Real Numbers

Any number that can be found in the real world is, literally, a real number. Counting objects gives a sequence of positive integers, or natural numbers, $\mathbb{N}.$ If you consider having nothing or being in debt as a number, then the set $\mathbb{Z}$ of integers, including zero and negative numbers, is in order. If you cut a cake into equal pieces, then you may have a piece which represents a rational number, which is a number that can be represented by an irreducible fraction of two integers. What is the next? We have the set $\mathbb{R}$ of real numbers, which is the union of the set $\mathbb{Q}$ of rational numbers and the set $\mathbb{I}$ of irrational numbers. The Venn diagram below depicts the relationship between these sets of numbers.

$\mathbb{R}$ is the set of numbers that can be measured, such as length or weight, which, of course, include $\mathbb{Q} ,$ which can be obtained from counting, subtracting, and dividing. What could be an example of number which is not in $\mathbb{Q}?$ Presumably, the first irrational number in history is $\sqrt{2},$ which was found by a follower of Pythagoras. $\sqrt{2}$ corresponds to the length of the diagonal of a square with side length 1, so $\sqrt{2}$ is a real number, which exists in the real world. However, $\sqrt{2}$ cannot be represented by an irreducible fraction of two integers. We can prove it here shortly.

Prove that $\sqrt{2}$ is not a rational number.

---

Let $\sqrt{2} = \frac{m}{n},$ where $m$ and $n$ are coprime integers. Then the square of $\sqrt{2}$ is $\left (\sqrt{2}\right)^2 = 2 = \frac{m^2}{n^2} ,$ which implies that $m^2$ is a multiple of 2. If the square of a number is a multiple of 2, then the number is also a multiple of 2. Therefore, using the substitution $m=2m',$ where $m'$ is an integer, gives $\frac{m^2}{n^2} = 4\frac{m'^2}{n^2} = 2.$ Then $\frac{n^2}{m'^2}=2,$ and $n$ is also a multiple of 2. Now we have both $m$ and $n$ as multiples of 2, and this does not correspond with the condition that $m$ and $n$ are coprime integers. Hence we can conclude that $\sqrt{2}$ cannot be represented by an irreducible fraction of two integers, so it is an irrational number. $_\square$

Likewise, measuring objects may give numbers other than rational numbers. Other common examples are $\pi,$ the ratio of a circle's circumference to its diameter and $e,$ Euler's number. Meanwhile, the exact definition or construction of $\mathbb{R}$ is quite difficult to understand, and it is taught in college-level math courses. For your information, $\mathbb{R}$ is commonly constructed by Dedekind cuts.

Clarify which set is a subset of another set among $\mathbb{I},$ $\mathbb{N},$ $\mathbb{Q},$ $\mathbb{R},$ and $\mathbb{Z}$--the sets of irrational numbers, natural numbers, rational numbers, real numbers, and integers, respectively.

---

The set $\mathbb{N}$ of natural numbers is the set of positive integers.
The set $\mathbb{Z}$ of integers is the set of rational numbers with denominator 1.
The set $\mathbb{Q}$ of rational numbers is the set of real numbers which can be represented by irreducible fractions of two integers.
The set $\mathbb{I}$ of irrational numbers is the set of real numbers which are not rational.

Therefore,

$$\mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R},$$

and

$$\mathbb{I} \subset \mathbb{R} .\ _\square$$

Is it true that $\sqrt{3}\in\mathbb{Q}?$

---

No, $\sqrt{3}$ is not an element of $\mathbb{Q},$ but is an irrational number. We can prove it similarly with the case of $\sqrt{2}.$

Let $\sqrt{3} = \frac{m}{n},$ where $m$ and $n$ are coprime integers. Then the square of $\sqrt{3}$ is $\left (\sqrt{3}\right)^2 = 3 = \frac{m^2}{n^2} ,$ which implies $m^2$ is a multiple of 3. If the square of a number is a multiple of 3, then the number is also a multiple of 3. Therefore, substituting $m=3m',$ where $m'$ is an integer, gives $\frac{m^2}{n^2} = 9\frac{m'^2}{n^2} = 3.$ Then $\frac{n^2}{m'^2}=3,$ and $n$ is also a multiple of 3. Now we have both $m$ and $n$ as multiples of 3, which does not correspond with the condition that $m$ and $n$ are coprime integers. Hence we can conclude that $\sqrt{3}$ cannot be represented by an irreducible fraction of two integers, so it is an irrational number. $_\square$

Is it true that $\sqrt{4}\in\mathbb{Q}?$

---

Yes, $\sqrt{4}$ is an element of $\mathbb{Q}.$

By definition, $\sqrt{4}$ is a positive number $x$ that satisfies $x^2 = 4.$ We know that $2^2 = 4,$ so $\sqrt{4}=2,$ which is an integer. All integers are rational numbers, so we can conclude that $\sqrt{4}\in\mathbb{Q}.$ $_\square$

The length of the longer side of an A4 paper is 29.7 cm. Is 29.7 rational?

---

We can represent 29.7 as an irreducible fraction of two integers:

$$29.7 = \frac{297}{10}.$$

Therefore, 29.7 is rational. $_\square$

The area of an equilateral triangle with side length 1 is $\frac{\sqrt{3}}{4} .$ Is $\frac{\sqrt{3}}{4}$ rational?

---

First, let's prove that the product of a nonzero rational number $\frac{m}{n},$ where $m$ and $n$ are coprime integers, and an irrational number $i$ cannot be rational. Assume that $\frac{m}{n} i = \frac{p}{q} ,$ where $p$ and $q$ are coprime integers. Then

$$\begin{aligned} i =& \frac{n}{m}\cdot \frac{p}{q} \\ =& \frac{np}{mq} \\ =& \frac{s}{t} , \end{aligned}$$

where $s=\frac{np}{\gcd(np,mq)}$ and $t=\frac{mq}{\gcd(np,mq)}.$ Now we have $i$ as a fraction of two integers, which contradicts with the premise that $i$ is irrational.

Therefore since $\sqrt{3}$ is irrational, a fourth of $\sqrt{3}$ is also irrational. $_\square$

The circumference of a circle with radius 1 is $2\pi.$ Is $2\pi$ rational?

---

As we have proved above, the product of a nonzero rational number and an irrational number cannot be rational. $\pi$ is known to be irrational. Hence twice $\pi$ is also irrational. $_\square$

• Modular Arithmetic

• Chinese Remainder Theorem