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Classifying Sums and Products as Rational or Irrational

For each sum or product, determine whether the result is a rational number or an irrational number. Then choose the appropriate reason for each.

\begin{tabular}{|c|c|c|}
\hline
& \begin{tabular}{l}
Result is \\
Rational
\end{tabular} & \begin{tabular}{l}
Result is \\
Irrational
\end{tabular} \\
\hline
(a) [tex]$34+\sqrt{7}$[/tex] & 0 & [tex]$\bigcirc$[/tex] \\
\hline
(b) [tex]$\frac{12}{17}+\frac{4}{21}$[/tex] & [tex]$\bigcirc$[/tex] & 0 \\
\hline
(c) [tex]$\sqrt{6} \times 23$[/tex] & 0 & [tex]$\bigcirc$[/tex] \\
\hline
(d) [tex]$8 \times \frac{13}{19}$[/tex] & [tex]$\bigcirc$[/tex] & 0 \\
\hline
\end{tabular}

Choose one reason for each:

1. The sum of two rationals is always rational.
2. The sum of a rational and an irrational is always irrational.
3. The sum of two irrationals is sometimes rational, sometimes irrational.
4. The product of two rationals is always rational.
5. The product of a nonzero rational and an irrational is always irrational.
6. The product of two irrationals is sometimes rational, sometimes irrational.

Sagot :

GinnyAnsweridn
Let's analyze each expression and determine whether it is rational or irrational.

### (a) [tex]\(34 + \sqrt{7}\)[/tex]
- Type of numbers involved:
- 34 is a rational number.
- [tex]\(\sqrt{7}\)[/tex] is an irrational number because the square root of a non-perfect square is irrational.
- Rule applied: The sum of a rational and an irrational number is always irrational.
- Conclusion:
- Result: Irrational
- Reason: The sum of a rational and an irrational is always irrational.

### (b) [tex]\(\frac{12}{17} + \frac{4}{21}\)[/tex]
- Type of numbers involved:
- Both [tex]\(\frac{12}{17}\)[/tex] and [tex]\(\frac{4}{21}\)[/tex] are rational numbers because they are ratios of integers.
- Rule applied: The sum of two rational numbers is always rational.
- Conclusion:
- Result: Rational
- Reason: The sum of two rationals is always rational.

### (c) [tex]\(\sqrt{6} \times 23\)[/tex]
- Type of numbers involved:
- [tex]\(\sqrt{6}\)[/tex] is an irrational number because the square root of a non-perfect square is irrational.
- 23 is a rational number.
- Rule applied: The product of a nonzero rational and an irrational number is always irrational.
- Conclusion:
- Result: Irrational
- Reason: The product of a nonzero rational and an irrational is always irrational.

### (d) [tex]\(8 \times \frac{13}{19}\)[/tex]
- Type of numbers involved:
- 8 is a rational number.
- [tex]\(\frac{13}{19}\)[/tex] is also a rational number.
- Rule applied: The product of two rational numbers is always rational.
- Conclusion:
- Result: Rational
- Reason: The product of two rationals is always rational.

### Final Table

[tex]\[ \begin{array}{|c|c|c|} \hline & \begin{array}{l} \text{Result is } \\ \text{Rational} \end{array} & \begin{array}{l} \text{Result is } \\ \text{Irrational} \end{array} \\ \hline (a) \ 34+\sqrt{7} & & \bigcirc \\ \hline (b) \ \frac{12}{17}+\frac{4}{21} & \bigcirc & \\ \hline (c) \ \sqrt{6} \times 23 & & \bigcirc \\ \hline (d) \ 8 \times \frac{13}{19} & \bigcirc & \\ \hline \end{array} \][/tex]

### Reasons Summary:
1. The sum of a rational and an irrational is always irrational.
2. The sum of two rationals is always rational.
3. The product of a nonzero rational and an irrational is always irrational.
4. The product of two rationals is always rational.
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