Converting recurring decimals to fractions new formula

In summary, a student discovered a faster way to convert recurring decimals in their math class. The traditional over 9 rule only works if the recurring numbers come directly after the decimal point. The student's method involves subtracting the non-recurring number from the whole number and placing the result on top, with one 9 for each recurring number and a 0 for each non-recurring number on the bottom. This method is not practical for computers, but the student found it useful for their exams. They also argue that being able to perform manual math is still important despite the existence of calculators.
  • #1
Charliepic
1
0
Okay so the name may be bigging it up however i found something in my maths class today that i don't think has previously been published or at least I've never been taught it, it makes converting reacuring decemals a lot quicker
I know about the over 9 rule however acording to my maths teacher this rule is only possible if the recurring number(s) come directly after the decimal point and not if there is a non reacuring number directly after the decimal point
I apologise if that's hard to understand not very good at explainging things via typing and i also want to stress that i have no idea if this is a thing alrady or not and if it is its handy and will help with my exams so a plus if its not then i guess i think it would be handy for people to know.
Please no negative stuck up replys i guess what I am asking is is it possible to use the over 9 rule with a reacurring decimal that has a non reacuring number directly after the decimal poing eg 0.2(nr)25(both reacuring •) i will post the "theory" if some one does not reply with it saying this is alrady a thing soz m8
 
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  • #2
0.225 = 1/10 * 2.25 = 1/10 (2+25/99) where bold means repeating.
No magic, very simple algebra. Way too simple for a publication.
 
  • #3
Okay answer appreciated however and this is according to my maths teacher my answer was correct 223/990 and i will post the theory just to show you but it was along the lines of
You take the non recurring number(s) in this case 2 then you take that number away from the whole number 225 leaving you with 223 you then place this on top and underneath you put one 9 for every recurring number and a zero for every non recurring number, so on top you have 223 and at bottom you have 990 one nine for the recurring 2 and one for the reoccurring 5 and one zero for the non recurring 2
This enabled me to do them all quicker than how we were shown which is why i thought it may be of some use ,its also worth noting that the formula does work for all the questions including ones such as 1.2(25) (brackets=recurring)
I appreciate that this may just be simple that just thought it was quite a handy thing to know like, however I am 15 in secondary school and wether its of any use or not is beyond me
 
  • #4
There is no practical application of making those transformations manually, and a computer doesn't care (computers store numbers differently).
It works, yes, in a similar way to the calculation in post 2, just a bit faster.
 
  • #5
Yes but practical application or not we are still taught it in school and require to be able to do so to pass our exams and if this way works faster and easier than the standard way then i just thought it would make sense for people to use it,as for computers not caring ok agreed however i could use a calculator to do most mathimatcal sums in our non calculator test in a matter of minutes but we are not allowed to use them so this is a manual form of doing the sum the fact that you have just stated there is no need for a way of manualy doing sums is wrong that's like saying no one needs to understand maths because they can use a calculator in every practical element of life, so there is no point in me doing maths at school? There is no point in peoples degrees and things in maths as a calculator can do it better ? No obliviously there is a point to being able to do manual maths and any way of making life easier is a good thing right
 
  • #6
Charliepic said:
Yes but practical application or not we are still taught it in school and require to be able to do so to pass our exams and if this way works faster and easier than the standard way then i just thought it would make sense for people to use it,as for computers not caring ok agreed however i could use a calculator to do most mathimatcal sums in our non calculator test in a matter of minutes but we are not allowed to use them so this is a manual form of doing the sum the fact that you have just stated there is no need for a way of manualy doing sums is wrong that's like saying no one needs to understand maths because they can use a calculator in every practical element of life, so there is no point in me doing maths at school? There is no point in peoples degrees and things in maths as a calculator can do it better ?
Off-topic, but the above is a very long sentence. Have you learned about the use of the period (.) yet?
 
  • #7
Charliepic said:
Okay answer appreciated however and this is according to my maths teacher my answer was correct 223/990 and i will post the theory just to show you but it was along the lines of
You take the non recurring number(s) in this case 2 then you take that number away from the whole number 225 leaving you with 223 you then place this on top and underneath you put one 9 for every recurring number and a zero for every non recurring number, so on top you have 223 and at bottom you have 990 one nine for the recurring 2 and one for the reoccurring 5 and one zero for the non recurring 2
This enabled me to do them all quicker than how we were shown which is why i thought it may be of some use ,its also worth noting that the formula does work for all the questions including ones such as 1.2(25) (brackets=recurring)
I appreciate that this may just be simple that just thought it was quite a handy thing to know like, however I am 15 in secondary school and wether its of any use or not is beyond me
I am more interested in why this works than I am in the process itself. Can you come up with an explanation of why this works?
 
  • #8
@Charliepic: The point of those exercises is to understand fractions, the decimal representation and so on. In particular, I think the more standard steps show more clearly what is happening.
If you understand why your method works (and not just by accident) then you can move on to the next topic anyway.

@Mark44: are you asking that in general, or specifically Charliepic?
 
  • #9
mfb said:
@Mark44: are you asking that in general, or specifically Charliepic?
Specifically Charliepic. I quoted what he wrote, which I thought would make it clear my comment was addressed to him.
 
  • #10
I just noted that this method has an ugly extra rule for numbers like 0.3412, where bold means repeating as above. You can still use it, but you need an additional rule to handle the carry as the obvious way does not work.
 
  • #11
Ymm I am in midde of my exams so haven't had a chance to look on hear sorry ymm il have a look as to how this works as for the number you put mfb could you place the reacuring in brakets so i can see no formating on my phone...[emoji58] - as for bad grammer il just apologies and leave it at that
 
  • #12
0.34(12)
0.34121212121212...
You could have quoted my post to see the formatting I guess.
 
  • #13
Off topic mfb you come across as a make every situation awkward t*** you could change that, i guess.
On topic i fail to see how that does not simply become
3412-34 = 3378
3378/9900
 
  • #14
3378/9900= 0.34(12)
 
  • #15
Please explain this extra rule because i did not use one and got a correct answer i get the feeling you don't like this but you don't need to be a dick
 
  • #16
Mark44 said:
Specifically Charliepic. I quoted what he wrote, which I thought would make it clear my comment was addressed to him.

Mark44 i think this works because the original method would be to times the number so that the repeat started directly after the decimal point eg.

0.341212121212 * 100 = 34.12121212

And then you would do the same but so one repeat is before the point eg.

0.341212121212*10000 = 3412.12121212

Then you take away the first number from the second leaving you with

9900 X= 3412.12121212-34.12121212 = 3378
X=3378/9900

My method is simply doing the minuses without the need to multiply hope this helps to understand [emoji4]
 
  • #17
Sorry mfb post #13 was meant for mark44 post #15 was for you
 
  • #18
Hmm it works without a special rule, right. My mistake.
 
  • #19
[emoji111]️[emoji263]
 

1. What is the new formula for converting recurring decimals to fractions?

The new formula for converting recurring decimals to fractions is numerator = recurring decimal - non-recurring decimal and denominator = 9s, where s is the number of repeating digits.

2. How is the new formula different from the old formula?

The old formula for converting recurring decimals to fractions was numerator = recurring decimal and denominator = 9s, which only worked for one-digit repeating decimals. The new formula takes into account both the recurring and non-recurring decimals, making it applicable to all recurring decimals.

3. Can the new formula be used for both terminating and non-terminating recurring decimals?

Yes, the new formula can be used for both terminating and non-terminating recurring decimals. It is a more comprehensive formula that covers all types of recurring decimals.

4. Are there any limitations to the new formula?

The new formula may not work if the recurring decimal has a pattern that is not a repeating sequence. In such cases, a different method may need to be used to convert the decimal to a fraction.

5. How can I check if my answer is correct when using the new formula?

You can simply convert the resulting fraction back to a decimal and see if it matches the original recurring decimal. If it does, then your answer is correct. You can also use a calculator to verify your answer.

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