In Exercises 45–50, express each repeating decimal as a fraction ...

Question

In Exercises 45–50, express each repeating decimal as a fraction in lowest terms.0.257 ̅ (repeating 257)

Accepted Answer

Hello everyone in this video we're going to be looking at this practice problem where we want to write and simplify the decimal 0.5-3, it's a repeating decimal as a fraction. So first let's see what this looks like as a fraction. So 0.5 - three is the same thing as 523 divided by 1000. And because it's a repeating decimal with three digits I have to add 523 divided by one million. Notice how I added three Zeros because I have a three digit number on top and I can continue like this forever just adding three Zeros to my denominator. But let's try to rewrite this using our knowledge of sequences and writing this as a infinite sum. So the first thing that we're gonna want to do is to try and find the common ratio which all due note as our and to find our I'm going to divide this second term 523 divided by one million divided by the first term divided by 1000. And from here recall that I can change this division into multiplication if I switch the denominator and the numerator Of my second fraction. So my second fraction will now be 1000 divided by 523. And from here you can see that the 520 three's cancel out to one and I can get rid of three zeros from the denominator of the first reaction and three zeros from the numerator of the second fraction And that leaves me with a common ratio of one over 1000. So now I have a common ratio. And because this is a geometric sequence, So note that it's a geometric sequence because when I multiply the first term, 523 divided by 1000 with one over 1000, I get the second term. So I have that common ratio. And the sum of an infinite geometric sequence is given with the formula. The sum of an infinite number of terms is equal to a One divided by -2. Where a one is the first term and R is the common ratio. And we have both of those values. So we can determine what this is. So us or the sum of the infinite sum is equal to a one which is 523 divided by divided by one minus R, which is one over 1000. And if we simplify this I still have 523 divided by divided by I'm going to change the one to be 1000 over and I need to subtract one from the 1000. So this becomes 523, divided by 1000, divided by Divided by 1000. So notice I just subtracted one from the 1000 in my numerator And now I'm going to do the same switcheroo where I'm going to change this division into multiplication by switching the denominator and numerator. So now I have a numerator of 1000 in my second fraction And a denominator of 999. And from here you can see that the 1000s cancel out leaving me with one. So the infinite sum is actually Divided by 999. And this means that I can write the decimal 0.5-3, repeating as the fraction divided by 999. So this becomes my final answer. And you can see that this aligns with answer choice D. So hopefully this video opt and see you next time.

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In Exercises 45–50, express each repeating decimal as a fraction in lowest terms.0.257 ̅ (repeating 257)

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