I know that real number system is combination of rational and irrational number.
Q: Is there any other number except rational and irrational number in real number system?
Please anyone help me.
Thanks for any input.
I know that real number system is combination of rational and irrational number.
Q: Is there any other number except rational and irrational number in real number system?
Please anyone help me.
Thanks for any input.
On
There are lots of useful subsets of real numbers, and many of them overlap.
Rational numbers: any number that can be expressed as a ratio of two integers.
Irrational numbers: any number that is not rational.
Algebraic numbers: roots of polynomials with integer coefficients
Transcendental numbers: numbers that are not algebraic.
These two pairs of types of numbers are two different ways of splitting the reals into two different groups. So every real number is either rational or irrational, and every real number is either algebraic or transcendental.
There are more types of numbers that might be useful for a variety of reasons:
Computable numbers: numbers that can be computed by a Turing machine to arbitrary precision in a finite number of steps.
Non-computable numbers: well, not computable
Definable numbers: numbers whose properties can be defined in a finite number of symbols
Undefinable numbers: I can't give you an example of one of these, but I know they exist
Possibly the type of number used the most is IEEE 754 floating point numbers, which is what actual computers do arithmetic with.
I'm sure there are many more types of numbers, but that's all I can think of off the top of my head.
The answer (to the question in the title) is that yes, that covers all of them, by definition.
We take the real numbers $\mathbb{R}$ and the rational numbers $\mathbb{Q}$. We note that $\mathbb{Q}$ is a proper subset of $\mathbb{R}$, so we take all the real numbers that aren't rational and call them irrational numbers, i.e. we define $\mathbb{I} = \mathbb{R} \setminus \mathbb{Q}$.
There are other splits possible. In particular, we can take all the real numbers that are roots of polynomials with integer coefficients, and call them the algebraic numbers (sometimes notated as $\mathbb{A}$, although that also sometimes includes the complex algebraic numbers). So we can also consider the set of what's left, and that's what we call the transcendental numbers $\mathbb{R} \setminus \mathbb{A}$. Since any rational number $\frac{p}{q}$ is the solution to $qx - p = 0$, we can soon confirm that $\mathbb{Q} \subset \mathbb{A}$, i.e. the rational numbers form a proper subset of the algebraic numbers, and conversely the transcendental numbers form a proper subset of the irrational numbers.