# Video: AQA GCSE Mathematics Higher Tier Pack 2 • Paper 1 • Question 25

Convert the decimal 0.2253 recurring to a fraction in its simplest form.

05:11

### Video Transcript

Convert the decimal 0.2253 recurring to a fraction in its simplest form.

First, let’s just be clear about exactly which digits are recurring. The first digit after the decimal point, a two, doesn’t have a dot. So this digit is not recurring. The second and fourth digits after the decimal point each have a dot over them. This means that they are the first and last digits of the recurring section. So the two, the five, and the three are all recurring. This means that the decimal written out in longhand is 0.2253253253, and so on.

Now we’ve been asked to convert this decimal to a fraction in its simplest form. And there’s a standard process that we need to follow. First, we’ll just assign a letter to this decimal. So we can choose to refer to it as 𝑥. What we’re going to do is multiply this decimal by powers of 10, so that’s 10, 100, 1000, and so on, until we create two numbers which have exactly the same pattern of repeating digits after their decimal point.

To begin with, we can multiply both sides by 10. 𝑥 multiplied by 10 gives 10𝑥. And to multiply a decimal by 10, we keep the decimal point fixed and move all of the digits one place to the left. So we now have the decimal 2.253253253, and so on. Notice though that the pattern of repeating digits after the decimal point does not agree between the two numbers. So we need to multiply by 10 again.

Now you can either think of this as multiplying 10𝑥 by 10 or multiplying the original number 𝑥 by 100. And it gives 100𝑥 equals 22.532532532, and so on. Again, notice though that the pattern of repeating digits after the decimal point for 100𝑥 doesn’t agree with the pattern for either 10𝑥 or 𝑥. So we need to multiply by 10 again.

So now we have 1000𝑥 equals 225.325325325, and so on. We still haven’t created two numbers which have an identical pattern of repeating digits after the decimal point. So we need to multiply by 10 another time. This gives 10000𝑥 equals 2253.253253253, and so on. This time, if we look carefully, we see that, for 10000𝑥, the pattern of repeating digits after the decimal point is the same as it is for 10𝑥. In each case, it’s 253, 253, 253, and so on.

This is what we were hoping to achieve. And the reason for this is we can now subtract our value for 10𝑥 from our value for 10000𝑥. I’ve written the value for 10𝑥 out again directly below the value for 10000𝑥 for ease. On the left, we have 10000𝑥 minus 10𝑥, which is 9990𝑥.

Now we always start a column subtraction from the right-hand side. So we’re starting after the decimal point. But because all of the numbers after the decimal point in 10𝑥 agree exactly with the corresponding numbers after the decimal point in 10000𝑥, when we subtract, we just get an infinitely long stream of zeros. This is why we set out trying to find two multiples of 𝑥 which had an identical pattern of repeating digits after the decimal point, because then we knew they would cancel when we subtract.

To the left of the decimal point, we can perform our column subtraction as usual. But it’s actually just 2253 subtract two, which is 2251. We don’t need to include that string of zeros. So we now have that 9990𝑥 is equal to 2251.

To find the value of 𝑥 then, we need to divide both sides of this equation by 9990. On the left, this will cancel with the 9990 in the numerator. And on the right, we’ll have 2251 over 9990. So we found this recurring decimal as a fraction.

Now the question does specify that it needs to be a fraction in its simplest form. But in fact, this fraction is already in its simplest form, as 2251 is a prime number, although probably not one that you’re familiar with. 2251 isn’t a factor of 9990, which means the fraction can’t be reduced any further. And so our final answer is that, as a fraction in its simplest form, the recurring decimal 0.2253 recurring is equal to 2251 over 9990.