Question

opinion please ...



One possible Proof of Collatz’s Conjecture

Andrew John Gábor
Gizella út 6/b
1143 Budapest - Hungary
gbrndrs1968@yahoo.com


,,...Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1....

Answer 1

One possible Proof of Collatz’s Conjecture

Andrew John Gábor
Gizella út 6/b
1143 Budapest - Hungary
gbrndrs1968@yahoo.com


,,...Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called "Half Or Triple Plus One", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.

Paul Erdős said, allegedly, about the Collatz conjecture: "Mathematics is not yet ripe for such problems" and also offered $500 for its solution....”

n1=1 --- 3x1+1 -- 4/2 -- 2/2 = 1
n2=2 --- 2/2 = 1
n3=3 --- 3x3+1 = 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n4=4 --- 4/2 -- 2/2 = 1
n5=5 --- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n6=6 --- 6/2 -- 3x3+1 -- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n7=7 --- 7x3+1 -- 22/2 -- 11x3+1 -- 34/2 -- 17x3+1 -- 52/2 -- 26/2 -- 13x3+1 -- 40/2 -- 20/2 -- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n8=8 --- 8/2 -- 4/2 -- 2/2 = 1
n9=9 --- 9x3+1 -- 28/2 -- 14/2 -- 7x3+1 -- 22/2 -- 11x3+1 -- 34/2 -- 17x3+1 -- 52/2 -- 26/2 -- 13x3+1 -- 40/2 -- 20/2 -- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 --
-- 2/2 = 1
n10=10 --- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
….........................................................................................................................



2.



for n=(n2)n
(n2)n=2n --- ... -- 5n2/2 -- (2n2 + 1)x3 + 1 -- 6n2 + 4 -- 8n2/2 -- 4n2/2 -- 2n2/2 -- 2/2 = 1
(n2)n+1=2n+1 --- ... -- 5n2/2 -- (2n2 + 1)x3 + 1 -- 6n2 + 4 -- 8n2/2 -- 4n2/2 -- 2n2/2 -- 2/2 = 1

for n = k suppose that is true
(n2)n=(k2)n=2k --- ... -- 5k2/2 -- (2k2 + 1)x3 + 1 -- 6k2 + 4 -- 8k2/2 -- 4k2/2 -- 2k2/2 -- 2/2 = 1
(n2)n+1=(k2)n+1=2k+1 --- ... -- 5k2/2 -- (2k2 + 1)x3 + 1 -- 6k2 + 2k2 -- 8k2/2 -- 4k2/2 -- 2k2/2 -- 2/2 = 1

for k = k + 1
k2n=(k2n +1) --- … -- (5(k2+1))x3 +1 -- 15k2+15 +1 -- 15k2+16 -- 15k2+8k2 -- 23k2/2 -- (11k2+1)x3 +1 -- 33k2+2k2-- 35k2/2 -- (17k2+1)x3 +1 -- 51k2+ 2k2 -- 53k2/2 -- (26k2+1)x3 +1-- 78k2+2k2-- 80k2/2 -- 40k2/2 -- 20k2/2 -- 10k2/2 -- 5k2/2 -- (2k2+1)x3 +1 -- 6k2+2k2 -- 8k2/2 -- 4k2/2 -- 2k2/2 -- 2/2 = 1
k2n+1=(k2n+1 +1) --- … -- (5(k2 +1) +1) -- 5k2 +6 -- 5k2 +3k2 -- 8k2/2 -- 4k2/2 -- 2k2/2 -- 2/2 = 1

q.e.d.

Answer 2

Together we can rule the galaxy!

Answer 3

interesting

Answer 4

what is for n=(n2)n ??

Answer 5

so n2 like a note wann being equal 2

Answer 6

so this mean 2n

Answer 7

sorry, can't undertand ... could you use latex??

Answer 8

n2 and k2 mean 2

ok ?

Answer 9

looks some people here might be interested in your work
http://mathoverflow.net/

Answer 10

thank you

Answer 11

@jhonyy9 Do you have an Erdős number? If not, you should.

Answer 12

too far above my head.

Answer 13

like reading Laos' language

Answer 14

You can post your findings here too, http://math.stackexchange.com/ , most of the advanced or professionals use it

Answer 15

yes i have posted it there and i have got for 1 reputation and i bronz badge with title of teacher

thank you for your idea and opinion

Answer 16

see in abc
ab=ac
ap=bc
a=20
to find pbc

Answer 17

to find x ie pbc

Answer 18

angle b is 80

Answer 19

Is that a proof by cases?

Answer 20

do you see that just for cases ?

by mathinduction we get for these cases from what we can deducting that is true for all numbers

or not ?