Week 10

Hi all,

Before I prove something I want to highlight the space race at Dropbox for U of T https://www.dropbox.com/spacerace?r=NTQ4NDE0MjI3OQ. Go to this link to get a lot of space on Dropbox. As of this post everyone gets 15GB for two years.

Anyhow back to proving. Continuing on my problem. I am going to prove that once a power of 2 appears in the sequence we have a 1 in the sequence.

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Let P(n) be if the Collatz sequence starting at n has a x_m = 2^k, for some natural numbers m and k, then it contains a 1.

Assume x_m = 2^k.

x_m = 2*2^(k-1) so it is even. Following the sequence we divide by 2.
x_(m+1) = 2^(k-1) = 2*2^(k-2), we again divide by 2.

This procedure continues k times until x_(m+k) = 2^(k-k) = 2^0 = 1

Hence P(n).

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So now I know if I can show that the sequence at some point contains a power of 2 then I am done. I know at that point the sequence contains a 1.

More next time

Wendy

Week 9

Hi all,

So I've decided on a new problem, the Collatz Problem.

This is the one where there is a sequence that starts with a positive non-zero natural number and then does the following

• if its even: divide by 2
• if its odd: multiply by 3, then add 1

The point is to show that this sequence will always returns to 1. This problem is unsolved.

My natural instinct is to first show that in the sequence before 1 there will be a previous sequence that we have seen before and know that it returns to 1. After seeing on Wolfram the sequence written out, it seems that all the sequences end the same way.

  It also seems that it might be useful to see how these operations combine for any one number. And especially for one that is prime.

So heres my first part, I shall prove that when the odd operation is done a even operation immeadiately follows.

let a_n be 2k + 1 for some k in natural numbers. then P(n) is that 3a_n + 1 = 2m for some m in natural numbers.

So direct substitution:

3(2k + 1) +1 = 6k + 3 + 1 = 6k+4 = 2(3k + 2) = 2m

where m = 3k + 2

because a_n is an arbitrary odd number. it is true for all odd numbers.


Thats a start.

Wendy