Repeating decimals. We can use the formula for the sum of an infinite geometric series to express a repeating decimal as simple as possible of a fraction.
Consider the successive quotients that we obtain in the division of 10 by 3 at different steps of division. In this process we get 3, 3.3, 3.33, 3.333, … and so on. These quotients also form a sequence. The various numbers occurring in a sequence are called its terms. We denote the terms of a sequence by a1, a2, a3, …, an, the subscripts denote the position of the term. The nth term is the number at the nth position of the sequence and is denoted by an The nth term is also called the general term of the sequence.
Thus, the terms of the sequence of the example of successive quotients mentioned above are:
The formula for the sum of an infinite geometric series can be used to write a repeating decimal as a fraction. Remember that decimals with bar notation such as 0.2 and 0.47 represent 0.222222… and 0.474747…, respectively. Each of these expressions can be written as an infinite geometric series.
📌 Example 1: Write a Repeating Decimal as a Fraction with two methods.
Write 0.39 as a fraction.
💎 Method 1
Label the given decimal.
Multiply each side by 100.
💎 Method 2
Write the repeating decimal as a sum.
In this series, a1=0.39 and r=0.01.
💎 Another method. This is made quite easy with the following observations:
So we can see that fractions with a denominator of 9, 99, 999, 9999, are very useful for making repeating decimals. All we have to do is to reduce the fraction to its lowest terms.
For example, let’s look at the repeating decimal 0.027027027… .
Clearly this is 27/999=1/37 (having divided top and bottom of the fraction by 27)
Now let’s try something more tricky. Take a look at 0.4588888888… This isn’t simply something divided by 99, since the 45 bit doesn’t repeat. What we need to do is move the decimal point over to the start of the repeating bit. In this case we multiply by 100 to get 45.888888… .
Now we know the fraction part is 8/9. In total we have 45+8/9=413/9 (changing into an improper fraction will make things easier for us). So 45.88888…=413/9
Now just divide both sides by 100 to get: 0.458888…=413/900
💎 Thus, the third method to solve the problem of example 1 is as follows.
📌 Example 2: Writing a Repeating Decimal as a Fraction
Express 0.555555… as a fraction.
0.555555… can be written
where a=5/10=.5 and r=1/10=.1. Using equation (5.0) for sum to infinity
📌 Ex3. A rational number is a number that can be expressed as a quotient of two integers. Show that 0.6=0.666… is a rational number.
This is an infinite geometric series with a1=6/10and r=1/10; therefore,
📌 Ex4. Express 0.7 as a fraction in lowest terms.
Observe that 0.7 is an infinite geometric series with first term 0.7 and common ratio 0.1. Because r=0.1, the condition that |r|<1 is met.Thus, we can use our formula to find the sum. Therefore,
An equivalent fraction for 0.7 is 7/9.
📌 Example 5: Using an Infinite Geometric Series to Solve a Problem
Determine a fraction that is equal to 0.49.
The repeating decimal 0.49 can be expressed as:
The repeating digits form an infinite geometric series. The series converges because -1<r<1. Use the rule for S∞.
Substitute: t1=0.09, r=0.1, or 0.009/0.09=0.1
Add 0.1 to 0.4, the non-repeating part of the decimal:
📌 Ex6. Use the same method even if your number begins as a non-repeating number. For instance, write 5.13333333333… as a fraction. Writing this as the sum of its fractional parts yields 51/10+3/100+3/1000+3/10000+… . You should recognize that the repeating part is a geometric series and apply the formula
Add this to the non-repeating part of the number (51/10) and you get 51/10+1/30=154/30=77/15.
📌 Example 7: Writing a Repeating Decimal as a Fraction
Write 0.181818… as a fraction.
(Write rule for sum.)
(Substitute for a1 and r.)
(Write as a quotient of integers.)
The repeating decimal 0.181818… is 2/11 as a fraction.
📌 Example 8. Write decimal 0.2727272… as a fraction.
We know that 0.2727272… can be written as the sum of 0.27+0.0027+0.000027+0.00000027+⋯.
These series 0.27+0.0027+0.000027+0.00000027+… can be written as the sum using summation notation ∑∞n=1 27(1/100)n . To find an infinite sum, we use the formula for the sum (S) of an infinite geometric sequence:
Divide 3/11 on your calculator. 3/11= 0.2727272727… .
Changing Repeating Decimals to Fractions:
📌 Ex9. Show that the repeating, non-terminating decimal 0.2727… is equal to 3/11.
✍ Solution: The decimal can be expanded and written as 0.27+0.0027+0.000027+… . The expanded decimal looks like an infinite geometric series.
Writing the decimal as a fraction gives
The series of numbers really is an infinite geometric series, since there is a common ratio, r=1/100, with a1=27/100. So solving for the sum, gives
So it has been shown that 0.2727… =3/11.
This example shows how to change repeating, non-terminating decimals to fractions. Actually all repeating, non-terminating decimals can be changed to fractions using this method. Other examples are shown so this method gets used to solving this type of problem.
📌 Example 10: Expressing a Repeating Decimal as a Fraction
Represent the repeating decimal 0.454 545…=0.45 as the quotient of two integers. Recall that a repeating decimal names a rational number and that any rational number can be represented as the quotient of two integers.
The right side of the equation is an infinite geometric series with a1=0.45 and r=0.01. Thus,
Hence, 0.45 and 5/11 name the same rational number. Check the result by dividing 5 by 11.
📌 Example 11: Repeating decimal
Write 1.2454545…(=1.245) in terms of an infinite geometric series, then use Equation (5) to express 1.245 in the form p/q, where p and q are integers.
The terms following 12/10 form an infinite geometric series with a=0.045=45/1000 and r=0.01=1/100 Since r is between -1 and 1, we may use Equation (5) to express the sum as
Therefore, 137/110 and 1.245 represent the same number.
📌 Ex12. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015, … .
📌 Ex13. Change 5.135135… to fraction in lowest terms.
The sequence of numbers after 5, forms an infinite geometric series with a1=135/1000 and r=1/1000. So that
Solution to a problem like this looks so hard. Well, notice that every detail has been put here so one gets to understand each step better. But when one solves on his/her own, shortcuts may be used. Look method 2, and method 3(Another method) above.