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Sagot :
Let's analyze each expression step-by-step and determine which sets they belong to. We'll place an [tex]\( X \)[/tex] in the correct columns accordingly.
### 1. Expression: [tex]\(\frac{2}{9}\)[/tex]
- Natural Numbers: These are positive integers starting from 1 onwards (1, 2, 3, ...). [tex]\(\frac{2}{9}\)[/tex] is not a natural number.
- Whole Numbers: These include all natural numbers plus zero (0, 1, 2, 3, ...). [tex]\(\frac{2}{9}\)[/tex] is not a whole number.
- Integers: These include all positive and negative whole numbers and zero (..., -3, -2, -1, 0, 1, 2, 3, ...). [tex]\(\frac{2}{9}\)[/tex] is not an integer.
- Rational Numbers: These numbers can be expressed as the quotient of two integers, where the denominator is not zero. [tex]\(\frac{2}{9}\)[/tex] can be expressed as [tex]\(\frac{2}{9}\)[/tex], so it is a rational number.
- Irrational Numbers: These cannot be expressed as the quotient of two integers. [tex]\(\frac{2}{9}\)[/tex] is a rational number, not irrational.
- Real Numbers: These include both rational and irrational numbers. Since [tex]\(\frac{2}{9}\)[/tex] is rational, it is also a real number.
Thus, for [tex]\(\frac{2}{9}\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & \text{Expression} & \text{Natural} & \text{Whole} & \text{Integer} & \text{Rational} & \text{Irrational} & \text{Real} \\ \hline 1. & \frac{2}{9} & & & & X & & X \\ \hline \end{array} \][/tex]
### 2. Expression: [tex]\(\sqrt{2}\)[/tex]
- Natural Numbers: [tex]\(\sqrt{2}\)[/tex] is not a whole number, nor is it a positive integer.
- Whole Numbers: [tex]\(\sqrt{2}\)[/tex] is not a whole number.
- Integers: [tex]\(\sqrt{2}\)[/tex] is not an integer.
- Rational Numbers: [tex]\(\sqrt{2}\)[/tex] cannot be expressed as a ratio of two integers, so it is not rational.
- Irrational Numbers: [tex]\(\sqrt{2}\)[/tex] is an irrational number, as it cannot be expressed as the quotient of two integers.
- Real Numbers: [tex]\(\sqrt{2}\)[/tex] falls within the set of real numbers since it is either rational or irrational.
Thus, for [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & \text{Expression} & \text{Natural} & \text{Whole} & \text{Integer} & \text{Rational} & \text{Irrational} & \text{Real} \\ \hline 2. & \sqrt{2} & & & & & X & X \\ \hline \end{array} \][/tex]
### 3. Expression: [tex]\(\sqrt{(3)^2 - 4(1)(2)}\)[/tex]
First, simplify the expression inside the square root:
[tex]\[ (3)^2 - 4(1)(2) = 9 - 8 = 1 \][/tex]
Then, take the square root of the result:
[tex]\[ \sqrt{1} = 1 \][/tex]
Since [tex]\(1\)[/tex] is obtained, let's see which sets it belongs to:
- Natural Numbers: [tex]\(1\)[/tex] is a natural number.
- Whole Numbers: [tex]\(1\)[/tex] is a whole number.
- Integers: [tex]\(1\)[/tex] is an integer.
- Rational Numbers: [tex]\(1\)[/tex] can be expressed as [tex]\(\frac{1}{1}\)[/tex], so it is a rational number.
- Irrational Numbers: [tex]\(1\)[/tex] is rational, not irrational.
- Real Numbers: Since [tex]\(1\)[/tex] is rational, it is also a real number.
Thus, for [tex]\(\sqrt{(3)^2 - 4(1)(2)}\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & \text{Expression} & \text{Natural} & \text{Whole} & \text{Integer} & \text{Rational} & \text{Irrational} & \text{Real} \\ \hline 3. & \sqrt{(3)^2-4(1)(2)} & X & X & X & X & & X \\ \hline \end{array} \][/tex]
To summarize everything in one table:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
& Expression & Natural & Whole & Integer & Rational & Irrational & Real \\
\hline
1. & [tex]\(\frac{2}{9}\)[/tex] & & & & X & & X \\
\hline
2. & [tex]\(\sqrt{2}\)[/tex] & & & & & X & X \\
\hline
3. & [tex]\(\sqrt{(3)^2 - 4(1)(2)}\)[/tex] & X & X & X & X & & X \\
\hline
\end{tabular}
\end{center}
### 1. Expression: [tex]\(\frac{2}{9}\)[/tex]
- Natural Numbers: These are positive integers starting from 1 onwards (1, 2, 3, ...). [tex]\(\frac{2}{9}\)[/tex] is not a natural number.
- Whole Numbers: These include all natural numbers plus zero (0, 1, 2, 3, ...). [tex]\(\frac{2}{9}\)[/tex] is not a whole number.
- Integers: These include all positive and negative whole numbers and zero (..., -3, -2, -1, 0, 1, 2, 3, ...). [tex]\(\frac{2}{9}\)[/tex] is not an integer.
- Rational Numbers: These numbers can be expressed as the quotient of two integers, where the denominator is not zero. [tex]\(\frac{2}{9}\)[/tex] can be expressed as [tex]\(\frac{2}{9}\)[/tex], so it is a rational number.
- Irrational Numbers: These cannot be expressed as the quotient of two integers. [tex]\(\frac{2}{9}\)[/tex] is a rational number, not irrational.
- Real Numbers: These include both rational and irrational numbers. Since [tex]\(\frac{2}{9}\)[/tex] is rational, it is also a real number.
Thus, for [tex]\(\frac{2}{9}\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & \text{Expression} & \text{Natural} & \text{Whole} & \text{Integer} & \text{Rational} & \text{Irrational} & \text{Real} \\ \hline 1. & \frac{2}{9} & & & & X & & X \\ \hline \end{array} \][/tex]
### 2. Expression: [tex]\(\sqrt{2}\)[/tex]
- Natural Numbers: [tex]\(\sqrt{2}\)[/tex] is not a whole number, nor is it a positive integer.
- Whole Numbers: [tex]\(\sqrt{2}\)[/tex] is not a whole number.
- Integers: [tex]\(\sqrt{2}\)[/tex] is not an integer.
- Rational Numbers: [tex]\(\sqrt{2}\)[/tex] cannot be expressed as a ratio of two integers, so it is not rational.
- Irrational Numbers: [tex]\(\sqrt{2}\)[/tex] is an irrational number, as it cannot be expressed as the quotient of two integers.
- Real Numbers: [tex]\(\sqrt{2}\)[/tex] falls within the set of real numbers since it is either rational or irrational.
Thus, for [tex]\(\sqrt{2}\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & \text{Expression} & \text{Natural} & \text{Whole} & \text{Integer} & \text{Rational} & \text{Irrational} & \text{Real} \\ \hline 2. & \sqrt{2} & & & & & X & X \\ \hline \end{array} \][/tex]
### 3. Expression: [tex]\(\sqrt{(3)^2 - 4(1)(2)}\)[/tex]
First, simplify the expression inside the square root:
[tex]\[ (3)^2 - 4(1)(2) = 9 - 8 = 1 \][/tex]
Then, take the square root of the result:
[tex]\[ \sqrt{1} = 1 \][/tex]
Since [tex]\(1\)[/tex] is obtained, let's see which sets it belongs to:
- Natural Numbers: [tex]\(1\)[/tex] is a natural number.
- Whole Numbers: [tex]\(1\)[/tex] is a whole number.
- Integers: [tex]\(1\)[/tex] is an integer.
- Rational Numbers: [tex]\(1\)[/tex] can be expressed as [tex]\(\frac{1}{1}\)[/tex], so it is a rational number.
- Irrational Numbers: [tex]\(1\)[/tex] is rational, not irrational.
- Real Numbers: Since [tex]\(1\)[/tex] is rational, it is also a real number.
Thus, for [tex]\(\sqrt{(3)^2 - 4(1)(2)}\)[/tex]:
[tex]\[ \begin{array}{|c|c|c|c|c|c|c|c|} \hline & \text{Expression} & \text{Natural} & \text{Whole} & \text{Integer} & \text{Rational} & \text{Irrational} & \text{Real} \\ \hline 3. & \sqrt{(3)^2-4(1)(2)} & X & X & X & X & & X \\ \hline \end{array} \][/tex]
To summarize everything in one table:
\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|}
\hline
& Expression & Natural & Whole & Integer & Rational & Irrational & Real \\
\hline
1. & [tex]\(\frac{2}{9}\)[/tex] & & & & X & & X \\
\hline
2. & [tex]\(\sqrt{2}\)[/tex] & & & & & X & X \\
\hline
3. & [tex]\(\sqrt{(3)^2 - 4(1)(2)}\)[/tex] & X & X & X & X & & X \\
\hline
\end{tabular}
\end{center}
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