Express each repeating decimal as a fraction in lowest terms. 0.6...

Question

Express each repeating decimal as a fraction in lowest terms. 0.6 (repeating 6)

Accepted Answer

Welcome back. I'm so glad you're here. We've got a repeating decimal 8.333 in those threes, they just keep going all the way to infinity, we want to express this as a fraction by adding the whole number. Okay, we know the whole number. That's eight to the sum of a geometric series. Okay, we'll write that as s so all we have to do is eight plus s Now let's figure out s okay to figure out s some of the geometric series. First we got to know do we have a geometric series going on here? We're not gonna worry about that. Eight. We're just looking at that 80.3 repeating. Is that a geometric series? Let's rewrite it that .3 in the 10th place we could say well that is added to This three, that's in the hundreds place. So that could be rewritten as . and well that's being added to this next three in the 1000th place. So that's being added 2.003. And yeah those will just continue going to infinity looking like a series. Let's change this to fractions because we're going to be working with fractions. So that again we said that was in the 10th place. So 3/10 plus this one this three is in the hundreds place. So 3 1/100 and this three is in the 1000th place. So 3 1/1000 and how do we know for sure if we've got a geometric series Well we can test the ratio, we can pick two different sets of numbers that are sequential and we can see if they have the same ratio. So for the first set of numbers that we're going to test, let's pick the first two terms. And we're going to take the second term sub two and divide it by the term immediately preceding it. A sub one. So a sub two, that is 3 1/100 divided by the first term. That's 3/10. I'm going to rewrite this as 31 hundred's divided by 3/10 because that reminds me that we're just going to multiply by the reciprocal of the second term. And then we get to do some beautiful cancelations. The three's cancel and some of these zeroes cancel the first one. So that just leaves us with 1/10. Okay, we've got one ratio between the 1st and 2nd terms. Now to confirm that it's a geometric series, we pick two others. And so let's pick the 2nd and 3rd terms. Again we take the second of the two that we've chosen in this case. That is the third term. A sub three divided by the one immediately preceding it which is a sub two. So we've got 3 1/1000 divided by 3. 1/100. Going to rewrite that so that we can see we're just going to be multiplying by the reciprocal of the second term. And then we get to do some beautiful cancelations. Those threes are gone two zeroes from each of those. And yes, we can confirm that this is a geometric series because the ratios are the same. The ratio is 1/10. Great. So the next step we know that it's an infinite geometric series because it keeps going and going and going and we want to know if we can in fact find the sum of it because that our value has to meet certain criteria. The criteria is if our Is between -1 and one then we can keep going. So is 1/ Between -1 and one. Yes it is. So we can keep going with our formula. We recall from previous lessons that the some oh that's what we were looking for 1000 years ago. The sum of an infinite geometric series is a sub one. The value of our first term divided by one minus r. As long as our meets those criteria, our has to be between negative one and one. And it is so now we can fill in our a sub one. That was that first term here. So a sub one Is 3/10. I can write that better. A sub one equals 3/10. Plugging that in up here divided by one minus Our is 1/10. Great draw a line here starting to get busy. So let's simplify. We've got 3/10 divided by well 1 -1/10. We know that 1 -1/10 is going to be 9/10. So rewriting this is 3/10 times the reciprocal of the second term or 10/9. Can do some beautiful cancelation here. The tens are gone. The three in the nine can reduce each other to one and three so are some Is going to be 1/3. We've got our s value. That's what we were trying to find. So now we plug this in. We can go eight plus one third. Well these two terms don't have a common denominator. So let's multiply eight times 3/3. That will give us 24 3rd plus one third. Ad straight across for the enumerators. And that's 25 3rd. We look at our answer choices and that matches with answer choice. C Well done. We'll catch on the next one.

Express each repeating decimal as a fraction in lowest terms. 0.6 (repeating 6)

Key Concepts

Repeating Decimals

Conversion of Decimals to Fractions

Lowest Terms

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