Convert the following recurring decimals into rational numbers :
(i) `0.4overset(cdot)3overset(cdot)7` (ii) `1.7overset(cdot)2overset(cdot)3`
(iii) `0.overset(cdot)2overset(cdot)3overset(cdot)1` (iv) `0.4overset(cdot)5overset(cdot)6`

Explanation

To convert the given recurring decimals into rational numbers, we will follow a systematic approach for each case. Here’s the step-by-step solution for each part of the question:

### (i) Convert `0.4̇3̇7` into a rational number.

1. **Let x = 0.437373737...** (where 37 is repeating)
2. **Multiply by 10:**
\( 10x = 4.37373737... \)
3. **Multiply by 1000:**
\( 1000x = 437.37373737... \)
4. **Set up the equations:**
- Equation 1: \( x = 0.437373737... \)
- Equation 2: \( 1000x = 437.37373737... \)
5. **Subtract Equation 1 from Equation 2:**
\( 1000x - 10x = 437.37373737... - 4.37373737... \)
\( 990x = 433 \)
6. **Solve for x:**
\( x = \frac{433}{990} \)

### (ii) Convert `1.7̇2̇3` into a rational number.

1. **Let y = 1.723232323...** (where 23 is repeating)
2. **Multiply by 10:**
\( 10y = 17.23232323... \)
3. **Multiply by 1000:**
\( 1000y = 1723.23232323... \)
4. **Set up the equations:**
- Equation 1: \( y = 1.723232323... \)
- Equation 2: \( 1000y = 1723.23232323... \)
5. **Subtract Equation 1 from Equation 2:**
\( 1000y - 10y = 1723.23232323... - 17.23232323... \)
\( 990y = 1706 \)
6. **Solve for y:**
\( y = \frac{1706}{990} \)
Simplifying gives \( y = \frac{853}{495} \).

### (iii) Convert `0.0̇2̇3̇1` into a rational number.

1. **Let z = 0.023123123...** (where 231 is repeating)
2. **Multiply by 1000:**
\( 1000z = 23.123123123... \)
3. **Multiply by 10:**
\( 10z = 0.231231231... \)
4. **Set up the equations:**
- Equation 1: \( z = 0.023123123... \)
- Equation 2: \( 1000z = 23.123123123... \)
5. **Subtract Equation 1 from Equation 2:**
\( 1000z - 10z = 23.123123123... - 0.231231231... \)
\( 990z = 23 - 0.023 = 23.1 \)
6. **Solve for z:**
\( z = \frac{23.1}{990} \)
Converting \( 23.1 \) to a fraction gives \( z = \frac{231}{990} \) and simplifying gives \( z = \frac{77}{330} \).

### (iv) Convert `0.4̇5̇6` into a rational number.

1. **Let p = 0.456565656...** (where 56 is repeating)
2. **Multiply by 10:**
\( 10p = 4.56565656... \)
3. **Multiply by 1000:**
\( 1000p = 456.56565656... \)
4. **Set up the equations:**
- Equation 1: \( p = 0.456565656... \)
- Equation 2: \( 1000p = 456.56565656... \)
5. **Subtract Equation 1 from Equation 2:**
\( 1000p - 10p = 456.56565656... - 4.56565656... \)
\( 990p = 452 \)
6. **Solve for p:**
\( p = \frac{452}{990} \)
Simplifying gives \( p = \frac{226}{495} \).

### Summary of Results:
1. \( 0.4̇3̇7 = \frac{433}{990} \)
2. \( 1.7̇2̇3 = \frac{853}{495} \)
3. \( 0.0̇2̇3̇1 = \frac{77}{330} \)
4. \( 0.4̇5̇6 = \frac{226}{495} \)