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Subject: Fifty divided by Seven rss

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Michael
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Is pretty fun. It goes out to be
07.142856142856 and so on.

If you write it like
07. 14 28 57 14 28 57
you could see that
its just 7*0.02^0 + 7*0.02^1 + 7*0.02^2 + 7*0.02^3...
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Jim Cote
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The fractional part is simply 1/7.

1000000x = 142857.142857 repeating
x = 0.142857 repeating
==================================
999999x = 142857
x = 142857/999999
x = 1/7
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Walt
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You need to go here: http://www.wolframalpha.com/
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...sure...
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This isn't WeirdMathGeek! Go away. You're scaring me!





O, it's ChitChat. Sorry. Carry on.
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Dave
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mylittlepwny wrote:
Is pretty fun. It goes out to be
07.142856142856 and so on.

If you write it like
07. 14 28 57 14 28 57
you could see that
its just 7*0.02^0 + 7*0.02^1 + 7*0.02^2 + 7*0.02^3...


I think you've just opened up a portal somewhere... goo ... surprise !!!
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Blorb Plorbst
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mylittlepwny wrote:
Is pretty fun. It goes out to be
07.142856142856 and so on.

If you write it like
07. 14 28 57 14 28 57
you could see that
its just 7*0.02^0 + 7*0.02^1 + 7*0.02^2 + 7*0.02^3...


There's nothing significant about that sequence beyond a little happenstance.

First: 50/7 = 7.142857 repeating

Second: It doesn't work beyond 7*0.02^3 and even 7*0.02^3 is .000056, an approximation of the actual value.
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Michael
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CrankyPants wrote:
mylittlepwny wrote:
Is pretty fun. It goes out to be
07.142856142856 and so on.

If you write it like
07. 14 28 57 14 28 57
you could see that
its just 7*0.02^0 + 7*0.02^1 + 7*0.02^2 + 7*0.02^3...


There's nothing significant about that sequence beyond a little happenstance.

First: 50/7 = 7.142857 repeating

Second: It doesn't work beyond 7*0.02^3 and even 7*0.02^3 is .000056, an approximation of the actual value.


It does work actually, shame on you!

7*0.02^3 is .000056, but 7*0.02^4 is .00000112
making that .000056 into the appropriate .000057
Next .00000012 needs to become 14 for the pattern to continue, which it will when it gains the 2 from .0000000224. And so on.

And regards to the happenstance, that's what makes it fun.


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Walt
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As Wolfram-Alpha and Cranky note, this is a repeating decimal, usually written as:
______ ______
7.142857 or 0.714285 × 10^1

Of course, if we switch to base 7, you no longer have a repeating decimal but just 10.17 or, back in base 10, 1 × 7 + 0 × 1 + 1 × 1/7, which corresponds to the left repeating decimal, above.

If you like this kind of stuff, you should read Martin Gardner, long time (and apparently irreplaceable) author of Scientific American's Mathematical Games column. I have many of his books, and I most recently got his collected columns on CD. Sadly, he died last year; still, at 95 years old. We had him for a long an productive time.
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Ron Parker
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It's not even happenstance. It's just fun with numbers.

Consider:

(n^2+1)/n = n + 1/n

Therefore,

1/n = (n+1/n) / (n^2+1)

so

1/n = (n+ (n+1/n)/(n^2+1))/(n^2+1) = n/(n^2+1) + (n+1/n)/(n^2+1)^2

and you can keep expanding that last term until your arms fall off.

So if you take that reformulation of 1/n and plug it back into the original equation, you get this:

(n^2+1)/n = n/(n^2+1)^0 + n/(n^2+1)^1 + n/(n^2+1)^2 + n/(n^2+1)^3 + ...

which is just the general version of the case for n=7 presented here.

In my opinion, the truly weird thing about 1/7 is this:
1/7 = .142857...
2/7 = .285714...
3/7 = .428571...
4/7 = .571428...
5/7 = .714285...
6/7 = .857142...

(Of course, that's probably also a special case of a general mathematical principle, but that's enough math for me for today.)

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Not quite as interesting, when you take the decimal representations of 1/7 and 6/7 and add them together, you get .9 repeating. If you follow the nines all the way to the end, you fall off the universe.
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Walt
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parkrrrr wrote:
In my opinion, the truly weird thing about 1/7 is this:
1/7 = .142857...
2/7 = .285714...
3/7 = .428571...
4/7 = .571428...
5/7 = .714285...
6/7 = .857142...

We use base 10, which factors into the prime numbers 2 and 5 (2×5=10). An alternative to representing 50/7 as a repeating decimal or a fraction is as 7r1, seven with a remainder of 1. Obviously, there are 7 possible remainders 0 to 6. Zero is uninteresting. So, in the case of n/7 you cycle among the six "interesting" remainders, getting a period 6 repeating decimal fraction. But you don't always get n-1 periods; for example 1/11 is 0.090909....

When you divide something with factors in the base, you don't get repeating decimals: 1/5 = 0.2 exactly; that's it. (However, in base 2 it is a repeating decimal, 0.0011001100110011...; period 4.)

Some interesting fractions after 1/7 are:
________________
1/17=0.05882352941176470...
(period 16)
______________________
1/23=0.04347826086956521739130...
(period 22)
__________________________________________
1/49=0.0204081632653061224489795918367346938775510...
(period 42)
Note that 49=7×7: it's not prime, but it's relatively prime to the base, that is, neither 2 nor 5 are factors of 49. But why 42? Because 7/49=1/7, 14/49=2/7 etc. The period 6 repeating 1/7 fraction takes the other 6 possible remainders. So, you tend to get the longest periods with prime numbers, though as 1/11 shows, this isn't a guarantee.

Wolfram-Alpha makes it very easy to play with numbers like this.

And, BTW, your library will certainly have some Martin Gardner you can try before you buy the CD. I suspect MG has talked about repeating decimals in an article. I don't recall if a formula exists for the repetition period.
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Jim Cote
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Tall_Walt wrote:
I don't recall if a formula exists for the repetition period.

Not sure either, but you can always represent a repeating decimal as a fraction of the repeating part divided by an equal number of 9's. For example:

0.1234 12341 1234... = 1234/9999
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Ron Parker
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Tall_Walt wrote:
...a bunch more math...


What part of "enough math for me for today" did you miss?

Actually, I kinda like that 1/49 result, because for some reason it contains a bunch of powers of two:

2^1/10^2+2^2/10^4+2^3/10^6+2^4/10^8+2^5/10^10+2^6/10^12+...

Perhaps tomorrow I'll attempt to discover why.
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Michael
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parkrrrr wrote:
Tall_Walt wrote:
...a bunch more math...


What part of "enough math for me for today" did you miss?

Actually, I kinda like that 1/49 result, because for some reason it contains a bunch of powers of two:

2^1/10^2+2^2/10^4+2^3/10^6+2^4/10^8+2^5/10^10+2^6/10^12+...

Perhaps tomorrow I'll attempt to discover why.


isn't that just because it is the square of 1/7?

the formula is the same as the formula I made for 1/7th, just with the 7s taken out, because of course 1/7 divided by 7 is 1/49
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Ron Parker
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mylittlepwny wrote:
isn't that just because it is the square of 1/7?


I believe that it may be, yes. Way to take all the fun out of it.

By the way, in honor of this thread, I've promoted my first ever microbadge back to a place under my avatar.
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Michael
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parkrrrr wrote:
mylittlepwny wrote:
isn't that just because it is the square of 1/7?


I believe that it may be, yes. Way to take all the fun out of it.

By the way, in honor of this thread, I've promoted my first ever microbadge back to a place under my avatar.


Hmmm, which one though? My guess is "stained glass artist"
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Josh Jennings
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ekted wrote:
Tall_Walt wrote:
I don't recall if a formula exists for the repetition period.

Not sure either, but you can always represent a repeating decimal as a fraction of the repeating part divided by an equal number of 9's. For example:

0.1234 12341 1234... = 1234/9999


Very interesting! Except it doesn't work for 0.9 repeating... ;)
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Ron Parker
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mylittlepwny wrote:
Hmmm, which one though? My guess is "stained glass artist"


I'd claim that one was in honor of this thread but in reality it's been there since I bought it.
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Jim Cote
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thermogimp wrote:
ekted wrote:
Tall_Walt wrote:
I don't recall if a formula exists for the repetition period.

Not sure either, but you can always represent a repeating decimal as a fraction of the repeating part divided by an equal number of 9's. For example:

0.1234 12341 1234... = 1234/9999

Very interesting! Except it doesn't work for 0.9 repeating...

Of course it does. .9 repeating = 1. Unless you were joking.
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Andy Andersen
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mylittlepwny wrote:
Is pretty fun. It goes out to be
07.142856142856 and so on.

If you write it like
07. 14 28 57 14 28 57
you could see that
its just 7*0.02^0 + 7*0.02^1 + 7*0.02^2 + 7*0.02^3...


Funny - I always round up to 07.142856142858
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MisterCranky wrote:
Not quite as interesting, when you take the decimal representations of 1/7 and 6/7 and add them together, you get .9 repeating. If you follow the nines all the way to the end, you fall off the universe.

An infinite number of mathematicians walk into a bar.

"I'll have a pint of beer," says the first.

"I'll have half a pint," says the second.

"A quarter pint for me," says the third.

"I'll have one eighth," says the fourth.

"All right, all right, I get it," the bartender snarls, and pours two pints.
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Walt
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parkrrrr wrote:
Tall_Walt wrote:
...a bunch more math...

What part of "enough math for me for today" did you miss?

The part where you were the OP. laugh

parkrrrr wrote:
Actually, I kinda like that 1/49 result, because for some reason it contains a bunch of powers of two:

2^1/10^2+2^2/10^4+2^3/10^6+2^4/10^8+2^5/10^10+2^6/10^12+...

Perhaps tomorrow I'll attempt to discover why.

It's kind of an illusion: 100/49 is very close to 2/1, so you get 02, then 04, then 08 etc, until the 1/100 extra catches up to you at 65. Essentially, it's the number interacting with the base. 1/499 is:
____________________________________________________
0.002004008016032064128256513026052104208416833667334669
________________________________________________________
33867735470941883767535070140280561122244488977955911823
________________________________________________________
64729458917835671342685370741482965931863727454909819639
________________________________________________________
27855711422845691382765531062124248496993987975951903807
________________________________________________________
61523046092184368737474949899799599198396793587174348697
________________________________________________________
39478957915831663326653306613226452905811623246492985971
________________________________________________________
94388777555110220440881763527054108216432865731462925851
________________________________________________________
70340681362725450901803607214428857715430861723446893787
______________________________________________________
575150300601202404809619238476953907815631262525050100
(period 498)

As you can see, this maintains powers of two until 256, or further if you account for the numbers "leaking" from one triple to the next. (Though I suppose this also makes the 65 okay, since it's 64 plus the 1 from 128.)
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James Newton
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kuhrusty wrote:
MisterCranky wrote:
Not quite as interesting, when you take the decimal representations of 1/7 and 6/7 and add them together, you get .9 repeating. If you follow the nines all the way to the end, you fall off the universe.

An infinite number of mathematicians walk into a bar.

"I'll have a pint of beer," says the first.

"I'll have half a pint," says the second.

"A quarter pint for me," says the third.

"I'll have one eighth," says the fourth.

"All right, all right, I get it," the bartender snarls, and pours two pints.

That bartender is daft! He didn't have to pour a whole 2 pints because they won't all manage to drink their portion before closing time!

Unless they all have separate glasses - in which case the bartender won't have finished pouring them by closing time.
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Josh Jennings
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churchmouse wrote:
kuhrusty wrote:
MisterCranky wrote:
Not quite as interesting, when you take the decimal representations of 1/7 and 6/7 and add them together, you get .9 repeating. If you follow the nines all the way to the end, you fall off the universe.

An infinite number of mathematicians walk into a bar.

"I'll have a pint of beer," says the first.

"I'll have half a pint," says the second.

"A quarter pint for me," says the third.

"I'll have one eighth," says the fourth.

"All right, all right, I get it," the bartender snarls, and pours two pints.

That bartender is daft! He didn't have to pour a whole 2 pints because they won't all manage to drink their portion before closing time!

Unless they all have separate glasses - in which case the bartender won't have finished pouring them by closing time.


Actually, the glass has infinite straws of varying lengths that allow all of the mathematicians to drink at once. I know that it sounds preposterous, but I've seen someone drinking beer through a straw once.
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