are random.

I don't see the pertinence of the diagonal argument to whether or not transcendent numbers outnumber rationals?

If there were the same number of rationals as transcendentals, or if there were more rationals, then you could put all the transcendentals into a 1-to-1 correspondence with the rationals or with a subset of the rationals. (That's how it works with finite sets; that's how it works with ordinary infinite sets like the natural numbers. If |A| >= |B| then there exist a C and an M such that C is a subset of A and M is a 1-to-1 mapping and M(B) = C. For example, if the number of {lion, tiger, bear}=3 is greater or equal to the number of {goat, sheep}=2 then there exist C = {lion, bear} and M = {goat:bear, sheep:lion} such that {lion, bear} is a subset of {lion, tiger, bear} and {goat:bear, sheep:lion} is a 1-to-1 mapping and M({goat, sheep}) = {lion, bear}.)

So the relevance is that since the diagonal argument proves that you can't put all the transcendentals into a 1-to-1 correspondence with the rationals or with a subset of the rationals, it shows that the hypothesis that there are the same number of rationals as transcendentals, or that there are more rationals, implies a conclusion that we know isn't true. True premises don't imply false conclusions. Therefore the hypothesis that there are the same number of rationals as transcendentals, or that there are more rationals, must be false. Therefore transcendental numbers outnumber rationals.

You may be feeling this is wrong because the diagonal argument doesn't say anything about rationals -- it's all about setting up a 1-to-1 correspondence with the natural numbers. But that doesn't make any difference because we already know there are the same number of rationals as natural numbers. We know because we can make a 1-to-1 correspondence between them, as follows:

0 : 0
1 : 1/1
2 : -1/1
3 : 2/1
4 : -2/1
5 : 1/2
6 : -1/2
7 : 3/1
8 : -3/1
9 : 1/3
10: -1/3
11: 3/2
12: -3/2
13: 2/3
14: -2/3
15: 4/1
...

 

[Bomb#20]

It looks more like Q² > R(t)¹. IOW, the complex plane of rationals is greater in size than the transcendental reals (although you'd have to have the whole R₂ to have all the reals.... to get rationals from the function I mentioned, you need transcendentals). You do have infinite Q* infinite Q.... nicely defined as bigger than R(trans).

It's more of a "what's a bigger infinity defined by" thing. I think more dimensions>more "uncountable".

No, it's the other way around. More dimensions < more "uncountable". Extra dimensions don't get you a bigger infinity (unless you get to infinitely many dimensions! ).

The reason is bendy lines. You can take a one-dimensional infinite chain of numbers and bend it infinitely many times so it eventually goes everywhere in an N-dimensional space. For example,

21 22 23...
20  7  8  9 10 
19  6  1  2 11
18  5  4  3 12
17 16 15 14 13

Cantor's diagonalization "proof" is poofed by logic-

No, it's poofed by intuition, your intuition that QxQ is a bigger infinity than plain old Q.

Cantor presents a false argument by massaging infinites (assuming a starting depth from the left .000...0, .000...001, etc.) or reordering the set of naturals (0000, followed by 11111 infinity, starts in the middle somewhere, etc.).

I'm not following you. Where does he do this and why is it a problem?

Or simply not using binary to simplify the argument.

It could be done in binary too, but that actually complicates the argument, because then you have to deal with the complexity that 0.0111... = 0.1000..., as I described in post #19. Using a higher base allows us to just bypass that difficulty.

The columns in the below? 1/11, 1/101, 1/10001..... although technically they should all be 0... because they should have an infinite amount of preceding 0s....:

.00...000
.00...001
.00...010
.00...011
.00...100
.00...101
.00...110
.00...111

I assure you, you will find every pattern in the above (simplified binary distribution).

I'm sorry, I don't understand what you're getting at; and I won't find .010101... anywhere in the above simplified binary distribution. All the numbers in your table are terminating expansions, i.e., integer multiples of powers of 2. So 1/3 isn't in it.

if you thought Cantor was valid because AUTHORITY

Nobody thinks it's valid because Cantor's an authority. People think it's valid because we understand the argument and it makes sense to us. When he first presented it it was widely rejected because he wasn't an authority, and a lot of authority figures had intuition like yours that there was something wrong with it, because the conclusion was so contrary to conventional thinking. So mathematicians spent years trying to kick holes in it, and eventually realized they couldn't.

Bomb, did we have this conversation (about binary and proper ordering breaking Cantor) before?

Not that I recall; and I don't have anything like it in my notes, admittedly incomplete. You probably had the conversation with someone else.

Since varying rational n and x result in different trailing ends of transcendentals generated (with same starting digits), why would one ever assume that the rational complex plane (x,n) is not greater to or equal in size to the real transcendentals? Does this not prove that more dimensions implies greater size when dealing with infinities (almost definitely not...)?

Definitely not. We don't assume the rational complex plane (x,n) is not greater to or equal in size to the real transcendentals. We prove it, by exhibiting a 1-to-1 correspondence between the rational complex plane and the natural numbers.