Verified Answer :

  • Identify \(10.\overline{2}\) as a repeating decimal equal to \(10 + \frac{2}{9} = \frac{92}{9}\).
  • Identify \(\frac{5}{10}\) as the decimal \(0.5\).
  • Compare the two values and find they are not equal: \(\frac{92}{9} \approx 10.222 eq 0.5\).
  • Conclude that the statement is \(\boxed{False}\).

Explanation

  1. Analyzing the Notation

To determine the validity of the equation \(10.\overline{2} = \frac{5}{10}\), we first analyze the notation on both sides of the equation. On the left side, the expression is \(10.\overline{2}\). The bar notation over the digit \(2\) indicates that it is a repeating decimal, meaning \(10.\overline{2} = 10.2222...\). On the right side, the expression is \(\frac{5}{10}\), which is a simple fraction that simplifies to \(0.5\).

  1. Converting to Fractional Form

Next, we convert the repeating decimal \(10.\overline{2}\) into a fraction. We can represent it as \(10 + 0.\overline{2}\). The decimal \(0.\overline{2}\) is equivalent to \(\frac{2}{9}\). Adding these, we get:
\[10 + \frac{2}{9} = \frac{90}{9} + \frac{2}{9} = \frac{92}{9}\]
As a decimal, \(\frac{92}{9}\) is approximately \(10.222...\).

  1. Comparing and Concluding

We now compare the two values. The left-hand side is \(\frac{92}{9}\) (which is approximately \(10.222...\)), and the right-hand side is \(\frac{5}{10} = \frac{1}{2} = 0.5\). Since \(10.222... eq 0.5\), the equation given is mathematically incorrect. Therefore, the statement \(10.\overline{2} = \frac{5}{10}\) is false.

Examples

Understanding repeating decimals is important in fields like computer science and financial mathematics, where precise representations of numbers are necessary. For example, if you are calculating interest rates or recurring payments that are not simple fractions, using the correct fractional representation of a repeating decimal is essential to avoid compounding errors in financial models or software algorithms. Knowing how to convert between decimals and fractions ensures that your calculations remain accurate.