prove any number over the same amount of numbers equals a repeating decimal. example: 245/999=.245245?

Question

what ever number you have in the numerator, match the same amount of numbers in the denominator with 9's. such as: 22/99=.222222, or 34/99=.34343434, or 123/99=.123123123...etc.

Answer 1

I think you mean the same number of *nines* in the denominator.

Let x = .245245245...
1000x = 245.245245...

Subtracting you get:
999x = 245
x = 245/999

It works with any set of repeating digits that start at the decimal point.
y = .22222...
10y = 2.2222....

9y = 2
y = 2/9

Or:
z = .343434...
100z = 34.3434...

99z = 34
z = 34/99

Your last one should be:
123/999 = .123123123...

but this reduces to:
41/333

You can also go in reverse:
123/999 = x
123 = 999x
123 + x - x = 999x
123 + 0.123123123... - 0.123123123 = 1000x - x
123.123123123... - 0.123123123... = 1000x - x
x = 0.123123123...

Answer 2

how r u supposed to prove it? well, my first piece of advice would be to prove using the calc, but of course u cn't do that. :)
on the other hand, think about it. 22/99 is extremely close to 22/100, therefore .2222222... is an answer that is close to our estimate. also 1/9=.11111111 because if u try to make 9 go into 1 in anyway, u will get 1 as the remainder, and again u will contue to get .1111.. so doesn't it make sense that 2/9=2x.11111111...=.222222222...? also this will be the same for 22/99, 222/999, etc.
bye

Answer 3

Consider the Taylor series expansion of (1-x)^(-1) where |x| < 1 = 1+x+x^2+x^3......

Thus consider 99 = 100 - 1 = 100 * (1 - 0.01)

Now,
22/99
= 22/ { 100 * (1 - 0.01) }
= 0.22 * (1 - 0.01)^(-1)
= 0.22 * (1 + 0.01^2 + 0.01^3 + ........)
= 0.22 * (1 + 0.0001 + 0.000001 + ............)
= 0.22 + 0.0022 + 0.000022 + .....
= 0.222222...........

This method can then be generalized. I hope I don't need to provide a more proof but this is the basic idea.

Answer 4

One hint is attempt to be certain what fraction is going with a repeating decimal, is to certain it with 2 effortless fractions you already understand... case in point, you already know that one million/10 = 0.one million, and you already know that 0.1111 is extremely greater beneficial than 0.one million, which you will desire to attempt one million/9, which you will locate =0.111111. now attempt 0.33333, you already know that one million/4 is 0.25, and one million/2 is 0.5.... consequently the fraction it incredibly is comparable to 0.33333 would desire to be between one million/4 and one million/2. consequently you will possibly wager it incredibly is one million/3. 0.2 is two/10, or one million/5. by way of fact that 0.22222 is extremely greater beneficial than one million/5 you will possibly attempt one million/4, yet we already realize it incredibly is 0.25, that's in basic terms too great... this implies the fraction is between one million/4 and one million/5, so for that reason you ought to the two attempt a greater beneficial numerator till you get it, or you have to be conscious of that 0.22222 is precisely 2*0.11111. by way of fact that we already chanced on that one million/9 = 0.111111, all of us understand that 2*(one million/9) would desire to equivalent 0.22222. consequently, 2*(one million/9) = 2/9 = 0.2222. that very same technique can extremely be utilized to any repeating decimal... by way of fact that one million/9 = 0.1111111 2/9 = 0.22222222 3/9 = one million/3 = 0.33333333 4/9 = 0.44444444 etc. desire those innovations help.

Answer 5

that cannot be proven because 9/9 equals 1, a terminating decimal

Answer 6

False, since if you choose 1/2, it gives you 0.5 which is a non-repeating decimal number.

Answer 7

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