Converting Recurring Decimals into Fractions
To convert a recurring decimal into a fraction, we must first identify the pattern of repetition. We will then use this pattern to create an equation and solve it using simple algebraic processes.
Let us consider an example:
Convert 0.6666… into a fraction.
Step 1: Identify the Pattern of Repetition
In this case, the pattern of repetition is 6. The fraction we are looking for will have 6 as its repeating number in the numerator. Therefore, we can represent the given number as follows:
0.6666… = 0.6 + 0.06 + 0.006 + …
Note that the first term has one 6, the second term has two 6’s, the third term has three 6’s, and so on. Therefore, these terms form an infinite sequence where each term is ten times smaller than the previous term. This sequence can be represented as follows:
u1 = 0.6
u2 = 0.06
u3 = 0.006
…
Step 2: Create an Equation
To create an equation, we need to find the sum of the infinite sequence we identified in Step 1. We can do this by using the formula:
Sn = u1 / (1 – r)
Where Sn is the sum of the infinite sequence, u1 is the first term, and r is the common ratio between the terms.
In this case, the first term is 0.6, and the common ratio is 0.1 (i.e., ten times smaller). Therefore, we can substitute these values into the formula to get:
Sn = 0.6 / (1 – 0.1)
Sn = 0.6 / 0.9
Sn = 2 / 3
Step 3: Solve the Equation
We have found the sum of the infinite sequence of terms. However, this sum is not the fraction we are looking for. To find the fraction with 6 as its repeating number, we need to use a simple algebraic process.
Let x be the fraction we are looking for. Then:
x = 0.6666…
10x = 6.6666…
9x = 6
x = 6 / 9
x = 2 / 3
Therefore, we have successfully converted the recurring decimal 0.6666… into the fraction 2 / 3.
Conclusion
Converting recurring decimals into fractions can be quite simple if we follow the correct method. We must identify the pattern of repetition, create an equation using the sum of the infinite sequence, and solve for the fraction we are looking for. This method is reliable and can be applied in any situation. By using this method, we can avoid errors and obtain the correct answers when working with recurring decimals.