This question aims to find whether the **domain** of all the **rational numbers** is a set of all real numbers or not. We have to find whether this statement is **true or false**.

Any number that exists in the world and that can be seen falls in the category of real numbers. Real numbers include all **rational**, **irrational**, and **integers** except the complex numbers that are in the form of **iota**. Real numbers are the set of all infinite numbers that are **not complex**. **For example**: 4.0, 5, -8, 56.88 $ \sqrt 6 $ etc. Â The complex numbers like $ 2 + i $, $ \sqrt {6 } i â€“ 9 $

Real numbers are often written as R = $ Q \cup Qâ€™ $ which means the set of all the rational numbers **union** the set of all irrational numbers is called real numbers.

There are generally **two types** of real numbers as all numbers are either **rational** or **irrational**.

Rational numbers:

Any number represented as the **quotient** of numerator and denominator is called a rational number. Rational numbers often take the form of $ \frac { p } { q } $. The **p** in the quotient is the numerator while the **q** is the denominator which is always a **non-zero value**. The numerator can be in the form of any **integer**, **natural number**, **whole number**, or decimal. **For example**, 3.9, 0.8, 1.666, $ \frac { 2 } { 7 } $, $ \ frac { -8 } { 9 } $ etc

**Expert Answer**

Every **Rational numbe**r is a real number but the domain of the rational numbers is not always the set of all real numbers. The domain of the rational numbers is the **set** of **all real numbers** where the function is defined. If **zero** is included in the **denominator** then it is not the domain.

For example, if we take a function Â $ f ( x) $ and its domain is $ g ( \frac { 1 } { x } ) $ then it can be written as:

\[ Â f Â ( Â x Â ) Â = Â \fracÂ {Â 1Â }Â {Â xÂ }Â \]

If we put values of x in the function:

\[ f Â (Â 4Â ) = \fracÂ {Â 1Â }Â {Â 4 } \]

\[ f Â (Â 3Â ) = \fracÂ {Â 1Â }Â {Â 3 } \]

\[ f Â (Â 5Â ) = \fracÂ {Â 1Â }Â {Â 5 } \]

Then the **domains** of the functions are $ \fracÂ {Â 1Â }Â {Â 4 } $, $ \fracÂ {Â 1Â }Â {Â 3 } $ , $ \fracÂ {Â 1Â }Â {Â 5 } $ and above-mentioned statement becomes **false.**

**Numerical Results**

**The domain of all the rational numbers is a set of all real numbers that is not true; no vertical asymptote and hole is formed on the graph.Â **

## Example

If we put the following expressions in the function:

\[ Â fÂ (Â xÂ )Â =Â \fracÂ {Â 1Â } Â {Â xÂ }Â \]

\[Â fÂ (Â 1Â +Â 3Â xÂ )Â =Â \fracÂ {Â 1Â } Â {Â 1Â +Â 3Â xÂ }Â \]

The domain of all the rational numbers is a set of all real numbers that is not true as no vertical asymptote and hole is formed on the graph.Â

*Image/Mathematical drawings are created in Geogebra**.*