One has the field of rational functions with rational coefficients: Q(X). Here X is an abstract transcendental element. Well, Q(pi) is also a field, and is isomorphic with Q(X). The isomorphism just does f(pi) = X and preserves all rational numbers. The only difference is that pi is not an abstract transcendental quantity, but a concrete one. The reason of Q(X) to be countable is that (Q U QxQ U QxQxQ U ...)^2 is a countable set.
@Mihai Prunescu: "...The only difference is that pi is not an abstract transcendental quantity, but a concrete one" . Here is my question: consider the number seven: is 7 now an abstract integer quantity or a concrete one? In case you don't like this, take sqrt7.
I differ, pi is not a concrete one quantity, it is irrational and not rational number so it has no limit, now the question if Q (pi) is countable, there are infinite Q (pi), so it is not countable. It is not tue that f(pi) = X and preserves all rational numbers, becauese it is imposible to represent an irrational number equal to rational number.
1) "pi is not a concrete one quantity". pi is the area of a unit circle, 2pi is its circumference.
2) "because it is imposible to represent an irrational number equal to rational number": sqrt2 is the diagonal of the unit sqare.
This was geometrical. There a noumerous further representations of pi in series or definite integrals. Resume: As far as "quality" is concerned, the integers and the reals are abstract mathematical objects, the number 1 does'nt exist in nature.
Anton Schober, but the area of unit circle or area of any circle is not concrete one quantity.....always it is an approximationis give one concrete one quantity (value). The irrational number have infinite mantissa......geometrically, you can represent an irrational number, but not with "concrete one quantity".......
Q(pi) is the field of fractions of Q[pi] which is isomorphic (really only need a bijection) with the ordinary polynomial ring Q[x]. Now Q[x] is equal to the Union of the countable collection QuQxuQx^2uQx^3u... That is, it's a countable Union of countable sets (each Qx^n is in one to one correspondence with Q, and hence countable) and so Q[x] is countable. Now, if an integral domain is countable, it is not difficult to prove that it's field of fractions is countable. Hence Q(pi) is countable.
@Jose William Porras: this "concrete one quantity" exists eventually in your mind. I ask you: is an open set, subset of a topological space a "concrete one quantity"? Is the Lagrangian of the electromagnetic field a "concrete one quantity"? AND AGAIN: is the number 1 a "concrete one quantity"? Eventually you answer exactly these three questions.
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in the binary system there exist only zeroes(0) and ones (1). Now "concrete"?
next: also 1/3 = .333... is not "concrete" in your system
Anton Schober: In my mind not. Concrete quantity is a physical object or a collection of such objects. According with Definition in: http://www.finedictionary.com/Concrete%20quantity.html
Concrete quantity is a physical object or a collection of such objects. According with Definition in: http://www.finedictionary.com/Concrete%20quantity.html
abstract: Considered apart from concrete existence.
In Mathematics. Abstrct Quantity:
a number that does not designate the quantity of any particular kind of thing.
I am not sure whether this wiki page is correct. https://en.wikipedia.org/wiki/Concrete_number
It says the following:
In mathematics the term "number" is usually taken to mean an abstract number. A denominate number is a type of concrete number with a unit of measure attached with it. For example, 5 inches is a denominate number because it has the unit inches after it.
If it is correct, then pi is a concrete quantity. A circle with unit metre has pi metre^2 as area. So, it should be concrete number. If we consider a thread whose length is same as circumference of a circle of radius 1 inch, then half of its length is pi inches.
@Panchatcharam Mariappan: “Concrete quantity is a physical object or a collection of such objects” is like discrete topology (trivial = worthless). In categories we have objects and morphisms. These objects are the “abstract quantities”, and the morphisms are well defined mappings.
@dear Dr. Gabriel Thomas, "Q(pi) is included in R", sure, bur sqrt2 is not icluded in Q(pi)
"For Q(pi) : using an argument from Cantor on the countability of the polynomials with integral coefficients, you get the conclusion" I would like to see this polynomial! This should be an easy task!