06 Mar 08
Originally posted by David113I really want to see how this one is done. I tried for a little bit but couldn't get much done and for the moment I'm not seeing how it can be done either.
Prove:
0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
06 Mar 08
2 edits
Originally posted by David1131/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
Prove:
0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
it is actually not so.
The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
or if we show only the indices = 000 111 000 111 ...
Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 zeroes followed by 3 ones and again and again).
Just simple arithemtics.
If you want a proof that .000111000111... is really 1/1001 binary = 1/9 decimal, here it comes, again in binary arithemtics:
say that x = .000111000111...
then 1000000x = 111.000111...
and 1000000x-x = 111 exactly.
Now we safely can go back to decimal arithemtics again.
64x-x = 63x = 7
x = 7/63 = 1/9
And voilà we've shown that 1/9 is really .000111000111...
06 Mar 08
Originally posted by FabianFnas😕
1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
it is actually not so.
The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
or if we show only the indices = 000 111 000 111 ...
Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 ...[text shortened]... arithemtics again.
64x-x = 63x = 7
x = 7/63 = 1/9
And voilà We've shown that 1/9 is realy
- Joined
- 10 Dec 06
- Moves
- 8528
06 Mar 08
Originally posted by FabianFnasThe pattern that he is asked for is different than the one you have shown, the series for the numerators according to his stipulations are
1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
it is actually not so.
The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
or if we show only the indices = 000 111 000 111 ...
Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 ...[text shortened]... arithemtics again.
64x-x = 63x = 7
x = 7/63 = 1/9
And voilà We've shown that 1/9 is realy
0,0,0,1,1,0,1,1,2,1,0,1,2,2,2,3,4,1,1,3,4,3,1,5,5,2,5,2 for the first 28
that doesnt look like 0,0,0,1,1,1,0,0,0,1,1,1..........
06 Mar 08
1 edit
Originally posted by FabianFnasThe fact that there is another series whose sum is 1/9 does not mean that the sum of "my" series can't also be 1/9.
1/9, you say it is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
it is actually not so.
The correct is 0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 1/64 + 0/128 + ... = 1/9
or if we show only the indices = 000 111 000 111 ...
Go to the binary system, 1/9 (10) = 1/1001 (2), do the division 1/1001 and you get the result .000111000111... (a pattern of 3 ...[text shortened]... .
64x-x = 63x = 7
x = 7/63 = 1/9
And voilà we've shown that 1/9 is really .000111000111...
Your series is the only one in which the denominators are 2,4,8,... and whose sum is 1/9 IF EACH NUMERATOR IS 0 OR 1.
This doesn't mean that if the numerators are not restricted to being 0 or 1 there isn't another series whose sum is 1/9.
06 Mar 08
Originally posted by David113A hint on how it's done. Puhlease. Nothing too giving just a little hint! I love solving this type of things.
The fact that there is another series whose sum is 1/9 does not mean that the sum of "my" series can't also be 1/9.
Your series is the only one in which the denominators are 2,4,8,... and whose sum is 1/9 IF EACH NUMERATOR IS 0 OR 1.
This doesn't mean that if the numerators are not restricted to being 0 or 1 there isn't another series whose sum is 1/9.
07 Mar 08
2 edits
Originally posted by David113wait. I'm confused. If you add anything positive to 1/2, it goes up. Since 1/9 is less than 1/2, it's mathematically impossible, isn't it?
Prove:
0/2 + 0/4 + 0/8 + 1/16 + 1/32 + 0/64 + 1/128 + ... = 1/9
Where the denominators are the powers of 2, and each numerator equals the number of odd digits in the denominator.
07 Mar 08
I don't know if this is a good way to go about it, but I'd like to present a different viewing angle to the problem.
In any doubling series:
A digit is even if the digit immediately after it in the previous number is 0, 1, 2, 3, or 4. [Since the digit is doubled, and doesn't carry a one, the digit must become even].
A digit is odd if the digit immediately after it in the previous number is 5, 6, 7, 8, or 9. [The digit is doubled, and it carries a one, so it must become odd].
Therefore the number of odd digits in a power of 2 is equal to the number of digits greater than 4 in the previous number. Maybe that can help ('?'😉