The collatz conjecture

I get it.
But what do they mean it’s not proven?

Surely timesing any uneven number by 3 and adding 1 will make it an even number.
And deviding an an even number by 2 will half its size.
So, ultimately the number will get smaller and smaller, and, eventually reach 1.

What I don’t get is that they say it’s unproven. What do they mean by that?

Cheers.

@shavixmir said
I get it.
But what do they mean it’s not proven?

Surely timesing any uneven number by 3 and adding 1 will make it an even number.
And deviding an an even number by 2 will half its size.
So, ultimately the number will get smaller and smaller, and, eventually reach 1.

What I don’t get is that they say it’s unproven. What do they mean by that?

Cheers.

The point is: It is trivial to prove that if the values go down they will end in 8,4,2,1, there is no other way.

But the problem is that in principle the values are not bound to go down, are they?
If you take an odd number and multiply by three you just add a three to the prime factorization, but by adding one you alter the whole factorization, and you can't prove (yet) that it will always alter to a lot of two's in the end when things will go down.

@shavixmir said
Surely timesing any uneven number by 3 and adding 1 will make it an even number.
And deviding an an even number by 2 will half its size.
So, ultimately the number will get smaller and smaller, and, eventually reach 1.

That does not (necessarily!) follow. You overlook that the odd case increases the number more than the even case decreases it. To be precise, the increase is slightly over (because of the +1) 150% as the decrease: times three plus a bit vs. divided by two. This means that, if there were always one even case for every odd one, the numbers would rise indefinitely if jerkily: three steps forward, two back, as in Echternach.

However, that is not the case. An odd case must inevitably lead to an even one, but an even case may lead to an odd one or more even ones. Therefore, there will be more even cases than odd ones. The crucial question is now: are there sufficiently more even cases to counterbalance the larger odd ones?

Collatz' conjecture says: yes, for every number.
Experience says: yes for every number we have tried, but for some only by a small margin, and we haven't tested them all yet so there may be a few, rare numbers for which the answer is no.
Proven mathematical reasoning says: we don't know, and we don't even really have any idea how to tackle the problem definitively.
We do have a proof that it is true for 'almost all' numbers, meaning that it is not possible for the conjecture to be mostly true up to a large number but false above that. However, there are other conjectures we knew to be almost always true, believed to be always true, and then found a large counter-example, for instance the Pólya conjecture.

As I stated in my OP: it seems to me logical that the numbers would go down.

5
5x3 = 15+1 = 16
16/2 = 8
8/2 = 4
Etc.

So ever larger numbers, even though many more steps, will eventually, end up smaller.

I don’r get the problem.

@shavixmir said
As I stated in my OP: it seems to me logical that the numbers would go down.

It may seem logical to you, but 'stands to reason, dunnit?' does not count as a mathematical proof. You need to be a lit more rigorous than that.

As I said, mathematics has seen conjectures before which 'made sense', seemed correct up to one gajillion, but proved false for one gajillion and umpty. Mathematics demands proof, not trial and error.

@shavixmir said
As I stated in my OP: it seems to me logical that the numbers would go down.

5
5x3 = 15+1 = 16
16/2 = 8
8/2 = 4
Etc.

So ever larger numbers, even though many more steps, will eventually, end up smaller.

I don’r get the problem.

Around 6500 there is a number requiring about 250 steps... and it goes up quite a bit.

As I tried to say before with 3n+1 the prime factorization alters quite a bit and even if you half the number after each rise you can't be sure that it comes down each and every time.

@ponderable said
Around 6500 there is a number requiring about 250 steps... and it goes up quite a bit.

As I tried to say before with 3n+1 the prime factorization alters quite a bit and even if you half the number after each rise you can't be sure that it comes down each and every time.

Every single number they’ve tried, it always comes down at the end.

If I throw a hammer in air from the ground, it comes down. If I throw a hammer in the air from a tree top it comes down.
Now, I’ve never thrown a hammer into the air from a bridge before… but, you know, it will come down.

It’s still not clear to me what sort of proof they need.

@shavixmir said
It’s still not clear to me what sort of proof they need.

I can't explain it any more clearly than before: this would not be the first theory which had been tested for enormous amounts of numbers and yet proven false for an even larger one.

Mathematics demands proof that the Collatz sequence must go down to 1 for all numbers, not that it will probably do so until we get to 10^140.

Look up the Mertens conjecture for another example of how 'It Stands To Reason' can fail in mathematics. And consider that the proof for the infinity of prime numbers doesn't involve finding ever larger and larger primes, either.