Converting Repeating Decimals to Vulgar Fractions

The problem asks us to convert the given repeating decimal \(0.02\overline{45}\) into a vulgar fraction in its simplest form. A vulgar fraction is a fraction of the form \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).

The notation \(0.02\overline{45}\) means that the digits '45' repeat infinitely after the '02'. So the number is \(0.0245454545...\).

To convert a mixed repeating decimal like this (where there are non-repeating digits after the decimal point but before the repeating block) into a fraction, we can use an algebraic method:

1. Let the given decimal be equal to a variable, say \(x\).
2. Multiply \(x\) by a power of 10 such that the decimal point moves just before the repeating block.
3. Multiply \(x\) by another power of 10 such that the decimal point moves past one full repeating block from the original position.
4. Subtract the equation from step 2 from the equation in step 3. This will eliminate the repeating part.
5. Solve the resulting equation for \(x\).
6. Simplify the fraction obtained to its simplest form.

Step-by-Step Conversion of \(0.02\overline{45}\)

Let's apply the steps to \(x = 0.02\overline{45}\):

1. Let \(x = 0.02454545...\)
2. The non-repeating part after the decimal is '02', which has two digits. To move the decimal point just before the repeating part ('45'), we multiply \(x\) by \(10^2 = 100\).
   \(100x = 2.454545...\)
3. The repeating block is '45', which has two digits. To move the decimal point past one full repeating block from the original position \(0.024545...\), we need to move it 2 (non-repeating digits) + 2 (repeating digits) = 4 places to the right. So, we multiply \(x\) by \(10^4 = 10000\).
   \(10000x = 245.454545...\)
4. Now, subtract the equation from step 2 from the equation in step 3:
   \(10000x - 100x = 245.454545... - 2.454545...\)
   \(9900x = 243\)
5. Solve for \(x\):
   \(x = \frac{243}{9900}\)
6. Simplify the fraction \(\frac{243}{9900}\). We look for common factors for both the numerator and the denominator. The sum of the digits of 243 is \(2+4+3=9\), which is divisible by 3 and 9. The sum of the digits of 9900 is \(9+9+0+0=18\), which is divisible by 3 and 9.
   Let's divide both by 9:
   \(243 \div 9 = 27\)
   \(9900 \div 9 = 1100\)
   So, \(x = \frac{27}{1100}\).

Now, we check if the fraction \(\frac{27}{1100}\) can be simplified further. The prime factors of 27 are \(3 \times 3 \times 3\). The prime factors of 1100 are \(11 \times 100 = 11 \times 10 \times 10 = 11 \times (2 \times 5) \times (2 \times 5) = 2^2 \times 5^2 \times 11\). There are no common prime factors between 27 and 1100. Thus, the fraction \(\frac{27}{1100}\) is in its simplest form.

Resulting Vulgar Fraction

The repeating decimal \(0.02\overline{45}\) written as a vulgar fraction in its simplest form is \(\frac{27}{1100}\).

Revision Table: Decimal to Fraction Conversion

Additional Information: Understanding Decimal and Fraction Forms

Numbers can be represented in different forms, including decimals and fractions. Decimals can be terminating (like 0.5, 1.25) or non-terminating. Non-terminating decimals can be repeating (like \(0.\overline{3}\), \(0.\overline{142857}\), \(0.02\overline{45}\)) or non-repeating and non-terminating (like \(\pi \approx 3.14159...\), \(\sqrt{2} \approx 1.41421...\)).

• Terminating and repeating decimals are rational numbers, meaning they can be written as a simple fraction \(\frac{p}{q}\), where \(p\) and \(q\) are integers and \(q \neq 0\).
• Non-repeating, non-terminating decimals are irrational numbers and cannot be written as a simple fraction.

Converting between decimal and fraction forms is a fundamental skill in mathematics. For repeating decimals, the algebraic method demonstrated for \(0.02\overline{45}\) is a reliable way to find the equivalent vulgar fraction.