A Conversation for The 3n+1 Conjecture - Proof Needed!
It works with every number
Arthur Dent Started conversation Jun 12, 2001
i cant prove it does not work because you cant. any odd number times by another odd number + any odd number is an even number, and all even numbers are in the 2 times table so it will always reach one, you will also notice the last few numbers are always the same and are all even.
It works with every number
Gnomon - time to move on Posted Jun 12, 2001
If you can either prove that it always works, or that there is a number for which it doesn't work, then you get the money, the glory and the fame!
It works with every number
Ruppinger ~ zaphodista ~ former keeper of vegan affairs ~ new keeper of rainbows, until the old one shows up again Posted Jun 25, 2001
Proof:
If you multiply an odd number with an odd number you get an odd number of odd numbers, which is odd.
And because
odd = even + 1
and
even = divisible with 2
--> odd*odd+odd (which is (even+1)+(even+1)) = even+even+2 = even
(I don't want the glory, give me the money 
)
It works with every number
Ruppinger ~ zaphodista ~ former keeper of vegan affairs ~ new keeper of rainbows, until the old one shows up again Posted Jun 25, 2001
BTW, I didn't do the last part of the last line. I wrote , but what I wanted to say was 
, so:
and odd is of course even + 1 OR 1 
It works with every number
Gnomon - time to move on Posted Jun 26, 2001
OK Ruppinger, what you have proved is that if the first number is even, then the second will be odd and the third even. Now what?
It works with every number
Gnomon - time to move on Posted Jun 26, 2001
No that's not what you've shown. You've shown that any odd number will be transformed into an even number by multiplying by 3 and adding 1. Since this is even, the next step will be to divide this by two.
So what?
This new number will be greater than the original odd number, and can be either even or odd. So the process may go up or down from here. Is there any starting number for which it will always go up (on average)?
Or can you prove that there isn't?
It works with every number
Ruppinger ~ zaphodista ~ former keeper of vegan affairs ~ new keeper of rainbows, until the old one shows up again Posted Jun 26, 2001
Sorry Gnomon,
I posted just after reading Arthurs post on this page, without reading your previous guide entry 
Furthur more, IRL disturbed my thinking.
I will read your entry thoroughly and be back within the next days.
It works with every number
ZiGGY^ Posted Feb 3, 2002
For something with so many exceptions is there really any point trying to disprove it? its a standard numerical shell decay with 0 and infinity our of its range
It works with every number
ZiGGY^ Posted Feb 3, 2002
For something with so many exceptions is there really any point trying to disprove it? its a standard numerical shell decay with 0 and infinity our of its domain
It works with every number
ZiGGY^ Posted Feb 3, 2002
damn 3am typing :P moderator you know what to do man
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It works with every number
- 1: Arthur Dent (Jun 12, 2001)
- 2: Gnomon - time to move on (Jun 12, 2001)
- 3: Ruppinger ~ zaphodista ~ former keeper of vegan affairs ~ new keeper of rainbows, until the old one shows up again (Jun 25, 2001)
- 4: Ruppinger ~ zaphodista ~ former keeper of vegan affairs ~ new keeper of rainbows, until the old one shows up again (Jun 25, 2001)
- 5: Gnomon - time to move on (Jun 26, 2001)
- 6: Gnomon - time to move on (Jun 26, 2001)
- 7: Ruppinger ~ zaphodista ~ former keeper of vegan affairs ~ new keeper of rainbows, until the old one shows up again (Jun 26, 2001)
- 8: ZiGGY^ (Feb 3, 2002)
- 9: ZiGGY^ (Feb 3, 2002)
- 10: ZiGGY^ (Feb 3, 2002)
- 11: ZiGGY^ (Feb 3, 2002)
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