Recurring decimals take a fraction of effort
A. Students sometimes get confused and wrongly think that "recurring decimals" are irrational numbers (numbers that can't be written as a fraction). A rational number is a number that can be written in the form a/b where "a" and "b" are integers. Recurring decimals are actually rational numbers and practice in converting from recurring decimals to fractions helps to reinforce this definition.
A good introduction to the topic would be to do some quick-fire sums such as: 2.5025025025... - 0.5025025025... = 2.
At this stage you should also reinforce the notation for recurring decimals as 2.502 - 0.502, (these are sometimes also written with a line above the recurring part: 2.502 - 0.502). Vary these to include examples such as: 512.12 - 5.12.
This leads to the more difficult understanding of how to use a technique in algebra to get a fraction.
We'll start with a simple case: changing 0.5555555... = 0.5 into a fraction. You could ask students to do this by writing 0.5 as a recurring decimal in the form a/b, where "a" and "b" are integers.
We begin by letting n = 0.5, then multiplying 10n = 5.5. We have multiplied both sides of the equation by 10. Clever, isn't it? Our decimal parts now match.
If you then subtract 10n - n = 5.5 - 0.5, now 9n = 5.
Divide both sides by 9 and we have n = 5/9, which means that 0.5 = 5/9.
For recurring decimals that have a longer recurring sequence the method is nearly the same, except you multiply by a different power of 10.
An example might be: 0.12121212... = 0.12 Let n = 0.12.
Multiply by 100 this time to give 100n = 12.12 Subtracting we have 100n - n = 12.12 - 0.12, which gives 99n = 12.
Divide both sides by 99 and we have n = 12/99. Then find the simplest equivalent fraction, which is n = 4/33. Divide the top and bottom of fraction by 3.
However, is does become more challenging. What about a problem such as 0.5121212... = 0.512. This time we have to add an extra step. Let n = 0.512 and then 10n = 5.12. As you can see it would be hard to work out 5.12 - 0.512, so we also let 1000n = 512.12. Can you see that now the decimal parts match in 10n and 1000n?
Now we can work out 1000n - 10n = 512.12 - 5.12. This gives 990n = 507.
Divide both sides by 990 gives n = 507/990 and n = 169/330. Then find the simplest equivalent fraction.
The brilliant thing about doing these is that students can easily check them on their calculators using the fraction button. Try teaching the method to a friend to boost your confidence before doing it with a class.
Q. I vaguely remember seeing a way of finding the lowest common multiple of a set of numbers, which involved a single division sum. Do you know the one I mean?
A. Yes. This is done by an unusual kind of division. With a set ofnumbers such as 8, 3 and 6, for example, use the lowest prime number that will divide into at least one of them, then the next lowest prime number and so on until the last line is all ones.
In this case begin with 2, like this: 2 |8, 3, 6 2 |4, 3, 3 Three is not divisible by 2, so remains the same.
2 |2, 3, 3 3 |1, 3, 3 Now we are left with a 1, so we try division by 3.
|1, 1, 1 We can now calculate the lowest common multiple as 2 x 2 x 2 x 3 = 24.
The lowest common multiple is often used in adding and subtracting fractions. For example:
"Two, 4, 6, 8, who do we appreciate? Ten, 12, 14, 16, those who are really keen. Eighteen, 20, 22, 24, they're the ones who'll score for sure."
Wendy Fortescue-Hubbard is a teacher and game inventor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses. Email your questions to Mathagony Aunt at email@example.comOr write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX