A complete collection of 20 practice examples on number sets: natural numbers, integers, rational numbers, irrational numbers, and real numbers. Each exercise shows how to correctly identify the sets a number belongs to, with a step-by-step explanation of the reasoning.
Recall the fundamental chain of inclusions:
\[ \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset \mathbb{R} \]
Furthermore, the set of irrational numbers is:
\[ \mathbb{R}\setminus\mathbb{Q} \]
Exercise 1 — level ★☆☆☆☆
Classify the following number:
\[ 7 \]
Answer
\[ 7 \in \mathbb{N},\quad 7 \in \mathbb{Z},\quad 7 \in \mathbb{Q},\quad 7 \in \mathbb{R} \]
Solution
Analysis of the number
The number \(7\) is a positive integer. Since it belongs to the set of natural numbers, we have:
\[ 7 \in \mathbb{N} \]
Membership in larger sets
Every natural number is also an integer, so:
\[ 7 \in \mathbb{Z} \]
Moreover, every integer can be written as a fraction with denominator \(1\):
\[ 7=\frac{7}{1} \]
Therefore \(7\) is also rational:
\[ 7 \in \mathbb{Q} \]
Finally, every rational number is a real number:
\[ 7 \in \mathbb{R} \]
Exercise 2 — level ★☆☆☆☆
Classify the following number:
\[ -3 \]
Answer
\[ -3 \in \mathbb{Z},\quad -3 \in \mathbb{Q},\quad -3 \in \mathbb{R} \]
\[ -3 \notin \mathbb{N} \]
Solution
Exclusion from the natural numbers
The number \(-3\) is negative. The natural numbers are the numbers used for counting:
\[ \mathbb{N}=\{0,1,2,3,\dots\} \]
Therefore:
\[ -3 \notin \mathbb{N} \]
Membership in the integers
The set of integers contains the natural numbers, their negatives, and zero:
\[ \mathbb{Z}=\{\dots,-3,-2,-1,0,1,2,3,\dots\} \]
Thus:
\[ -3 \in \mathbb{Z} \]
Membership in the rationals and the reals
Since:
\[ -3=\frac{-3}{1} \]
the number \(-3\) is rational, and consequently it is also real.
Exercise 3 — level ★☆☆☆☆
Classify the following number:
\[ \frac{5}{2} \]
Answer
\[ \frac{5}{2} \in \mathbb{Q},\quad \frac{5}{2} \in \mathbb{R} \]
\[ \frac{5}{2} \notin \mathbb{N},\quad \frac{5}{2} \notin \mathbb{Z} \]
Solution
Verifying the rational form
A number is rational if it can be written in the form:
\[ \frac{a}{b},\quad a,b\in\mathbb{Z},\quad b\neq 0 \]
The given number is already expressed as a ratio of two integers:
\[ \frac{5}{2} \]
therefore:
\[ \frac{5}{2}\in\mathbb{Q} \]
Exclusion from the naturals and the integers
Computing the decimal value:
\[ \frac{5}{2}=2.5 \]
The number is not an integer, so it belongs to neither \(\mathbb{N}\) nor \(\mathbb{Z}\).
Exercise 4 — level ★★☆☆☆
Classify the following number:
\[ \sqrt{2} \]
Answer
\[ \sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Analysis of the square root
The number \(\sqrt{2}\) is the square root of \(2\). Since \(2\) is not a perfect square, its square root is not an integer.
Irrational nature
The number \(\sqrt{2}\) is a classic example of an irrational number: it cannot be expressed as a ratio of two integers.
Its decimal expansion is infinite and non-repeating:
\[ \sqrt{2}=1.4142135\dots \]
Therefore:
\[ \sqrt{2}\notin\mathbb{Q} \]
However, \(\sqrt{2}\) is a real number, so:
\[ \sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 5 — level ★★☆☆☆
Classify the following number:
\[ 0 \]
Answer
\[ 0\in\mathbb{N},\quad 0\in\mathbb{Z},\quad 0\in\mathbb{Q},\quad 0\in\mathbb{R} \]
Solution
The role of zero
Under the convention most widely adopted in mathematics education, zero belongs to the set of natural numbers:
\[ 0\in\mathbb{N} \]
Membership in the other sets
Zero is also an integer:
\[ 0\in\mathbb{Z} \]
It can also be written as a fraction:
\[ 0=\frac{0}{1} \]
so it is rational:
\[ 0\in\mathbb{Q} \]
Being rational, it is also real.
Exercise 6 — level ★★☆☆☆
Classify the following number:
\[ -\frac{7}{4} \]
Answer
\[ -\frac{7}{4}\in\mathbb{Q},\quad -\frac{7}{4}\in\mathbb{R} \]
Solution
Fractional form
The given number is a fraction with integer numerator and denominator:
\[ -\frac{7}{4} \]
Since the denominator is non-zero, the number is rational:
\[ -\frac{7}{4}\in\mathbb{Q} \]
Why it is not an integer
Computing the decimal value:
\[ -\frac{7}{4}=-1.75 \]
The number is not an integer, so it belongs to neither \(\mathbb{Z}\) nor, consequently, \(\mathbb{N}\).
Exercise 7 — level ★★☆☆☆
Classify the following number:
\[ \pi \]
Answer
\[ \pi\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
The nature of \(\pi\)
The number \(\pi\) is a real number of fundamental importance in geometry, defined as the ratio of a circle's circumference to its diameter.
Irrationality
The number \(\pi\) cannot be expressed as a ratio of two integers. Its decimal expansion is infinite and non-repeating:
\[ \pi=3.14159265\dots \]
Thus:
\[ \pi\notin\mathbb{Q} \]
Since \(\pi\) lies on the real number line, we conclude:
\[ \pi\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 8 — level ★★☆☆☆
Classify the following number:
\[ \sqrt{16} \]
Answer
\[ \sqrt{16}=4 \]
\[ 4\in\mathbb{N},\quad 4\in\mathbb{Z},\quad 4\in\mathbb{Q},\quad 4\in\mathbb{R} \]
Solution
Computing the square root
Before classifying the number, it must be simplified:
\[ \sqrt{16}=4 \]
Indeed:
\[ 4^2=16 \]
Classification
Since \(4\) is a natural number, it also belongs to all subsequent sets:
\[ 4\in\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R} \]
Exercise 9 — level ★★★☆☆
Classify the following number:
\[ 0.\overline{3} \]
Answer
\[ 0.\overline{3}\in\mathbb{Q},\quad 0.\overline{3}\in\mathbb{R} \]
Solution
Repeating decimal
The number \(0.\overline{3}\) is a repeating decimal, because the digit \(3\) recurs indefinitely:
\[ 0.\overline{3}=0.3333\dots \]
Converting to a fraction
Every terminating or repeating decimal is rational. In this case:
\[ 0.\overline{3}=\frac{1}{3} \]
Therefore:
\[ 0.\overline{3}\in\mathbb{Q} \]
Being rational, it also belongs to \(\mathbb{R}\).
Exercise 10 — level ★★★☆☆
Classify the following number:
\[ 3+\sqrt{2} \]
Answer
\[ 3+\sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Analysis of the terms
The number \(3\) is rational, since:
\[ 3=\frac{3}{1} \]
The number \(\sqrt{2}\), on the other hand, is irrational:
\[ \sqrt{2}\notin\mathbb{Q} \]
Sum of a rational and an irrational
The sum of a rational number and an irrational number is always irrational.
Indeed, if \(3+\sqrt{2}\) were rational, subtracting the rational number \(3\) would give:
\[ (3+\sqrt{2})-3=\sqrt{2} \]
which would make \(\sqrt{2}\) rational — a contradiction.
Therefore:
\[ 3+\sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 11 — level ★★★☆☆
Classify the following number:
\[ 2+\frac{1}{2} \]
Answer
\[ 2+\frac{1}{2}=\frac{5}{2} \]
\[ \frac{5}{2}\in\mathbb{Q},\quad \frac{5}{2}\in\mathbb{R} \]
Solution
Adding the terms
Write \(2\) as a fraction with denominator \(2\):
\[ 2=\frac{4}{2} \]
Then:
\[ 2+\frac{1}{2}=\frac{4}{2}+\frac{1}{2}=\frac{5}{2} \]
Classification
The number \(\frac{5}{2}\) is a fraction of two integers with a non-zero denominator. Therefore:
\[ \frac{5}{2}\in\mathbb{Q} \]
It is not an integer, however, because:
\[ \frac{5}{2}=2.5 \]
Exercise 12 — level ★★★☆☆
Classify the following number:
\[ \sqrt{18} \]
Answer
\[ \sqrt{18}=3\sqrt{2} \]
\[ \sqrt{18}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Simplifying the radical
Factor \(18\) by extracting a perfect square:
\[ 18=9\cdot 2 \]
Then:
\[ \sqrt{18}=\sqrt{9\cdot 2}=\sqrt{9}\sqrt{2}=3\sqrt{2} \]
Classification
The number \(\sqrt{2}\) is irrational. Multiplying an irrational number by the non-zero rational \(3\) yields another irrational number.
Therefore:
\[ 3\sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 13 — level ★★★☆☆
Classify the following number:
\[ \frac{\sqrt{4}}{2} \]
Answer
\[ \frac{\sqrt{4}}{2}=1 \]
\[ 1\in\mathbb{N},\quad 1\in\mathbb{Z},\quad 1\in\mathbb{Q},\quad 1\in\mathbb{R} \]
Solution
Computing the square root
First, evaluate the square root:
\[ \sqrt{4}=2 \]
Substituting:
\[ \frac{\sqrt{4}}{2}=\frac{2}{2}=1 \]
Classification
The number \(1\) is a natural number. Consequently, it belongs to the integers, the rationals, and the reals as well:
\[ 1\in\mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R} \]
Exercise 14 — level ★★★★☆
Classify the following number:
\[ \sqrt{5}+\sqrt{3} \]
Answer
\[ \sqrt{5}+\sqrt{3}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Analysis of the radicals
The numbers \(5\) and \(3\) are not perfect squares, so:
\[ \sqrt{5}\notin\mathbb{Q},\quad \sqrt{3}\notin\mathbb{Q} \]
Caution: the sum of two irrationals can be rational
A subtle point deserves attention: the fact that both \(\sqrt{5}\) and \(\sqrt{3}\) are irrational is not sufficient to conclude that their sum is irrational. A simple counterexample shows why:
\[ (1+\sqrt{2})+(1-\sqrt{2})=2\in\mathbb{Q} \]
Proving that \(\sqrt{5}+\sqrt{3}\) is irrational requires a proof by contradiction.
Proof by contradiction
Suppose, for a contradiction, that there exists a rational number \(q\in\mathbb{Q}\) such that:
\[ \sqrt{5}+\sqrt{3}=q \]
Squaring both sides:
\[ \left(\sqrt{5}+\sqrt{3}\right)^2=q^2 \]
Expanding the left-hand side:
\[ 5+2\sqrt{15}+3=q^2 \]
that is:
\[ 8+2\sqrt{15}=q^2 \]
Isolating the radical:
\[ \sqrt{15}=\frac{q^2-8}{2} \]
The right-hand side is rational, as it is obtained from \(q\in\mathbb{Q}\) using only operations that stay within \(\mathbb{Q}\). It would follow that \(\sqrt{15}\) is rational.
Irrationality of \(\sqrt{15}\)
However, \(15\) is not a perfect square, so:
\[ \sqrt{15}\notin\mathbb{Q} \]
This is a contradiction, so the original assumption is false.
Conclusion
Therefore:
\[ \sqrt{5}+\sqrt{3}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 15 — level ★★★★☆
Classify the following number:
\[ \left(\sqrt{2}\right)^2 \]
Answer
\[ \left(\sqrt{2}\right)^2=2 \]
\[ 2\in\mathbb{N},\quad 2\in\mathbb{Z},\quad 2\in\mathbb{Q},\quad 2\in\mathbb{R} \]
Solution
Simplifying the expression
For every non-negative real number \(a\), the following identity holds:
\[ \left(\sqrt{a}\right)^2=a \]
Applying this with \(a=2\):
\[ \left(\sqrt{2}\right)^2=2 \]
Classification
Even though \(\sqrt{2}\) is irrational, its square is the natural number \(2\). The result therefore belongs to all sets:
\[ 2\in\mathbb{N},\quad 2\in\mathbb{Z},\quad 2\in\mathbb{Q},\quad 2\in\mathbb{R} \]
Exercise 16 — level ★★★★☆
Classify the following number:
\[ \frac{1}{\sqrt{2}} \]
Answer
\[ \frac{1}{\sqrt{2}}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Rationalizing the denominator
Rationalize the denominator:
\[ \frac{1}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}}{2} \]
Classification
The number \(\sqrt{2}\) is irrational. Dividing an irrational number by the non-zero rational \(2\) yields another irrational number.
Therefore:
\[ \frac{\sqrt{2}}{2}\in\mathbb{R}\setminus\mathbb{Q} \]
and consequently:
\[ \frac{1}{\sqrt{2}}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 17 — level ★★★★☆
Classify the following number:
\[ \sqrt{8}-\sqrt{2} \]
Answer
\[ \sqrt{8}-\sqrt{2}=\sqrt{2} \]
\[ \sqrt{8}-\sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Simplifying the first radical
Factor \(8\):
\[ 8=4\cdot 2 \]
Thus:
\[ \sqrt{8}=\sqrt{4\cdot 2}=2\sqrt{2} \]
Reducing the expression
Substituting back into the original expression:
\[ \sqrt{8}-\sqrt{2}=2\sqrt{2}-\sqrt{2}=\sqrt{2} \]
Classification
Since \(\sqrt{2}\) is irrational, the given expression is irrational as well:
\[ \sqrt{8}-\sqrt{2}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 18 — level ★★★★★
Classify the following number:
\[ \frac{3+\sqrt{2}}{3} \]
Answer
\[ \frac{3+\sqrt{2}}{3}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Splitting the fraction
Separate the two terms in the numerator:
\[ \frac{3+\sqrt{2}}{3}=\frac{3}{3}+\frac{\sqrt{2}}{3} \]
therefore:
\[ \frac{3+\sqrt{2}}{3}=1+\frac{\sqrt{2}}{3} \]
Nature of the term \(\frac{\sqrt{2}}{3}\)
We show that \(\frac{\sqrt{2}}{3}\) is irrational. Suppose for a contradiction that it is rational, meaning there exists \(q\in\mathbb{Q}\) such that:
\[ \frac{\sqrt{2}}{3}=q \]
Multiplying both sides by \(3\):
\[ \sqrt{2}=3q \]
But the product of two rational numbers is rational, so \(3q\in\mathbb{Q}\). This would imply \(\sqrt{2}\in\mathbb{Q}\), which is a contradiction. Therefore:
\[ \frac{\sqrt{2}}{3}\in\mathbb{R}\setminus\mathbb{Q} \]
Sum of a rational and an irrational
As established in Exercise 10, the sum of a rational number and an irrational number is always irrational.
Since \(1\in\mathbb{Q}\) and \(\frac{\sqrt{2}}{3}\in\mathbb{R}\setminus\mathbb{Q}\), we conclude:
\[ 1+\frac{\sqrt{2}}{3}\in\mathbb{R}\setminus\mathbb{Q} \]
and therefore:
\[ \frac{3+\sqrt{2}}{3}\in\mathbb{R}\setminus\mathbb{Q} \]
Exercise 19 — level ★★★★★
Classify the following number:
\[ \sqrt{2}\cdot\sqrt{8} \]
Answer
\[ \sqrt{2}\cdot\sqrt{8}=4 \]
\[ 4\in\mathbb{N},\quad 4\in\mathbb{Z},\quad 4\in\mathbb{Q},\quad 4\in\mathbb{R} \]
Solution
Product of radicals
Since both radicands are non-negative, we may apply the property:
\[ \sqrt{a}\sqrt{b}=\sqrt{ab} \]
Therefore:
\[ \sqrt{2}\cdot\sqrt{8}=\sqrt{16} \]
and:
\[ \sqrt{16}=4 \]
An important observation
Even though both factors \(\sqrt{2}\) and \(\sqrt{8}\) are irrational, their product can be rational. In this case the result is even a natural number.
Exercise 20 — level ★★★★★
Classify the following number:
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} \]
Answer
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}}=1+\frac{\sqrt{6}}{2} \]
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}}\in\mathbb{R}\setminus\mathbb{Q} \]
Solution
Splitting the fraction
Divide each term of the numerator by the denominator:
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}} = \frac{\sqrt{2}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}} \]
The first term equals:
\[ \frac{\sqrt{2}}{\sqrt{2}}=1 \]
therefore:
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}}=1+\frac{\sqrt{3}}{\sqrt{2}} \]
Rationalizing the denominator
Rationalize the second term:
\[ \frac{\sqrt{3}}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{6}}{2} \]
Therefore:
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}}=1+\frac{\sqrt{6}}{2} \]
Final classification
The number \(\sqrt{6}\) is irrational, since \(6\) is not a perfect square. Consequently, \(\frac{\sqrt{6}}{2}\) is also irrational.
The sum of the rational number \(1\) and the irrational number \(\frac{\sqrt{6}}{2}\) is irrational.
Therefore:
\[ \frac{\sqrt{2}+\sqrt{3}}{\sqrt{2}}\in\mathbb{R}\setminus\mathbb{Q} \]
Summary
To classify a number correctly, it is best to work through the following steps in order:
- simplify the expression, if possible;
- check whether the result is a natural number;
- check whether it is an integer;
- check whether it can be written as a fraction;
- if it is not rational but lies on the real number line, classify it as irrational.
The fundamental relationship among the number sets is:
\[ \mathbb{N}\subset\mathbb{Z}\subset\mathbb{Q}\subset\mathbb{R} \]
while the irrational numbers are precisely those real numbers that are not rational:
\[ \mathbb{R}\setminus\mathbb{Q} \]