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## Integral Calculus

### Course: Integral Calculus>Unit 5

Lesson 2: Infinite geometric series

# Infinite geometric series word problem: repeating decimal

See how we can write a repeating decimal as an infinite geometric series. Created by Sal Khan.

## Want to join the conversation?

• What is the trick he used at
• What if you have a number like 0.76383838383838?

It's starts with 0.76, then repeats the 38. This is really stumping me in my homework. Having trouble finding the common ratio to use.
• I don't use Sal's method, but find the following more practical. This is mathematically equivalent to what Sal does, but it combines a bunch of steps and eliminates the subtraction.

The example you gave is particularly difficult to do, so it is a good one.

Step 1: Multiply and divide by a sufficient power of 10 to move the decimal place to the first of the repeating digits:
= (1/100)*(100)(0.7638383....)
= (1/100)(76.38383....)
Step 2: split the number into whole number and decimal portions
= (1/100)(76+ 0.38383....)

Step 3: Multiply and divide by as many 9s as there are repeating digits. One repeating digit means multiply by 9, two repeating digits means multiply by 99, three repeating digits means multiply by 999, etc.
= (1/100)(1/99) [99(76+ 0.38383....)]
= (1/9900) [(99*76) + 99(0.38383...)]

This uses a mathmatical trick so that the repeating portion multiplied by 9s will equal one set of the repeating digits. So 99(0.383838...) = 38
= (1/9900) [(99*76) + 38]
= (1/9900) [7525 + 38]
= (1/9900) [7525 + 38]
= (1/9900) [7562]
= 7562 / 9900
This is the repeating decimal as a fraction.

Step 4: Simplify if possible (not always easy). The GCF of 7562 and 9900 is 2, Thus
= 3781/4950
So, your example as a fraction in simplest form is 3781/4950.
• According to Sal's method, any repeating decimal can be expressed as an infinite geometric series with `r = 0.1 or 0.01 or 0.001 or 0.0001 or so on`. Is it safe for us to conclude that any repeating decimal can be expressed as a fraction with the denominator with this format `9 or 99 or 999 or 9999 or so on`?
• I guess it depends on what you mean by a "repeating decimal." If the decimal consists completely of a repeating sequence, then I'd say the answer is yes, since your r values will be as you stated and the first term can be taken as the repeating sequence.

However, what if you have a decimal such as 0.12323232323232323..., where just the 23 repeats? Would you consider this a repeating decimal? Probably so (and it's definitely a rational number), but it can't be strictly represented as a geometric sequence. The 0.1 at the start throws it off a bit. So it ends up looking like 122/990 in fractional form, which doesn't follow your pattern of 9, 99, 999, etc. What do you think?
• So, is that actually a proof that 9.9999... = 10?
I could write it as sum 9*(0.1)^k, from k = 0 to inf, which would result in
9/(1-0.1) = 10
• 10-9.999... gives an answer that is closer to 0 by any other number you can give - this is why the two numbers are equal (allthough written in different ways)
• @. Why not put the k back into 10^-4 = .4008(10^-4k). Or am I missing something entirely.
• They are the same. (10^-4)^k = (10^-4k) = 1/(10^4k)
(1 vote)
• What is .016 with the 16 repeating as a fraction in simplest form?
• Okay: Let x = .0161616.... Then 100x = 1.6161616.... What would 100x - x =?
100x -x = 99x = (1.6161616.... - .0161616....) = 1.6 Therefore x = 1.6/99 = 16/990 = 8/495
I can't think of any endlessly repeating real number that can't be expressed as a fraction by this method.
P.S. 1.6/99 = 16/990 by multiplying by 10/10. 16/990 = (2*8)/(2*495) = 8/495 by noting 2/2 = 1
• Doesn't |r| have to be less than 1? In the video, it is equal to 10.
(1 vote)
• It is not equal to 10, it is equal to 10^-4, which is the same as 0.0001, which is less than 1.
• Why does Sal do anything with the denominator at ? I dont understand why he would expand those like that. It certainly gives a different answer if you do that. I was going through the questions and didn't use fractions like that but kept the decimals and my answer was 1.96 instead of 196. Can someone explain this to me?
• From his example: 1 - 0.0001 = 0.9999, which is also equal to (9999/10,000). He probably just does this to remove decimals, since 0.4008/(9,999/10,000) = (0.4008*10,000)/9,999 = 4,008/9,999. This is still equal to 0.4008/0.9999.
• This seems to obviously generalize to any repeating decimal, and so all repeating decimals are rational - because they are all equal to the ratio of two integers. No?
(1 vote)
• Yep! You're on point. Any repeating decimal can be represented as a/b, where a and b are integers. (with nonzero b)
• I got the sum from k=1 to infinity (4008/10^4k)
Is this equivalent to sal's answer?
Should I write it from k=0 to infinity because then i can check it's sum using a/1-r?

EDIT
My reasoning is because when k = 1, 4008/10^4 = .4008
k = 2, 4008/10^4(2) = .00004008
using inductive reasoning,
.400840084008....