A Maths Problems

Im having problem with this question.. could anyone help?

"Prove that 0.58 (recurring) = 58/99"

Thanks.

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What do i put here again?

You have to express 0.5858585858... as an infinite sum. Lets call the number x for simplicity.

0.58 = 58/100
0.5858 = 58/100 + 58/10000 = 58/100 + 58/(100²)
0.585858 = 58/100 + 58/10000 + 58/1000000 = 58/100 + 58/(100<sup>2<sup>) + 58(100³)

We use this pattern to express x as the sum of an infinite number of values.

x = (sum from n = 1 to infinity) 58/(100ⁿ)

From there, use properties of summation to simplify the expression until you end up with 58/99.

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"If you talk at all during this lesson, you have detention. Do you understand?"

• My yr11 Economics teacher

Ok thx

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What do i put here again?

Without taking off your shoes, prove that 10+10 = 20. :razz:

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The best things in life, aren't things.

The problem with these sorts of things is that its not entirely clear what number is meant by ".58585858 repeating". As Wilson hints at, the key is not to think of .585858... as a number with that particular decimal expansion, but rather to think of .58585858... as the the limit of partial sums. You can think of that as the definition of .585858 repeating.

I'm not going to do your problem for you (thats your job), but heres something very very similar: .9999... = 1.

.999999... = lim [n->infinity] (sum [i=1..n] (9/10ⁱ) )

Our goal here is to show that this limit of partial sums converges to 1. Proving that a sequence converges to a limit is like playing a game: You pick any number, nomatter how small, and I have to come up with an element of the sequence for which that element, and every later element, is closer to the limit than the number you chose. If I can "win" the game for any number you choose, then the sequence converges to the limit.

Suppose you choose the number .5. Thats a pretty easy one. Then I can pick N=1 and so 1-sum[i=1..N](9/10ⁱ) = 1-sum[i=1..1](9/10ⁱ) = 1-.9 = .1. For every integer n that is bigger than N (here bigger than 1), then the difference between 1 and the sum will be even smaller than that (n=2 yields 1-.99=.01, n=3 yields 1-.999=.001, etc) Since .1 is less than .5, I win that one.

Now what if you choose the number .00002? Thats a pretty small difference, but I can still find a value of N for which 1 minus the sum is smaller than .00001. Specifically, N=5 gives the following result: 1-sum[i=1..5](9/10ⁱ) = 1-.99999 = .00001 which is less than .00002. Again, if there is another number n that is greater than N (5 here), then the difference between 1 and the sum will be even smaller than that. I win again.

You choose .00000000008? I pick N=11. You choose 10^-99? I choose N=100.

Hopefully (?) it should be clear that I will always be able to "win" this game, but to prove that the limit converges to 1 nomatter what number you choose, we have to carefully write down the above process in rigorous mathematical language.

Let epsilon be a real number that is greater than zero (this is the number that you choose in the "game"). Pick an integer N such that N>log[base 10] (1/epsilon) (this is the number I choose in the "game"). We know that log[base 10] (1/epsilon) exists since epsilon is greater than 0, and so an integer bigger than that exists (N), and we have N>log[base 10] (1/epsilon). If n>=N, then n>log[base 10] (1/epsilon). Raising 10 to the power of each side, we have 10ⁿ > 1/epsilon. Inverting, we have 10^-n < epsilon.

But 10^-n = 1-sum[i=1..n](9/10ⁱ), so 1-sum[i=1..n](9/10ⁱ) < epsilon. Since this holds for any epsilon greater than 0, and for any n greater than N, we can conclude: for any epsilon > 0, there exists an integer N such that for all n>=N, 1-sum[i=1..n](9/10ⁱ) < epsilon. Thus lim [n->infinity] (sum [i=1..n] (9/10ⁱ) ) = 1. (end of proof)

Hopefully that helps.

There are some other ways to prove it, but they depend on some theorems that you may not know yet. For example, the partial sums are bounded above with supremum 1, and monotonically increasing, and real numbers, so their limit must be 1.

[edit: whoops, had a couple typos, and slightly changed how N is chosen. Fixt.]

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http://maze5.net

Wow, that's some reply mazey. These questions are only for the Standard tests at 16.. but any help is help I guess!

I'm glad I never had to prove this and probably never will :biggrin:
Good luck with your test(?).

Im having problem with this question.. could anyone help?
"Prove that 0.58 (recurring) = 58/99"

Thanks.

Haven't the time to read mazemasters and wilsons solutions, but try this alternate method...

Let x = 0.58[rep]

then 100x = 58.58[rep]

And therefore

(100x - x) = (58.58(rep) - 0.58[rep])

And the required result follows.

I suspect this is the so-called and oft-hated expected answer, if the
british education system at your level is anything like what ours used
to be at your level.

I would go with whatever Mazemaster is using, I think it's called a proof by induction - it's pretty awesome.

Most days, I like hanging out around here. The atmosphere is near perfect, as far as website forums go but...

People who post like Nick does, scare the living s**t out of me. To much smarts all wrapped up in one noggin has got to be stuff that they make movies about. :eek:

Go ahead Nick. Rub it in. Orph is to stupid to follow the conversation.

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The best things in life, aren't things.

Heheh, I'm with Fraggard here :biggrin: Good answers from the other two, but I doubt GCSE's expect anything more than that.

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[img]http://card.mygamercard.net/sig/Default/reno84.png[/img]
Designer @ Haiku Interactive | ReNo-vation.net

Looking back at that I can't believe I understood it :o

$loth said:
Looking back at that I can't believe I understood it :o

Which part? The part where you cannot take off your shoes? Thats the only piece I got.. Good thing too, since I posted it. :heee:

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The best things in life, aren't things.

$loth said:
Looking back at that I can't believe I understood it :o

Which part? The part where you cannot take off your shoes? Thats the only piece I got.. Good thing too, since I posted it. :heee:

:razz:

Still, looking back at the maths I was doing, even if it was just GCSE, looks mind boggling now.

I would go with whatever Mazemaster is using, I think it's called a proof by induction - it's pretty awesome.

Not quite. Its an epsilon-N proof actually. I agree that induction kicks ass though.

Let x = 0.58[rep]
then 100x = 58.58[rep]

And therefore
(100x - x) = (58.58(rep) - 0.58[rep])

All of the manipulations in that "proof" will only work if the infinite sum .58 + .0058 + .000058 + ... is absolutely convergent. For many infinite sums you can't do that. For example, subtracting 2 infinite sums term-by-term does not always produce the correct answer.

Of course the sum is absolutely convergent so the math you did is OK, but you still have prove it.

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http://maze5.net

God I hate sums, Especially the infinite ones. I was very happy when I learned that the definite intergral did the same thing as a riemans sum.

rs6 said:
God I hate sums, Especially the infinite ones. .

I'd advise you to never get married then. You'd be amazed at how many times a woman can remember your inadequacies. Men seem to screw up an infinite amount of times, and to balance that out, women can recall any one of them, no matter how many years have been in between.

A woman can sum up a man in moments but it lasts forever. :rolleyes:

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The best things in life, aren't things.