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Hi
While solving the two proceding problems, you may have discovered that quotient digits (and remainders) became periodic :
_______
7  100000..(twenty zeros)

Answer :1428571428571...
Is it just a coincidence, or will this pattern repeat?
Well, no it's not a coincidence and yes the pattern will repeat. I'm not sure this is the complete reason, but anyway :
Let's take 100 to simplify things
_______
7  100
 answer : 14,2857...
10070=30
3028=2
2,01,4=0,6
0,60,56=0,04
0,040,035=0,005
0,00500,0049=0,0001
As we can see, a new cycle (Or a new patern in our quotient) begins when the digit "1" is the highest value in our number obtained after a substraction(The second period or cycle begins at 0,0001), and wether 7 for 10, 70 for 100 , etc. will always enter one time in it, and this patern continue on for an infinite time.(Well, If it doesn't, it goes for a longlong time, for sure)
Now, like I said, I feel it's lacking something. I don't want any help for now, besides answering this question : I'm not sure if the question in the quote is telling to find the exact cause to this problem, or it only wants a yes or no answer. WHat do you think ? Thank you
*sorry, I did an error in my calculations. Now it should be fine
Last edited by AlAllo (20130710 11:58:18)
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Either a fraction terminates such as 1 / 2 or 1 / 5 or it has some repeating pattern 1 / 3 = .3333333333...
For any fraction 1 / p this pattern is supposed to be less than p
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
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Other question, this is going to infinity, isn't it ? And could you elaborate what you just said?
Last edited by AlAllo (20130710 11:02:37)
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Yes, that is what the 3 dots at the end mean
1 / 3 = .333333333333...
When p is a prime, 1 / p will have a period of p  1.
1/7 = 0.142857 ; 6 repeating digits
1/17 = 0.05882352 94117647 ; 16 repeating digits
1/19 = 0.052631578 947368421 ; 18 repeating digits
1/23 = 0.04347826086 95652173913 ; 22 repeating digits
1/29 = 0.0344827 5862068 9655172 4137931 ; 28 repeating digits
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
Ok, but what do you think of my answer, should I elaborate, did I answer what was expected ?
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What it is lacking is an argument of why this will always happen.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
Yes, that is what the 3 dots at the end mean
1 / 3 = .333333333333...
When p is a prime, 1 / p will have a period of p  1.
Wikipedia wrote:1/7 = 0.142857 ; 6 repeating digits
1/17 = 0.05882352 94117647 ; 16 repeating digits
1/19 = 0.052631578 947368421 ; 18 repeating digits
1/23 = 0.04347826086 95652173913 ; 22 repeating digits
1/29 = 0.0344827 5862068 9655172 4137931 ; 28 repeating digits
What about 5? Or 11? Or 13? They are not p1 then.
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
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bobbym wrote:Yes, that is what the 3 dots at the end mean
1 / 3 = .333333333333...
When p is a prime, 1 / p will have a period of p  1.
Wikipedia wrote:1/7 = 0.142857 ; 6 repeating digits
1/17 = 0.05882352 94117647 ; 16 repeating digits
1/19 = 0.052631578 947368421 ; 18 repeating digits
1/23 = 0.04347826086 95652173913 ; 22 repeating digits
1/29 = 0.0344827 5862068 9655172 4137931 ; 28 repeating digits
What about 5? Or 11? Or 13? They are not p1 then.
That's what I also saw...
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What it is lacking is an argument of why this will always happen.
But isn't this correct :
10070=30
3028=2
2,01,4=0,6
0,60,56=0,4
0,40,35=0,5
0,500,49=0,1
Is it contradictable ?
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No, it is not. But it will be a little bit tedious if I said
1 / 97.
You did well, I am just saying that you could argue the point in another way.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
Why would it be a little tedious with 1/99 ????
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Sorry, I meant 1 / 97. How would you do it?
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
WEll, for now I'm not sure. I need to get my imagination working on it....
Edit: I edited my numbers in my first post, a part of the calculations were wrong.
Last edited by AlAllo (20130710 11:59:39)
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Very good. Post when you have an answer you like.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
I'll try to see.
Btw, here are the real numbers :
10070=30
3028=2
2,01,4=0,6
0,60,56=0,04
0,040,035=0,005
0,00500,0049=0,0001
It was my fault, sorry. (If you ever saw it ^^)
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That is okay, do not worry about it.
What about 5? Or 11? Or 13? They are not p1 then.
The period of the repeating decimal of 1/p is equal to the order of 10 modulo p. If 10 is a primitive root modulo p, the period is equal to p − 1; if not, the period is a factor of p − 1.
I know that but did not want to get into this. I wanted to point AlAllo into the direction that the period will always be p1 or less. I was trying to make the statement that it can be determined easily by a simple argument rather than the long division method which is tedious for 1 / 97 by hand.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
hi AlAllo
You want the complete reason for this. I suggest you start by looking at the fractions that terminate. What property causes them to terminate? then you can state which fractions will produce recurring digits.
How many in a recurring cycle? Well consider how many possible remainders there can be. 1/7ths are a good example of this. If there is a limit to the number of remainders then what happens when you 'run out' of 'new' remainders?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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Ok, I'll also keep searching for that. I think that by understanding how to predict if they are terminate or not, it will help help me a lot in other related problems.
Last edited by AlAllo (20130711 09:53:00)
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Ok, well I'm not sure of this theory but let's go :
Just so you know, I was working with fractions like 1/2, 1/3, etc. So what is being said may not be completly true for other types of fractions ?
Also, I was using long divisions to work out the fraction.
In a fraction like 1/2, or any other(the denominator being higher), we know that it will always begin by 0.
(
2 ( 10
(
Answer 0,
As we can see, 10 (Which I'm totally aware is not the real value, but I wasn't sure which terminology to use if I had worked with decimals) has in it's prime factorization : 2 and 5. Every fractions will have to work with 10 and it's prime factors and this is exactly what will determinate if the decimal representation will be terminate or recurring.(Which will also depend on the divisor's prime factors.)
(
2 ( 10
( 10
Answer 0,5
Now, what conclusion can we make from this ? If we know that 10=2*5, we can say that a fraction will always? be terminate if it has divisor with a prime factorisation of 2,5 or both of them (Like 4,8,16,25,etc.) On the contrary, if you had 10 objects, 20 objects, 40 objects (All numbers whose prime factors are 2 and 5's or only one them) and you tried to complete it with an odd like 3, it would never work. So if we said 1/6
2*5/3*2, for the same reasons, 6, which is composed of two 3's, won't fit in.
What do you think ? I know that there is still a lot which is missing, and I don't know much about number theory, but am I on the right path ? (Btw, bobbym, I think that I will be able to answer your question if I know how terminate and recurring number appears)
And I only want critiques on what is said, no spoilers or anything else. (I haven't finished, some more could appear if I have new ideas)
Last edited by AlAllo (20130712 08:58:19)
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Hi AlAllo;
I had the same idea about factors of 10.
You always were able to answer the question about my fraction.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
Hi AlAllo;
I had the same idea about factors of 10.
You always were able to answer the question about my fraction.
You mean what I'm saying is correct ? (Just to be sure... xD) About your fraction, maybe I was able, it's just I didn't give it much thought, don't worry, I'll try to spend more time on it. I'll be adding my ideas on this topic anyway.
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Hi;
As near as I can understand it, yes.
In mathematics, you don't understand things. You just get used to them.
I have the result, but I do not yet know how to get it.
All physicists, and a good many quite respectable mathematicians are contemptuous about proof.
Online
hi AlAllo
Yes, factors of 2 and 5 are the key to this. You seem to have got the terminating ones sorted out.
Now: why does 1/7 cycle with 6 digits recurring and all the others 2/7, 3/7 etc follow the same cycle?
Bob
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
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hi AlAllo
Yes, factors of 2 and 5 are the key to this. You seem to have got the terminating ones sorted out.
Now: why does 1/7 cycle with 6 digits recurring and all the others 2/7, 3/7 etc follow the same cycle?
Bob
Lol, the funny thing is, I had the exact problem asked with 1/7, 2/7, etc. before you asked me to search for it. I began it, but didn't finish it.
The number 142857 seems to have some interesting properties. Anyway, I'll continue searching
Last edited by AlAllo (20130712 14:28:12)
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hi AlAllo
You want the complete reason for this. I suggest you start by looking at the fractions that terminate. What property causes them to terminate? then you can state which fractions will produce recurring digits.
How many in a recurring cycle? Well consider how many possible remainders there can be. 1/7ths are a good example of this. If there is a limit to the number of remainders then what happens when you 'run out' of 'new' remainders?
Bob
I just wanted to answer that last question, before answering to the one with 1/7,2/7,etc.
We know that there is a direct equality between the number of remainders and the number of digits in your quotient (The cycle)
For example : 1/7
7*0,1=0,7 >> 10,7=0,3 1
7*0,04=0,28 >> 0,30,28=0,02
7*0,002=0,014 >> 0,020,014=0,006 3
7*0,0008=0,0056 >> 0,0060,0056=0,0004 4
7*0,00005=0.00035 >> 0,00040.00035=0,00005 5
7*0,000007=0.000049 >> 0,000050.000049=0,000001 6
Which gives us : 0,142857... ( we have exactly 6 digits in one cycle)
As we can see, there is always remainders(The quotient going to infinity), but there is a limit to the kinds of remainders. We always have new old ones the moment we hit a certain point (This certain point in this example corresponds to 0,000001, but I'm not sure if it always needs to be of one unit of difference to have a new cycle, I would need to check it) and this is when the cycle continues again, being periodic
Anyway, any opinion on what I said ? If you think something is missing, please tell me. The only thing that is irritating me is, I haven't discoverred why certain numbers have longer periods than others, I think that by knowing this, it will help me a lot. If you know the answer, please don't tell me, I want to be able to answer it myself. Thank you again!
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