Can you prove that all never ending but repeating decimals can be expressed as the quotient of two integers?

Question

Like 1/3 is .33333....
Prove that all can be or that some can't be.

I'm working on a different approach, but some help would be nice, a straight answer would be even better :-)

-David

"I'm sorry, it's too long to explain and it's like elementary math."

So what you are trying to say is that you don't know what the **** you are talking about, and don't want to sound retarded. Right?

"assumption is sometimes the best approach you know"

I'll assume that you don't know what you are talking about either :-) Have a nice day.

"edit: No, what I'm saying is that you, mr. retarded f***tard, are too stupid to even realize that what you are asking is taught in elementary. I'm sorry, I tried to be polite but your stupidity and arrogance in answering me just pissed me off. Thanks for the two points anyway"

Lol... I will simply report you and make you lose your 2 points and 8 more :-)

Anyway, before I do that I would like to say that if it was so simple, then it really shouldn't take too long to explain, since the reasoning couldn't be anymore complicated than simple operations you could easily give a simple outline of the idea. SInce you didn't even do that, I still believe that you are trying to cover up your own ignorance.

Have a nice day, and enjoy the -8 points :-)

-David

Answer 1

Yes. Consider any decimal which has a finite prefix followed by an infinitely repeating sequence of length n. Set the decimal equal to x. Multiply x by 10^n - this causes the decimal to shift by n places. Then subtract x from (10^n)x - this will cancel out everything past the point where the decimal starts repeating, thereby truning it into a terminating decimal, which is clearly a fraction. But if (10^n-1)x=p/q for some integers p and q, then x=p/(q*(10^n-1)), which is a quotient of two integers, as required.

Example:

.000432900432900[432900] (the brackets indicate infinite repetition)

x = .000432900432900[432900]
1000000x = 432.900432900432900[432900]

1000000x-x=
432.900432900432900[432900]
-00.000432900432900[432900]
=432.9
999999x = 432.9
9999990x = 4329
x = 4329/9999990 = 1/2310

x = 3.141592653595359[5359]
10000x = 31415.92653595359[5359]
9999x = 31412.7849433
99,990,000,000x = 314,127,849,433
x = 314,127,849,433 / 99,990,000,000

x=0.99[9]
10x=9.99[9]
9x=9
x=9/9=1/1

Answer 2

Certainly, by simply showing how it is done. First, subtract off the part of the number (if any) that does not include any of the repeats. This is obviously a quotient of integers. Mark off the first period of repetition, and draw a line under it. Under the line, put a 9 in each decimal position. The result is a fraction (i.e., a quotient of two integers), which can usually be reduced to lower terms. If there are zeroes in front (between the decimal point and the first repeating digit), put that number of zeroes in the denominator, and you're done.

Answer 3

Um, isn't this like, a definition of a set inside the real number set? Isn't this a postulate? Oh well, if it isn't one, then there is this method where you subtract the number from 1.0000/1.00000, depending on how long it takes before it repeats. I'm sorry, it's too long to explain and it's like elementary math.

edit: No, what I'm saying is that you, mr. retarded f***tard, are too stupid to even realize that what you are asking is taught in elementary. I'm sorry, I tried to be polite but your stupidity and arrogance in answering me just pissed me off. Thanks for the two points anyway.

Answer 4

I think the way to prove it is to think of it as an infinite sequence.

0.333333333... = 3/10 + 3/100 + 3/1000+ 3/10000 + ....

So any rational number of infinite sequence that starts as
0.(string of numbers that keep repeating) can be written as an infinite sum of the form:

a/10 + a/100 + a/1000 + a/10000 + ... where a is an integer.

now you have to show that the above sequence is equal to
a/(10^b - 1) where b is equal to the number of digits of a. Sorry to say but I don't know how to do that.... If you figure it out, let me know.

Answer 5

If a number has repeating decimals in the form of, say, 0.123412341234 then we have a geometric series with a=0.1234 (the repeating decimal) and r = 1/10000 (10^-(# of digits that repeat)). since |r| < 1, this series converges to a/(1-r), which in this case, is .1234(1-(1/10000)).

Answer 6

I shall explain with steps and then give an example
suppose it repeat after n digits(sequence of m digits)

multiply by 10^m
the digits after n digits of the 1st number and second number are identical and when we subtract they cancel out
now numerator is non repeating and denominator and that is the ratio. may be even simplified

for example say
a = .142857.....
10^6 a= 142857.142857 ...
subtarct 1st from 2nd 99999a = 142857

a = 142857/99999 = 1/7

Answer 7

let x=(any recurrung decimal number)d.eeeeeeeeeeeeee.... --->1
where 'd' is integral part and 'e' set of repeating digits,let that be n digits in one set(can be a single digit).
multipy 10^n to the eqn
10^nx=d.eeeeeeeeeeeee*10^n.......---->2
(here x,d,e are digits not variables)
2-1===> (10^n-1)x=deeeeeeeee...(upto-n digits)-d
x=(deeeeeeeeeee....-d)/(10^n-1)
here deeeeeeeeee... is a pure integer and also 10^n-1
so this is the solution

Answer 8

1/prime number=0.xxxxxxxxxxxxxxxxxxxxxxx....number of decimals b4 repeating usually is (prime number-1)

Answer 9

Go read the article at
http://mathforum.org/library/drmath/view/57041.html

I'm just too damned tired to type it all out ☺


Doug

Answer 10

I'd rather go have a beer than worry about this problem!