[hamdani yusuf (OP)]

That was an error in the method shown in that particular text (which was a PopSci version of Cantor's proof that ℜ is uncountable ).  However, it's possible to fix that and most texts describing the diagonalisation method do exactly that. 

What should be done to fix it?

[hamdani yusuf (OP)]

A / B     has  A as the numerator and B as the denominator.
I have no rules to know what to do, or what it means when A or B is infinite.

I said if numerator and denominator are allowed to be infinite, which means that they don't have to. For example, x=1/3, which can be approximated as 33/100, or more precisely, 3333/10000, or, if you want infinite precision, 333333.../1000000..., with both numerator and denominator have infinitely long digits.
The numerator can be expressed as , while the denominator can be expressed as
Both of them are integers, but as n approach infinity, they also approach infinity. But the ratio is finite, and can be expressed as



or

[hamdani yusuf (OP)]

https://en.wikipedia.org/wiki/Algebraic_number
An algebraic number is a number that is a root of a non-zero polynomial in one variable with integer (or, equivalently, rational) coefficients. For example, the golden ratio, (1+√5)/2, is an algebraic number, because it is a root of the polynomial x2 − x − 1. That is, it is a value for x for which the polynomial evaluates to zero. As another example, the complex number 1+i is algebraic because it is a root of x⁴ + 4.

All integers and rational numbers are algebraic, as are all roots of integers. Real and complex numbers that are not algebraic, such as π and e, are called transcendental numbers.

The set of algebraic numbers is countably infinite and has measure zero in the Lebesgue measure as a subset of the uncountable complex numbers. In that sense, almost all complex numbers are transcendental.

If our goal is to fill the gap in real numbers which are not covered by rational numbers (ratio of two finite integers), then algebraic numbers are not well suited for the job for two reasons. First, it can't cover all irrational numbers, such as 2^√2. Second, it covers non-real numbers instead, such as 1+i.

[Bored chemist]

When expressed as ratio, 1/2, 1/3, and 1/4 don't seem to need much different amount of information. But when expressed in decimal, 1/3 needs infinite number of digits. Although the decimal digits are repetitive, which make the information compressible.

Use base 12 rather than 10.
The problem goes away.

But this solution is not general, since it depends on the chosen base number, as well as the denominator. If the denominator has a prime factor not shared with the base number, the number of digits will be infinite.

A more general solution is using continued fraction, which is independent from base number selection. 

The general solution is to pick a base where the fraction doesn't recur.
The fact is that you can always do that (and, similarly, you can always choose one where it does).
So whether or not you get a recurring "decimal" isn't important.
It's just an artefact of the arbitrary choice of number base.

[Eternal Student]

Hi.

What should be done to fix it?      (the diagonalisation method discussed much earlier, e.g. post #14)

    Just don't allow the enumeration to contain numbers written with recurring 9 digits.    All such numbers can be written with recurring 0 digits instead, insist that this is done.
    Example:    0.123999999999.......   =   0.12400000000000......

Once you exclude either recurring 9 digits  (or recurring 0 digits) then a decimal expansion becomes unique - so that the problem shown in post #14 won't happen.

Best Wishes.

[hamdani yusuf (OP)]

Hi.

What should be done to fix it?      (the diagonalisation method discussed much earlier, e.g. post #14)

    Just don't allow the enumeration to contain numbers written with recurring 9 digits.    All such numbers can be written with recurring 0 digits instead, insist that this is done.
    Example:    0.123999999999.......   =   0.12400000000000......

Once you exclude either recurring 9 digits  (or recurring 0 digits) then a decimal expansion becomes unique - so that the problem shown in post #14 won't happen.

Best Wishes.

What if the number is written in binary?

[hamdani yusuf (OP)]

The general solution is to pick a base where the fraction doesn't recur.

That's an extra work. It's certainly not a practical way to build a computer.

[Eternal Student]

Hi.

What if the number is written in binary?

   That's fine.   Then just don't allow recurring 1 digits.

Example:   0.0111111111111111111....     ≡      0.1       ≡       0.10000000000000...   in binary.

In whatever number base, B, you choose,  you just can't allow recurring digits of (B-1).   Provided you do this then the representation of a real number as a string of digits is unique.

Best Wishes.

[hamdani yusuf (OP)]

In whatever number base, B, you choose,  you just can't allow recurring digits of (B-1).   Provided you do this then the representation of a real number as a string of digits is unique.

In binary code, the step would remove half of all numbers with  finite digits.

https://en.m.wikipedia.org/wiki/Cantor's_diagonal_argument
In set theory, Cantor's diagonal argument, also called the diagonalisation argument, the diagonal slash argument, the anti-diagonal argument, the diagonal method, and Cantor's diagonalization proof, was published in 1891 by Georg Cantor as a mathematical proof that there are infinite sets which cannot be put into one-to-one correspondence with the infinite set of natural numbers.[1][2]: 20– [3] Such sets are now known as uncountable sets, and the size of infinite sets is now treated by the theory of cardinal numbers which Cantor began.

Diagonalization arguments are often also the source of contradictions like Russell's paradox[7][8] and Richard's paradox.[2]: 27 

[hamdani yusuf (OP)]

https://en.m.wikipedia.org/wiki/File:Diagonal_argument_01_svg.svg

An illustration of Cantor's diagonal argument (in base 2) for the existence of uncountable sets. The sequence at the bottom cannot occur anywhere in the enumeration of sequences above.

The problem with infinite set of natural numbers is that there's no bottom. If we assume the existence of something that is not there, it's no wonder that we will get a contradiction.

[Bored chemist]

It's certainly not a practical way to build a computer.

Had anyone suggested that it was?

[Eternal Student]

Hi.

In binary code, the step would remove half of all numbers with  finite digits.

    Why?

Write down a number with finite digits,    e.g.    0.10101     .    Why would that be removed?
That representation is permitted.  (Although during the diagonalisation method you would consider it to have as many 0 digits at the end as required).

Best Wishes.

[hamdani yusuf (OP)]

Hi.

In binary code, the step would remove half of all numbers with  finite digits.

    Why?

Write down a number with finite digits,    e.g.    0.10101     .    Why would that be removed?
That representation is permitted.  (Although during the diagonalisation method you would consider it to have as many 0 digits at the end as required).

Best Wishes.

Any finite digits binary number can be written in 2 ways. Your restriction rejects one of them.

[hamdani yusuf (OP)]

It's certainly not a practical way to build a computer.

Had anyone suggested that it was?

No. It was to show that your solution can't be both simple and general to make it useful as a tool to solve the problem discussed here, which is the countability of a set of numbers.

[Bored chemist]

It's certainly not a practical way to build a computer.

Had anyone suggested that it was?

No. It was to show that your solution can't be both simple and general to make it useful as a tool to solve the problem discussed here, which is the countability of a set of numbers.

It's not so much as "solution" as an observation.

Because you can arbitrarily make any fraction either recur or not (by choosing the number base) there's nothing special about recurring  numbers.
So you can simply ignore them.
Ignoring the thing which, at first, looks like an additional complication does make it easier to calculate things.

[Eternal Student]

Hi again.

Any finite digits binary number can be written in 2 ways. Your restriction rejects one of them.

  Yes.   This is exactly what is required.  You want every real number to be uniquely represented.

Best Wishes.

[hamdani yusuf (OP)]

Here is a Venn diagram of real number:

Reference: https://www.physicsforums.com/threads/is-my-classification-of-transcendental-correct.921555

Here's a pattern.
There are infinitely many more integers than natural numbers.
There are infinitely many more rational numbers than integers.
There are infinitely many more algebraic numbers than rational numbers.
There are infinitely many more real numbers than algebraic numbers.

But somehow the state of countability changes only between real numbers and algebraic numbers.

[hamdani yusuf (OP)]

The "proof" in the video seems to skip algebraic numbers.
Moreover, at 2:07 rational numbers are classified as countable/list-able using two dimensional list. It's not clear why it stops there, instead of continuing to higher dimensional list/tensor.

[hamdani yusuf (OP)]

Out of curiosity, I put following questions in Wolfram Alpha.
is tan(2) transcendental?
the result is transcendental, while it's assumed to be in unit of radians.
the result is unknown, while it's assumed to be in unit of degrees.

I got the same result for sine and cosine.

log(2) is transcendental in base e, but unknown in base 10.
 

[hamdani yusuf (OP)]

There's also continued fraction to classify subsets of real numbers.

https://en.m.wikipedia.org/wiki/Continued_fraction
In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on.[1] In a finite continued fraction (or terminated continued fraction), the iteration/recursion is terminated after finitely many steps by using an integer in lieu of another continued fraction. In contrast, an infinite continued fraction is an infinite expression. In either case, all integers in the sequence, other than the first, must be positive. The integers {\displaystyle a_{i}}a_{i} are called the coefficients or terms of the continued fraction.[2]

Rational numbers have finite continued fraction. Irrational numbers, including transcendental numbers have infinite continued fraction.

We can make a new class of numbers which is a superset of algebraic numbers while being a subset of real numbers. Just like division was used to construct rational numbers, and algebraic functions were used to construct algebraic numbers, we can use a more general operation to construct this new class. Taylor series or generalized continued fraction can be used for this purpose.
https://en.wikipedia.org/wiki/Taylor_series
https://en.wikipedia.org/wiki/Generalized_continued_fraction#Transcendental_functions_and_numbers

A generalized continued fraction is an expression of the form

where the an (n > 0) are the partial numerators, the bn are the partial denominators, and the leading term b0 is called the integer part of the continued fraction.

For example,



This way, the examples for non-algebraic numbers in the Venn diagram below can be classified as analytic numbers, which means that they can be represented as a regular pattern in a generalized continued fraction. It means that they can be stated using a finite number of bits of information.