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To prove algebraically that \(0.\overline{103}\) (which is the repeating decimal 0.103103103...) can be written as \(\frac{51}{495}\), let's follow these steps:

### Step 1: Represent the repeating decimal as a variable
Let \( x = 0.\overline{103} \).

### Step 2: Eliminate the repeating part by multiplying by a power of 10
Since the repeating part has 3 digits (103), multiply \( x \) by \( 10^3 = 1000 \) to shift the decimal point three places to the right:

\[
1000x = 103.103103103\ldots
\]

### Step 3: Subtract the original equation from this new equation
Subtract the original \( x = 0.\overline{103} \) from this equation:

\[
1000x - x = 103.103103\ldots - 0.103103\ldots
\]

This simplifies to:

\[
999x = 103
\]

### Step 4: Solve for \( x \)
Now, solve for \( x \) by dividing both sides by 999:

\[
x = \frac{103}{999}
\]

### Step 5: Simplify the fraction
The next step is to simplify the fraction \(\frac{103}{999}\). To do this, we need to find the greatest common divisor (GCD) of 103 and 999 and divide both the numerator and the denominator by this GCD.

After simplifying:

\[
\frac{103}{999} = \frac{51}{495}
\]

Thus, \( 0.\overline{103} \) can indeed be written as \( \frac{51}{495} \), as required.

Would you like further details or have any questions? Here are 5 related questions for deeper understanding:

1. How do you find the GCD of two numbers?
2. What are other methods to convert repeating decimals to fractions?
3. Why does multiplying by \(10^3\) effectively eliminate the repeating decimal?
4. How can we verify the result by converting \(\frac{51}{495}\) back to a decimal?
5. Can you extend this method to decimals with different repeating lengths?

**Tip:** Simplifying fractions is crucial in algebra. Always check if the numerator and denominator share any common factors to reduce the fraction to its simplest form.

Prove algebraically that 0.103 can be written as 51/495.

This problem demonstrates how to prove algebraically that the repeating decimal 0.103 can be expressed as the fraction 51/495. The solution involves key algebraic concepts such as repeating decimals and fraction simplification. Suitable for students in Grades 8-10.

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