4,2,1

[mrgriffxy]

So there is this weird phenomenon that occurs when you follow these specific rules:

If ODD 3x+1

If EVEN x/2

The theory is that if you take a number through these rules as far as possible you will achieve a never ending cycle of 4,2,1. So say for instance you start with 5 follow the ODD function to get 16 then even function for 8 again for 4 again for 2 again for 1 then 4, 2, 1, 4, 2, 1... As far as I am aware this has not been proven to work for every number and it is unsure if there could be other sequences like this that appear from numbers untried. I guess my point is for a discussion to see if it can't be proven.

[Raider5678]

So there is this weird phenomenon that occurs when you follow these specific rules:

If ODD 3x+1

If EVEN x/2

The theory is that if you take a number through these rules as far as possible you will achieve a never ending cycle of 4,2,1. So say for instance you start with 5 follow the ODD function to get 16 then even function for 8 again for 4 again for 2 again for 1 then 4, 2, 1, 4, 2, 1... As far as I am aware this has not been proven to work for every number and it is unsure if there could be other sequences like this that appear from numbers untried. I guess my point is for a discussion to see if it can't be proven.

The idea is that when it multiplies the odd number by three, it returns it to a odd number of a higher power. By adding one, it turns it into an even number. Operating off of that, it's guaranteed to reach an even number that will eventually collapse into 4,2,1 fairly quickly. So since the multiply by three, add one, and divide by two when even can't work for any other numbers besides 4,2,1, it will always reach 4,2,1. Making this impossible to disprove.

[Delta1212]

The idea is that when it multiplies the odd number by three, it returns it to a odd number of a higher power. By adding one, it turns it into an even number. Operating off of that, it's guaranteed to reach an even number that will eventually collapse into 4,2,1 fairly quickly. So since the multiply by three, add one, and divide by two when even can't work for any other numbers besides 4,2,1, it will always reach 4,2,1. Making this impossible to disprove.

 

It's not actually yet been proven that this algorithm always collapses to 1 for any starting number. It's known as the Collatz Conjecture, and while it is generally considered that it is probably true, proving it is exceptionally difficult.

[fiveworlds]

proving it is exceptionally difficult

 

 

No it is isn't since all even numbers are divisible by 2. We just need to know what the last char is.

 

<script>

for(i=1;i<10;i=i+2){

if((i*3+1)%2!=0){

document.write("error");

}

}

</script>

 

Do you see what I am getting at?? What matters is the frequency of division by 2. Division by 2 occurs more frequently and therefore every number must reduce to 1.

Edited February 1, 2017 by fiveworlds

[Bignose]

No it is isn't

fiveworlds, if you think it is so easy, why don't you claim the prize for proving it:

 

http://thescienceexplorer.com/universe/earn-two-thousand-dollars-solving-simple-math-problem

[Raider5678]

fiveworlds, if you think it is so easy, why don't you claim the prize for proving it:

 

http://thescienceexplorer.com/universe/earn-two-thousand-dollars-solving-simple-math-problem

Every number up to 2^60(Frickin large ass number) has been tested.

Considering the numbers will eventually go lower then their original, and won't reach a point where they keep growing forever, they will eventually collapse down small enough to one of these numbers. Since collapsing down to any number below 2^60 is inevitable because all the mathematical patterns to there have been tested multiple times simply by the repeating 10,100,1000,10000,100000,1000000,10000000,100000000 etc, we know that it always eventually goes back down. There are only two types of numbers. Even and odd. There's only so many patterns that can be made with that.

[Bignose]

Every number up to 2^60(Frickin large ass number) has been tested.

...

Sure. But there are a lot more numbers above 2^60. An infinite number of them, to be exact.

 

That's why the proof for all numbers haven't been completed yet. It isn't easy. If you think your proof here is enough, why don't you claim the money?

Edited February 3, 2017 by Bignose

[Lord Antares]

But how can it then be proven if there are infinitely many numbers to test? The list cannot be completed.

 

Maybe you meant to say ''find one that doesn't match'', instead of ''prove that all match''?

[pzkpfw]

But how can it then be proven if there are infinitely many numbers to test? The list cannot be completed.

 

Maybe you meant to say ''find one that doesn't match'', instead of ''prove that all match''?

 

No, a proof like this doesn't mean that every possible number has been tested, it means that mathematical logic has been used to show that something must be true for any number in the problem space.

 

e.g. see the concept of induction: https://en.wikipedia.org/wiki/Mathematical_induction

 

The example there shows how it can be proven that 0 + 1 + 2 + ... + n = (n(n + 1)) / 2, for any natural number. There are infinite natural numbers, but the proof doesn't require each value of n to be plugged in and tested.

Edited February 3, 2017 by pzkpfw

[Bignose]

No, a proof like this doesn't mean that every possible number has been tested, it means that mathematical logic has been used to show that something must be true for any number in the problem space.

 

e.g. see the concept of induction: https://en.wikipedia.org/wiki/Mathematical_induction

 

The example there shows how it can be proven that 0 + 1 + 2 + ... + n = (n(n + 1)) / 2, for any natural number. There are infinite natural numbers, but the proof doesn't require each value of n to be plugged in and tested.

+1

 

This is what has to be done guys. And if someone does it, they win several thousand dollars. It's a hard problem.

[fiveworlds]

fiveworlds, if you think it is so easy, why don't you claim the prize for proving it:​

 

 

Okay.

 

So for collatz to be false. We must satisfy a simple function and show looping.

 

(3x+1)/2 =x

(3x + 1) = 2x

3x - 2x + 1 = 0

 

 

We can solve for all looping in the colatz conjecture by introducing a new variable y which is a multiple of 2.

 

3x - yx + 1 = 0

 

All except for 2 loops are looping fractions. Therefore no real number exists that falsifies colatz

[Acme]

fiveworlds, if you think it is so easy, why don't you claim the prize for proving it:

 

http://thescienceexplorer.com/universe/earn-two-thousand-dollars-solving-simple-math-problem

From your article:

The Collatz Conjecture, also known as the 3n+1 problem, has been driving mathematicians mad since the mid 1930s when it was it was first proposed by Lothar Collatz. Although it is simple to understand as it involves nothing more than addition and division it is not simple to prove.

 

Heres the problem:

Take a given integer n. If it is even, divide it by 2. If it is odd, multiply it by 3 and add 1. Continue the pattern taking the answer as your new n.

Lets look at some examples:

3, 10, 5, 16, 8, 4, 2, 1

7, 22, 11, 34, 17, 52, 26, 13, 40, 20, 10, 5, 16, 8, 4, 2, 1

12, 6, 3, 10, 5, 16, 8, 4, 2, 1

21, 64, 32, 16, 8, 4, 2, 1

You will notice that for each of these numbers, you end up at 1, which would lead to an infinite loop of 1, 4, 2, 1, 4, 2, 1

 

But is that the case for all integers? Thats the Collatz conjecture, and it has never been proven. It has been tested for absurdly large numbers for example, the largest number of steps it takes for a number below a billion to reach 1 is 986 but nobody has been able to say for sure that no number exists that wont ever reach 1.

H. S. M. Coxeter wrote of the difficulty of finding a counter-proof to the conjecture: I must warn you not to try this in your heads or on the back of an old envelope, because the result has been tested with an electronic computer for all x1 ≤ 500,000. This means that, if the conjecture is false, the prizewinner muse [sic] either find a sequence of this kind which he can prove to be divergent, or else find a cyclic sequence for this kind whose terms are all greater than half a million.

 

But, if you are still insistent, there is some prize money involved. Paul Erdos said of the conjecture: "Mathematics is not yet ready for such problems," but he offered $500 for a solution. Thwaites, after whom the problem is sometimes called the Thwaites conjecture, offered up £1000 (about $1500) as well.

Shall we calculate the odds that fivwolds collects a prize? Tries to claim the prize? Just asserts here he deserves the prize? Win, place, show?

[fiveworlds]

We could expand the equation out but all that happens the numbers become worse recurring fractions or negative numbers never positive whole numbers for loops and we can assume that no numbers cause it to increase forever.

[John Cuthber]

The idea is that when it multiplies the odd number by three, it returns it to a odd number of a higher power. By adding one, it turns it into an even number. Operating off of that, it's guaranteed to reach an even number that will eventually collapse into 4,2,1 fairly quickly. So since the multiply by three, add one, and divide by two when even can't work for any other numbers besides 4,2,1, it will always reach 4,2,1. Making this impossible to disprove.

This bit

"Operating off of that, it's guaranteed to reach an even number that will eventually collapse into 4,2,1 fairly quickly. "

is the problem there- you have stated it, but not proved it.

And it's pretty nearly a restatement of the original hypothesis.

Only numbers of the form 2^n are guaranteed to fall to 4-2-1 that way.

 

Most numbers are not of that form.

[Bignose]

... and we can assume that no numbers cause it to increase forever.

assumption =/= proof. That's kinda the whole point of proof, buddy.

[Raider5678]

This bit

"Operating off of that, it's guaranteed to reach an even number that will eventually collapse into 4,2,1 fairly quickly. "

is the problem there- you have stated it, but not proved it.

And it's pretty nearly a restatement of the original hypothesis.

Only numbers of the form 2^n are guaranteed to fall to 4-2-1 that way.

 

Most numbers are not of that form.

Good point. I made a hyperbole. My bad.

The idea is that the equation has always reached a n^2 eventually so far.

I haven't proven anything, but it can be assumed. Assumed that if I needed to place a bet, I would bet that all numbers reach 4,2,1.

[Bignose]

Assumed that if I needed to place a bet, I would bet that all numbers reach 4,2,1.

Sure, and I think most mathematicians agree with you. The Collatz Conjecture is generally believed to be true. But proving it has been difficult. That's where the fun in mathematics is.

[fiveworlds]

assumption =/= proof. That's kinda the whole point of proof, buddy.

Is it easier to see if I write it like this??

(3x+1)/2= x
y(3x+1)/z = x
3yx - xz + y = 0
x(3y - z) + y = 0
x = -y/(3y-z)
x = -10/(30-32)

As you can see if x is positive it is a fraction and it will not increase forever. We know it won't increase forever for certain values because it is a group of linear equations. Edited February 6, 2017 by fiveworlds

[Raider5678]

Is it easier to see if I write it like this??

(3x+1)/2= x
y(3x+1)/z = x
3yx - xz + y = 0
x(3y - z) + y = 0
x = -y/(3y-z)
x = -10/(30-32)

As you can see if x is positive it is a fraction and it will not increase forever. We know it won't increase forever for certain values because it is a group of linear equations.

 

What's z?

[mrgriffxy]

All even numbers are divisible by 2. So if you always make an even number you will eventually lead to 4,2,1.
3x+1=ODD+1=EVEN.

[fiveworlds]

I followed basic algebraic rules z is a positive whole number multiple of 2. Y is more complicated it is a black box polynomial we don't know anything about since each polynomial will be specific to what number you put in. We do know however that it will follow basic algebra so we don't need to know anything about it.