# Prove it

David113
Posers and Puzzles 05 Mar '08 22:51
1. 05 Mar '08 22:51
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.
2. 06 Mar '08 15:57
i take your word for itπ
Baby Gauss
06 Mar '08 16:33
Originally posted by David113
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.
I 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.
4. 06 Mar '08 18:221 edit
I think it has something to do with number divisibility with 9. But I can't remember what properties were required for the number to be divisible with 9. Eh, that was just a wild guess.

P.s. And, yeah, I'd also like to see this done.
5. 06 Mar '08 18:592 edits
Originally posted by David113
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.
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 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...
Baby Gauss
06 Mar '08 19:03
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
π
7. joe shmo
Strange Egg
06 Mar '08 19:11
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
The pattern that he is asked for is different than the one you have shown, the series for the numerators according to his stipulations are

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..........
8. 06 Mar '08 19:231 edit
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]... .
64x-x = 63x = 7
x = 7/63 = 1/9

And voilà we've shown that 1/9 is really .000111000111...
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.
Baby Gauss
06 Mar '08 19:29
Originally posted by David113
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.
A hint on how it's done. Puhlease. Nothing too giving just a little hint! I love solving this type of things.
10. 06 Mar '08 19:41
Originally posted by adam warlock
A hint on how it's done. Puhlease. Nothing too giving just a little hint! I love solving this type of things.
I'll wait a few days before giving a hint... maybe someone will solve it?
Baby Gauss
06 Mar '08 19:44
Originally posted by David113
I'll wait a few days before giving a hint... maybe someone will solve it?
how much number theory notion is needed? Can't you just tell what branch of math is more used? π΅
12. 07 Mar '08 01:532 edits
Originally posted by David113
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.
wait. 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?
13. joe shmo
Strange Egg
07 Mar '08 04:361 edit
Originally posted by kayakninja
wait. 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?
where do you see 1/2 in that series? what you should be seeing is 0/2 which = 0 ,
14. joe shmo
Strange Egg
07 Mar '08 04:40
Are there any 0's in the numerator after 2^11?
15. 07 Mar '08 05:44
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 ('?'π