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.

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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.

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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 \(\sqrt{3}\in\mathbb{Q}?\)

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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 \(\sqrt{4}\in\mathbb{Q}?\)

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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?

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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?

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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

\[\begin{align} i =& \frac{n}{m}\cdot \frac{p}{q} \\ =& \frac{np}{mq} \\ =& \frac{s}{t} , \end{align}\]

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?

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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