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Another Maths Lesson [Jan. 31st, 2007|04:16 pm]
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[Mood |curiouscurious]

Today we learnt about recurring decimals. Such as 0.444444... being 4/9 and so on. Me and Chunk were finding this easy, so he challenged me to find the equivalent fraction of 0.58666666... which at first I thought, is there such a fraction? Are there any decimal numbers which do not have an equivalent fraction? [Pi does spring to mind, but are there others?].

Following some thought I figured the answer would be 58/100 + 6.66/999 [which Chunk found was equal to 6/900]. Adding these gives 44/75. McMaffs then introduced such problems anyway with a different method:
let 0.5866666... = N
then 100N = 58.666666...
99N = 58.08
9900N = 5808
N = 5808/9900 = 44/75.

McMaffs then posed the problem 0.87444444....
Her solution was:
Let N = 0.8744444
then 100N = 87.444444
99N = 86.57
9900N = 8657
N = 8657/9900

Chunk then said "oh I have another method" and she let him go to the board to explain.
He said 0.874444 = 87/100 + whatever 0.004444444.... is.
He then said this was equal to 4/900.
McMaffs asked him to explain why.
He didn't know, neither do I, so this is my second question.
Chunk continues his working by adding the fractions and got:
783/900 + 4/900 = 787/900
At first McMaffs thought this was wrong until I pointed out it was equivalent to 8657/9900 [both numerator and denominator are 11 times smaller by the "Chunk theorem" method].

So yeah, this is the main point. Why is it 900? I assumed it would be like other recurring decimals [like 0.42424242... = 42/99] and feature denominators with all 9s, thats how I came up with 6.66/999 for the first problem in the first place. It did also occur to me that it was because it was like the normal ones [ie 4/9] but 100 times smaller, as its two decimal places to the right from where it should be, but I couldn't really explain it.
In other news, maths challenge tomorrow!

[User Picture]From: cupati
2007-01-31 06:32 pm (UTC)
It did also... is correct.
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[User Picture]From: yayworthy
2007-01-31 08:49 pm (UTC)
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[User Picture]From: stephib
2007-02-05 03:52 am (UTC)
We did that in year 9. I forgot all of it though :(.
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[User Picture]From: stephib
2007-02-09 09:29 pm (UTC)
On the main point - the no. of 00s after the 9 depends on how many digits come before the recurring ones. I think.
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From: (Anonymous)
2007-04-19 08:16 am (UTC)

I learnt that!!! XD

For some reason the Chuck guy's method works.

0.0123123123... = 123/9000
(3 recurring digits and one 0 in the front.)

0.111111... = 1/9
(1 recurring digit, no 0s in the front.)

So for every recurring digit there is a 9. And for every 0 digit in front is simply added to the back.

I might as well launch the debate on why 0.9999... = 1. You can prove it using the methods taught. You can test yourself by finding out my age:

Find the denominator of this recurring decimal. (My age!!!)

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From: patelcauesm
2013-02-17 02:37 pm (UTC)
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