/sci/ - Science & Math

[Pedro No.5868153]

0.101001000100001000001000000100000001000000001000000000100000000001000000000001000000000000100000000000001...

Convert to a fraction.

[Anonymous No.5868159]

can't be done

[Anonymous No.5868161]

>>5868153
0.101001000100001000001000000100000001000000001000000000100000000001000000000001000000000000100000000000001.../1

[Anonymous No.5868163]

>>5868161
Lol!

[Anonymous No.5868205]

x=0.101001000100001000001000000100000001...
1/(1/x)
Shift the decimal places in the numberator and denumberator over to infinity.

That's the best I can do.

[Anonymous No.5868232]

1/10.1001000100001000001000000100000001000000001000000000100000000001000000000001000000000000100000000000001...

right?

[Anonymous No.5868313]

>>5868153
0.^0.1/1

[Anonymous No.5868366]

Assuming there are n zeroes between the nth and the (n+1)th 1 then that number is irrational.

[Anonymous No.5868372]

>>5868153
sum from n=1 to infinity 10 to the minus n(n+1)/2

[Anonymous No.5868377]

>>5868372
10 year olds know what fractions are, why don't you?

[Anonymous No.5868384]

>2013
>still replying to this shitposting spic
This board deserves no mods for a reason.

[Anonymous No.5868412]

>nonterminating
>nonrepeating
HAY GUISE EXPRESS THIS QUANTITY AS THE RATIO OF TWO INTEGERS
I shiggy that diggy.

[Pedro No.5868478]

>>5868412
Where the denominator is a NON ZERO integer.
Fuck man.

[Anonymous No.5868521]

0.101001000100001000001000000100000001000000001000000000100000000001000000000001000000000000100000000000001 = S

S = 0.1 + 0.001 + 0.000001 + 0.0000000001
S = 1/10 + ( 1/10^3 + 1/10^6 + 1/10^9... )
well, this is a Geometric progression and the q = 1/10^3

The formula to calculate a infinite PG of 0<q<1
is equals A1/1-q where A1 is the first term and q is the "mutiplication factor of An to An+1.
So , (1/10^3)/1-1/10^3

Then

S = 1/10 +(1/10^3)/(1-1/10^3)