Uncountable Transcendental Numbers: Exploring the Mystery

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In summary, the set of all real numbers that are not algebraic is uncountable, while the set of all algebraic numbers is countable. The set of all transcendental numbers is also uncountable.
  • #1
murshid_islam
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can anyone tell me why the transcendental numbers are uncountable?
 
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  • #2
How many real numbers are there? How many algebraic numbers are there?
 
  • #3
All algebraic numbers, by definition, satisfy some polynomial equation with integer coefficients. Since a polynomial of degree n has exactly n coefficients, there are a countable number of such polynomials. Every possible such polynomial has at most n distinct zeros so there are a countable number of solutions to all possible nth degree polynomials. Since there are a countable number of possible degrees, the set of all possible solutions to all polynomials with integer coefficients, i.e. the set of all algebraic numbers, is countable. The set of all real numbers is uncountable so the set of all real numbers that are not algebraic, the set of all transcendental numbers, is uncountable.
 
  • #4
HallsofIvy said:
Since a polynomial of degree n has exactly n coefficients, there are a countable number of such polynomials.

i didn't understand this. could you elaborate a little?
 
  • #5
Here is an outline of the proof

Consider this definition of an algebraic number:

Let A denote the set of algebraic numbers. Then z is in A IFF z is a root of a ploynomial with integer coefficients (sometimes stated with rational coefficients, but this is equivalent).

That is z is in A IFF there exists integers a0,a1,a2,...,an not all zero such that the polynomial P(z):=sum(ak*z^k, k=0..n)=0.

You can prove A is countably infinite by:first defining the height of a polynomial as the sum of the absolute values of its coefficients and its degree, e.g., |P(z)|=|a0|+|a1|+...+|an|+n for the above polynomial.

Then prove that there are finitely many polynomials of a given height, and that each such polynomial has finitely many roots (use the fundamental theorem of algebra for the second part).
Let the (finite) union of the sets of the roots of polynomials of height m be denoted by Hm. Then A is the (countable) union of Hm over all positive integers m (why?) Therefore A is countable (why?)

The set of transdental numbers must then be uncountable (why?)
 
  • #6
benorin said:
You can prove A is countably infinite by:first defining the height of a polynomial as the sum of the absolute values of its coefficients and its degree, e.g., |P(z)|=|a0|+|a1|+...+|an|+n for the above polynomial.

Then prove that there are finitely many polynomials of a given height, and that each such polynomial has finitely many roots (use the fundamental theorem of algebra for the second part).

how? can you explain further? i understood the rest.
 
  • #7
murshid_islam said:
i didn't understand this. could you elaborate a little?

Do you know this theorem: If A and B are countable sets then AxB (Cartesian product: ordered pairs where first member is in A and second in B) is countable. To prove that, imagine that you make a table by listing all members of A horizontally (you can do that because A is countable) and listing all members of b vertically (you can do that because B is countable). The pair (a,b) is written in the column headed a along the top and the row headed b on the left. Now you can go through that table "diagonally"- you may have seen proofs that the set of all rational numbers is countable done that way.

The set of all pairs (a,b) where a and b are both integers is countable because it is NxN and N is countable. Of course, we can identify the linear polynomial a+ bx with (a,b).

It is then easy to show that the Cartesian product of three countable sets AxBxC is countable by thinking of this as (AxB)xC. AxB is the product of two countable sets and so is countable, then (AxB)xC is the product of two countable sets and so is countable.
But we can identify the polynomial a+ bx+ cx2 with (a, b, c).

In general, the Cartesian product of any finite number of countable sets is countable and we can identify a+ bx+ cx2+ ...+ zxn with the ordered n-tuple (a, b, c, ..., z) showing that the set of all possible nth degree polynomials with integer coefficients is countable.
 
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What are transcendental numbers?

Transcendental numbers are real numbers that are not algebraic, meaning they cannot be expressed as the root of a polynomial equation with rational coefficients.

How do transcendental numbers differ from algebraic numbers?

Transcendental numbers are different from algebraic numbers because they cannot be expressed as a finite combination of rational numbers using the basic arithmetic operations (addition, subtraction, multiplication, division) and taking roots. Algebraic numbers, on the other hand, can be expressed in this way.

Who discovered transcendental numbers?

The concept of transcendental numbers was first introduced by the mathematician Joseph Liouville in 1844. However, the first explicit example of a transcendental number was given by Johann Heinrich Lambert in 1761.

What is the most famous transcendental number?

The most famous transcendental number is pi (π), which is the ratio of a circle's circumference to its diameter. It is approximately 3.14159265358979323846 and has been studied and approximated by mathematicians for centuries. Another well-known transcendental number is e, the base of the natural logarithm.

How do transcendental numbers impact mathematics and science?

Transcendental numbers have a significant impact on mathematics and science as they provide a way to describe and understand the physical world with greater precision. They are used in various mathematical and scientific fields, such as physics, engineering, and cryptography. The discovery of transcendental numbers also led to the development of new mathematical concepts and techniques.

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