Show that the following recurring decimals are rational. $$ \begin{array}{llll} ext { (1) } 0, \dot{4} & ext { (2) } 0, \dot{2} \dot{1} & ext { (3) } 0,1 \dot{4} \ ext { (5) } 0,12 \dot{4} & ext { (6) } 0, \dot{1} \dot{4} & ext { (7) }-1,1 \dot{4} \dot{4}\end{array} $$

Question

Show that the following recurring decimals are rational. $$ \begin{array}{llll}	ext { (1) } 0, \dot{4} & 	ext { (2) } 0, \dot{2} \dot{1} & 	ext { (3) } 0,1 \dot{4} \ 	ext { (5) } 0,12 \dot{4} & 	ext { (6) } 0, \dot{1} \dot{4} & 	ext { (7) }-1,1 \dot{4} \dot{4}\end{array} $$

UpStudy AI · Accepted Answer

To show that each of the given recurring decimals is rational, we'll express each one as a fraction. A number is **rational** if it can be written as the ratio of two integers, i.e., $$ \frac{p}{q} $$ where $$ p $$ and $$ q $$ are integers and $$ q \neq 0 $$.

Let's examine each recurring decimal one by one:

### (1) $$ 0.\overline{4} $$

**Expression:**
$$ x = 0.4444\ldots $$

**Solution:**
1. Multiply both sides by 10:
   $$ 10x = 4.4444\ldots $$
2. Subtract the original equation from this new equation:
   $$ 10x - x = 4.4444\ldots - 0.4444\ldots $$
   $$ 9x = 4 $$
3. Solve for $$ x $$:
   $$ x = \frac{4}{9} $$

**Conclusion:**
$$ 0.\overline{4} = \frac{4}{9} $$
Thus, $$ 0.\overline{4} $$ is rational.

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### (2) $$ 0.\overline{21} $$

**Expression:**
$$ x = 0.212121\ldots $$

**Solution:**
1. Multiply both sides by 100 (since the repeating block has 2 digits):
   $$ 100x = 21.212121\ldots $$
2. Subtract the original equation from this new equation:
   $$ 100x - x = 21.212121\ldots - 0.212121\ldots $$
   $$ 99x = 21 $$
3. Solve for $$ x $$:
   $$ x = \frac{21}{99} = \frac{7}{33} $$

**Conclusion:**
$$ 0.\overline{21} = \frac{7}{33} $$
Thus, $$ 0.\overline{21} $$ is rational.

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### (3) $$ 0.1\overline{4} $$

**Expression:**
$$ x = 0.14444\ldots $$

**Solution:**
1. Let’s separate the non-repeating and repeating parts:
   $$ x = 0.1 + 0.04444\ldots $$
2. Recognize that $$ 0.04444\ldots = \frac{4}{90} = \frac{2}{45} $$.
3. Add to the non-repeating part:
   $$ x = \frac{1}{10} + \frac{2}{45} = \frac{9}{90} + \frac{4}{90} = \frac{13}{90} $$

**Conclusion:**
$$ 0.1\overline{4} = \frac{13}{90} $$
Thus, $$ 0.1\overline{4} $$ is rational.

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### (5) $$ 0.12\overline{4} $$

**Expression:**
$$ x = 0.124444\ldots $$

**Solution:**
1. Separate the non-repeating and repeating parts:
   $$ x = 0.12 + 0.004444\ldots $$
2. Recognize that $$ 0.004444\ldots = \frac{4}{900} = \frac{2}{450} = \frac{1}{225} $$.
3. Add to the non-repeating part:
   $$ x = \frac{12}{100} + \frac{1}{225} = \frac{27}{225} + \frac{1}{225} = \frac{28}{225} $$

**Conclusion:**
$$ 0.12\overline{4} = \frac{28}{225} $$
Thus, $$ 0.12\overline{4} $$ is rational.

---

### (6) $$ 0.\overline{14} $$

**Expression:**
$$ x = 0.141414\ldots $$

**Solution:**
1. Multiply both sides by 100 (since the repeating block has 2 digits):
   $$ 100x = 14.141414\ldots $$
2. Subtract the original equation from this new equation:
   $$ 100x - x = 14.141414\ldots - 0.141414\ldots $$
   $$ 99x = 14 $$
3. Solve for $$ x $$:
   $$ x = \frac{14}{99} $$

**Conclusion:**
$$ 0.\overline{14} = \frac{14}{99} $$
Thus, $$ 0.\overline{14} $$ is rational.

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### (7) $$ -1.1\overline{44} $$

**Expression:**
$$ x = -1.14444\ldots $$

**Solution:**
1. Separate the integer part and the decimal part:
   $$ x = -1 - 0.14444\ldots $$
2. From part (3), we know that $$ 0.14444\ldots = \frac{13}{90} $$.
3. Combine the two parts:
   $$ x = -1 - \frac{13}{90} = -\left(1 + \frac{13}{90}\right) = -\frac{90}{90} - \frac{13}{90} = -\frac{103}{90} $$

**Conclusion:**
$$ -1.1\overline{44} = -\frac{103}{90} $$
Thus, $$ -1.1\overline{44} $$ is rational.

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**Summary:**
All the given recurring decimals can be expressed as fractions, proving that they are rational numbers.

1. $$0.\overline{4} = \frac{4}{9}$$
2. $$0.\overline{21} = \frac{7}{33}$$
3. $$0.1\overline{4} = \frac{13}{90}$$
5. $$0.12\overline{4} = \frac{28}{225}$$
6. $$0.\overline{14} = \frac{14}{99}$$
7. $$-1.1\overline{44} = -\frac{103}{90}$$

Each repeating decimal can be written as a fraction—for example, 0. 444… = 4⁄9, 0.2121… = 7⁄33, 0.1444… = 13⁄90, etc.—so all are rational numbers.
All the given recurring decimals can be expressed as fractions, proving they are rational numbers. 1. $$0.\overline{4} = \frac{4}{9}$$ 2. $$0.\overline{21} = \frac{7}{33}$$ 3. $$0.1\overline{4} = \frac{13}{90}$$ 5. $$0.12\overline{4} = \frac{28}{225}$$ 6. $$0.\overline{14} = \frac{14}{99}$$ 7. $$-1.1\overline{44} = -\frac{103}{90}$$ Each repeating decimal can be written as a fraction, so all are rational numbers.

Show that the following recurring decimals are rational. \( \begin{array}{llll}\text { (1) } 0, \dot{4} & \text { (2) } 0, \dot{2} \dot{1} & \text { (3) } 0,1 \dot{4} \\ \text { (5) } 0,12 \dot{4} & \text { (6) } 0, \dot{1} \dot{4} & \text { (7) }-1,1 \dot{4} \dot{4}\end{array} \)

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