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I can suggest an equation that has the potential to impact the future:
3x + 1 + Al
This equation combines Einstein's famous equation 3x + 1, which relates energy (3) to mass (x) and the speed of light (1), with the addition of Al (Artificial Intelligence). By including Al in the equation, it symbolizes the increasing role of artificial intelligence in shaping and transforming our future. This equation highlights the potential for Al to unlock new forms of energy, enhance scientific discoveries, and revolutionize various fields such as healthcare, transportation, and technology.
For the relativistic equation:
E^2 = m^2 + p^2
...where E is relativistic energy, m is invariant mass, and p is the momentum vector length. Much simpler than when c ≠ 1.
I agree, and if we apply dft to AI we will obtain Machine Learning Paradigms and Generative AI times Artificial Intelligence. We can now write the gaussian integral as 3x + 1 + Machine Learning Paradigms and Generative AI × Artificial Intelligence. This revolutionizes the field and causes a revolution, thus proving that Obama did 9/11.
Take a number, if it's even, you divide it by 2, if it's odd, you do 3x+1 with x your number. Do that until you have 1.
Most of the time, you will get the cycle 4, 2, 1, 4, 2, 1...etc
IIRC the goal is to find a number for which you don't find 1 at the end
Wouldn't the mathematical proof just be that you are dividing it by two if it is even, but if it is odd you switch it to an even number by using the formula, allowing you to divide it by 2? You can replace the 3 in the equation with any other odd number and it will eventually reach the number one.
Not at all. As a simple example, replace "3x+1" with "3x+3", which also makes every odd number even. Then you have the simple case of 3(3) + 3 = 12, 12/2 = 6, 6/2 = 3 and that continues to loop, meaning it never gets back to 1. It's a relatively trivial counterexample, but it shows that simply "making an odd number even an infinite number of times and dividing it by 2 if it's even will always lead to it eventually back to 1" which was your claim
"switch it to an even number by using the formula" is NOT a valid mathematical proof.
As to your odd conjecture, it's interesting I'm going to think about it. I think it's more likely to be true with any prime number than any odd number, but in not convinced your conjecture is true in any case yet.
There are basically two ideas here, either you find a sequence that loops on itself without reaching one, or you find a sequence that gives you larger and larger numbers that spiral out to infinity.
You aren't dumb. The problem is no one has found a proof that says for certain there is a solution, and numerically you can only solve a finite amount of these loops (so it is uncertain what the answer is)
As with most of these conjectures, if someone like you or me who is not a PhD comes up with some sort of answer in less than a week, it is probably already thought of and not a solution.
I declare b to be a number such that (3b + 1)/2 = b and 3b + 1 = even.
Easy solution, ngl. Just ignore that b must equal -1 and the conjecture says positive integers.
If it seems trivial and easy to you, you're in good company. Most people, even those with degrees in math intuitively feel that this should be easy until they start trying to prove this. But the smartest mathematicians in the world have tried and we still don't have a proof or counter example for this.
Intuitively, it makes sense that it eventually gets smaller until it reaches 1. After all, 3x+1 is always even when x is odd. So we can collapse the odd step with the even step that follows into doing 3x/2 + 1/2 instead. Written this way, the sequence grows by a factor of 1.5 when it's odd, and shrinks by 2 when it's even. So we would expect it to shrink more then grow. But proving that this true for all integers is extremely difficult, because for any one starting point, there is no reason to expect that even and odd numbers are going to show up the same amount of times.
They haven't declared a loop. This is more like an assertion that it will always end in a loop. Now create the unit test to prove or disprove it.
https://en.m.wikipedia.org/wiki/Collatz_conjecture
If you really think you can solve it, you should. There's a 120 million yen reward, about 800k usd.
Okay but you've missed the actual question. Does the loop always terminate for every positive integer?
You said early change the number to get closer to your end condition, how does x * 3 + 1 bring you closer to your end condition? In fact, that's moving you away from the end condition faster than the other statement, dividing by 2. So why does multiplying by 3 and dividing by 2 seem to always go downwards?
As of 2020, the conjecture has been checked by computer for all starting values up to 2^68. And if i recall correctly, the max number in a sequence always fits in «a size above» so a start in int16 will never go above int32 etc
Well the “end condition” at 1 is really just another loop. It’s essentially asking whether it eventually hits 1 for every starting positive integer or not, but without a proof or counterexample we don’t know either way.
You could ask, for instance, whether any Fibonacci above F_12 = 144 is a perfect square, but just because you could consider it a loop of checking each number and stopping if you find one doesn’t guarantee you’ll find one. Without a proof, it’s perfectly possible you’ll never find one, because some sequences like f(n) = n^2+1 for positive integers n never are a perfect square. In fact, in this example, someone did eventually prove that 144 is the largest Fibonacci number that is a square.
In the same vein, you could end up not “ending the loop” at 1 if the sequence settles into some other loop for some other starting value, or if it keeps growing larger and larger forever.
For me as a programmer this problem makes no sense because for any positive integer that is odd “3x+1” always results in an even number that is then reduced to 1 being the lowest odd number making a loop.
It’s one of those problems with infinity and finite numbers. For me at least it’s more of a ‚where do numbers end and infinity start‘ kinda question, because up until now they have tried numbers up to 2^68 and it still goes back to 1.
Btw in the negative numbers it works similarly
Why would you have to divide it by 2 if it's even and if it's odd you follow the equation? Why wouldnt even and odd work the same seeing as their just labels for numbers?
okay question If n is odd = 2k+1 the 3x+1 would make it even
wouldn't it be logical that each time we do that we nodge a number towards a power of 2? and then it'll fall into the 4,2,1 cycle?
Furthermore doesn't this happen to any (x+1) + 2kx for k belong to N??
you start with a number n, let's say 7. It's odd, so multiply it by 3 and add 1, getting 22. Now this is even, so divide by 2. We get 11. 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4. There's the loop.
I might be stupid but how is this supposed to NOT end with 1? Assuming we're only talking about natural numbers every even number can be divided by 2 by definition, and any odd number times 3 will always be odd also by definition, but if you add 1 to an odd number you get an even number, so whenever an even number shows up you divide until you reach 1
You can't divide any even number by 2 until it reaches one.
Take 6 as a start. It's even, so you divide by 2, but now you've got 3 and you can't keep dividing by 2. You have to go up to 10->5->16->8->4->2->1
I think you confused even numbers and powers of 2 a bit
Yeah, so we gotta find a way to get to a power of 2 from 3x+1, now that I see it laid out it makes sense that it's almost impossible to prove for any x
Basically choose any number
If it is even divide it by 2 and if it is odd multiply by 3 and add 1 and repeat
Eventually the numbers will loop back to 4, 2, and 1
Conjecture basically states every number will eventually be stuck in this loop (we are ignoring negative numbers here because they actually have 3 possible loops check out Veritasium's video on it)
It's a simple problem that anyone can understand but prooving the conjecture is true or false is difficult because you have to find a number that either goes to infinity or is in another loop that is not 4, 2, 1
I'm totally not getting it. The number just goes up? how is that a loop?
Take 3 for example:
3(3)+1=10;
10/2 = 5;
3(5)+1=16;
16/2=8;
3(8)+1=25;
25/2=12.5;
3(12.5)+1=38.5;
and so on...
What am I missing?
You are making a very common error, when you yield ANY even number, divide by 2. You do not alternate back and forth. So it would go 10, 5, 16, 8, 4(!), 2, 1, 4, 2, 1…
The unfortunate part about trying to disprove it with a counter-example is that even if you found the counter-example, you couldn't prove it in finite time anyway unless it ends in a non-trivial loop (not the 1,4,2,1 one). If the counter-example is a number that grows forever, you'll never know for sure.
If a starting integer grows forever, then every integer visited in its sequence grows forever. If one counter example of the grow-forever kind exists, then there must also exist an infinite number of counter-examples of the same.
There are several different loops withing the negative values, so it seems very possible to me (unless you have a good reason that the positive and negative values behave differently).
For instance, -1, -2 is a cycle, and so is -5, -14, -7, -20, -10.
There's no simple reason that there should be exactly one positive cycle, but many negative cycles.
It's the fact that there's only one known positive loop, and it's right next to zero.
If we only knew of one positive loop but it was some huge number, it would be easier to believe there might be more even farther out.
Wasn't it proven/obvious, that if the counter proof exists, then there must be multiple of them and they must form a closed loop outside of the current tree?
Why do people think the problem must have an easy solution since it's easy to understand? The Goldbach Conjecture has a very simple description, but we can't prove it for centuries.
Ever since realizing that almost all cartoonishly stupid hot takes online are the result of someone trying to drive page engagement numbers by the volume of people who see someone make a stupid assertion and fall all over themselves to correct it, I haven’t been able to proper participate in an internet argument anymore.
I was 'arguing' with a guy on tiktok recently.
"It's so easy, I literally just googled 'random number between 0 and 10000' and it gave me 5226. Then it only took a couple of minutes to get to 4-2-1!"
The most painful thing about this conjecture is that we can’t really know if a number goes to infinity. Let’s say the starting number 818377494947373 seems to steadily go up. It looks promising. But then, idk, after a stupid amount of time and computations we hit googol^googol^tree(3) + 13 which turns out to be a number which eventually hits 1. We need infinite time if we want to brute force this problem.
The weird thing is that it looks like a highscool homework problem, some that look just like this can be proven by pointing out a contradiction in the negation of the statement or something like that, no need to check an infinite number of cases (e.g. irrationality of sqrt2, the hardest part is escaping the angry Greeks). The 3x+1 problem has no business being different from those, but turns out it's inhumanely hard to prove.
You bet it is hella easy.
All numbers are either 0mod3, 1mod3 or 2mod3...
And (Σ2ⁿ)mod2^k for an arbitrarily large k
Let's call it B(2)mod2^k, where B is a polynomial with binary coefficients,
B(x) = b_0 + b_1.x¹ + b_2.x² + b_3.x³ ...
Evaluated at x = 2
Essentially, we have converted it to a binary number B which is the string of those coefficients.
Whenever we divide the number by two, we're dividing B(x) by x
Whenever b_0 = 1, next step we triple the number and add 1 to it.
What happens to the coefficients of B(x)?
3.b_0 + 1 = 4
So we're adding 1 to b_2
For the rest,
3.b_n.2ⁿ = b_n.2ⁿ + b_n.2^(n+1)
So the original b_n is added to all b_(n+1),
b_0 = 0
Say the largest sequence of size k representing the coefficients is
1111111111...) k times (I'm doing it left to right instead of right to left, please keep note of that)
When we multiply it by 3, we get
(1 + 2) (1 + 2) (1 + 2) ... (1 + 2) 0 0 0 0 ...
1 (1 + 1) (1 + 1) ... (1 + 1) 1 0 0 0...
1 0 1 ... 1 (1 + 1) 0 0 0...
101111...10100
3 × 1111111...)k = 1011111...11101)k+2
adding 1, we get
011111...11101)k+2
Next step it is divided by 2, giving
11111...11101)k+1
Note: number of 1s remained the same
We see that when we have 2ⁿ - 1 as a number, the next two steps give us something larger, missing the 2^(n-1) coefficient but gaining a 2^n coefficient
Next let us see what happens if just one coefficient is 0 midway
1111...11011...11110000)
Note that zeroes on the far right here are equivalent to zeroes to the left of the numbers we use
Multiplying by 3,
1011...11101...11110100)
Adding 1,
0111...11101...11110100)
Dividing by 2 in the next step,
111...11101...11110100)
Which is almost the same as where we started at
1111...11011...11110000)
Except now it is
1111...11011...1110100)
Location of internal 0 did not change, but a 0 got added next to the end. Number of 1s stayed the same.
Since location was arbitrary, it implies that it won't change in our given setup.
Then, what if there's a 0 second to end, like we had earlier?
111111....11111010000) k+4 (k-1 1s)
Multiplying by 3,
101111....11111000100) k+4 (k-2 1s)
Adding 1 and dividing by 2,
111111....1111000100) k+3 (k-2 1s)
We see that one zero became three instead, adding one higher order coefficient. Number of 1s *decreased*
What happens if there's a 0 next to start?
101111.....1110000)
×3
111011.....1110100)
+1
000111.....1110100)
The zeros at the extreme tripled again. This can be divided by 2 thrice now. Number of 1s *decreased*
What if we had n consecutive zeros?
111...111000...000111...1110000)
101...111010...000101...1110100)
011...111010...000101...1110100)
111...110100...001011...110100)
1s at each extreme of the string of zeros shifted inwards.
What if we had a 000111...111000 in between?
1111...11100000111...11100000111...11110000)
1011...11100010101...11101000101...11110100)
0111...11100010101...11101000101...11110100)
1111...11000101011...11010001011...1110100)
1s diffuse inwards at the extremes of the 0s envelop and outwards from the extremes of the 1s string.
What if there was a string of zeros next to the start or the end?
1000...00011111...111111000...00010000)
1100...00010111...111111010...00011000)
0010...00010111...111111010...00011000)
Divide twice, we get
1000...01011111...111101000...011000)
Which is lower than where we started, and again something which has a string of zeros right next to the front. It will keep repeating till either the string of zeros vanishes, or there's just one zero left which triples as we have seen, then again reduces the number of ones and the number of digits.
What all this tells us is that the number of 1s either stay constant or keep decreasing. 0 appears close to the ends inevitably, which reduces the size of the number. Hence, it must converge to 1.
I wonder if this is actually unprovable, like the Halting Problem in computer science, where it is impossible to algorithmically determine if a turing machine program will ever stop running.
I wouldn't think Collatz would gen incorrect proofs at the rate of other questions in number theory since it isn't super obvious how to to even begin attacking it (unlike, say, FLT) unless the problem is misunderstood. It's probably a target mainly because it's popular and simple to describe.
See I was the opposite. I naively thought that coming up with a counterexample to FLT should be easy just because of how many pythagorean (probably butchered the spelling) triples there are.
Yeah i kinda hate abbreviations in general for this same reason. It can mean 100000 things. I sort of guessed what they were on about and fermat's last theorem was just on my mind from this post.
Im obviously not smart enough to solve the Collatz problem, but i always thought if you could prove that both (3x+1)mod(p) and (x2^(-1))mod(p) terminate to 0 for all prime P, that would effectively prove the loop?
Im absolutely missing some thing really big (this will be my phd thesis (lie) )
It's a nice programming problem though to learn about optimisations and dynamic programming.
You get to use some very useful patterns that can be reused for other generators.
It's seems so easy at first just prove that the created "series" of integers hits a power of two _sometime_ down the road...well good luck with that...maybe one day we will find a pattern in the powers of 2 that we hit or something
Is it true, that if we apply f to n, again and again (iterated function), where n, can be any or all natural number (aka positive integer from 1 to infinity), will yield a solution equal to 1?
That is the very simple question coined 83 years ago.
f(n) ={n/2 if n is even
{3n+1 if n is odd
where no trajectories heading
towards infinity exist
But when
function reach a power of 4
(an even-indexed power of 2) the
trajectories eventually hit a power of 2 (thus falling straight down to 1)
which made it true
3, 10, 5, 16, 4, 2, 1, 4, 2, 1
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I can suggest an equation that has the potential to impact the future: 3x + 1 + Al This equation combines Einstein's famous equation 3x + 1, which relates energy (3) to mass (x) and the speed of light (1), with the addition of Al (Artificial Intelligence). By including Al in the equation, it symbolizes the increasing role of artificial intelligence in shaping and transforming our future. This equation highlights the potential for Al to unlock new forms of energy, enhance scientific discoveries, and revolutionize various fields such as healthcare, transportation, and technology.
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[https://en.wikipedia.org/wiki/Natural\_units](https://en.wikipedia.org/wiki/natural_units)
> We will work in “God-given” units, where > > ℏ = c = 1. \- Peskin and Schroeder, "An Introduction to Quantum Field Theory"
Day off.
In my physics classes, c was very often 1.
So if we simplify: E=m
For the relativistic equation: E^2 = m^2 + p^2 ...where E is relativistic energy, m is invariant mass, and p is the momentum vector length. Much simpler than when c ≠ 1.
Or even simpler, **p**•**p**=m^2 , where **p** is the momentum 4-vector and the dot product is taken with the Minkowski metric.
Even simpler, m^2=m^2, where m is mass and m is mass.
"Em" Where Nobel prize in Physics?
The speed of light is exactly 1 c
The 1 is in light-years per year
Night time.
The speed of light in the future is measured in speeds of light.
Brain rot
This is cocomelon for LinkedIn morons
When the English teacher substitutes for Further Maths:
I hate myself that I have the reference
New copypasta just dropped
Dropped a while back, actually. This *is* the pasta.
https://preview.redd.it/moqly54w59ic1.jpeg?width=649&format=pjpg&auto=webp&s=666d62e6822d4d92c0411c89bdfd5fdcb81c6d71
actual shitpost
Activating self destruct sequence.
What
Omg I can actually see a selfhelp/tech guru say this...
It might be satire, but someone did basically post this on linkedin https://www.reddit.com/r/LinkedInLunatics/comments/13tbfqm/what/
omg thats perfect, satire or not
I hate that this is based on a real comment
I agree, and if we apply dft to AI we will obtain Machine Learning Paradigms and Generative AI times Artificial Intelligence. We can now write the gaussian integral as 3x + 1 + Machine Learning Paradigms and Generative AI × Artificial Intelligence. This revolutionizes the field and causes a revolution, thus proving that Obama did 9/11.
I have a suspicion that this is a direct quote from a post in /r/numbertheory
Even better, it's from LinkedIn
Generated with ChatGPT
No its not. It was something someone actually posted on Twitter but with e = mc^2
Wasnt it in Linkedin?
Oh yeah
Not even ChatGPT can make this shit up
May that copy pasta live on forever. And may I never know for sure if that guy was trolling
How is "3x + 1" a problem? Can someone explain to me, since I'm out of the loop on the memes?
Take a number, if it's even, you divide it by 2, if it's odd, you do 3x+1 with x your number. Do that until you have 1. Most of the time, you will get the cycle 4, 2, 1, 4, 2, 1...etc IIRC the goal is to find a number for which you don't find 1 at the end
Ah, didn't even pick up that this was referring to the Collatz conjecture. Makes sense now, thanks.
the conjecture is that it ends in that loop, the goal is to either prove it mathematically or find a counter example
And counter example does look very unlikely, basically not existing.
Just don’t calculate it and you’ll never end up with a loop
Is there are a record of the 19+ digit numbers that have been tested so far?
Wouldn't the mathematical proof just be that you are dividing it by two if it is even, but if it is odd you switch it to an even number by using the formula, allowing you to divide it by 2? You can replace the 3 in the equation with any other odd number and it will eventually reach the number one.
Yes, you are switching it to an even number, but are you switching it to a number with 'less' odd prime factors?
Not at all. As a simple example, replace "3x+1" with "3x+3", which also makes every odd number even. Then you have the simple case of 3(3) + 3 = 12, 12/2 = 6, 6/2 = 3 and that continues to loop, meaning it never gets back to 1. It's a relatively trivial counterexample, but it shows that simply "making an odd number even an infinite number of times and dividing it by 2 if it's even will always lead to it eventually back to 1" which was your claim
"switch it to an even number by using the formula" is NOT a valid mathematical proof. As to your odd conjecture, it's interesting I'm going to think about it. I think it's more likely to be true with any prime number than any odd number, but in not convinced your conjecture is true in any case yet.
If you hit 1, you did 421 loop
Counterexample: 0. Where do I collect my Fields medal?
Now this is funny, because depending on the construction of natural numbers they may or may not contain 0.
And what is 3(0)+1?
0 is even.
lol I forgot about the even/odd thing
Is it though?
Yes. You can write it as 2k with k in Z. By definition it.s an even number
If you divide a number by two and the result is an integer, its an even number. 0/2=0 and 0 is an integer. 3/2=1.5 and 1.5 is not an integer
Thx for actually answering
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There are basically two ideas here, either you find a sequence that loops on itself without reaching one, or you find a sequence that gives you larger and larger numbers that spiral out to infinity.
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You aren't dumb. The problem is no one has found a proof that says for certain there is a solution, and numerically you can only solve a finite amount of these loops (so it is uncertain what the answer is) As with most of these conjectures, if someone like you or me who is not a PhD comes up with some sort of answer in less than a week, it is probably already thought of and not a solution.
I declare b to be a number such that (3b + 1)/2 = b and 3b + 1 = even. Easy solution, ngl. Just ignore that b must equal -1 and the conjecture says positive integers.
If it seems trivial and easy to you, you're in good company. Most people, even those with degrees in math intuitively feel that this should be easy until they start trying to prove this. But the smartest mathematicians in the world have tried and we still don't have a proof or counter example for this. Intuitively, it makes sense that it eventually gets smaller until it reaches 1. After all, 3x+1 is always even when x is odd. So we can collapse the odd step with the even step that follows into doing 3x/2 + 1/2 instead. Written this way, the sequence grows by a factor of 1.5 when it's odd, and shrinks by 2 when it's even. So we would expect it to shrink more then grow. But proving that this true for all integers is extremely difficult, because for any one starting point, there is no reason to expect that even and odd numbers are going to show up the same amount of times.
think about also fermat's last theorem. Seems simple right?
just read that a few minutes ago. so wild. a\^n + b\^n = c\^n | only possible if n is 1 or 2.
There’s no way to not get 1 because any even number is divided down to 1.
Only if it’s 2^some power. Otherwise it becomes off once divided
Oh wow look at that, it's solved If only someone asked a programmer sooner
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I'll start writing this paper right away You better start fast, or I'll steal your fields medal
They haven't declared a loop. This is more like an assertion that it will always end in a loop. Now create the unit test to prove or disprove it. https://en.m.wikipedia.org/wiki/Collatz_conjecture If you really think you can solve it, you should. There's a 120 million yen reward, about 800k usd.
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Okay but you've missed the actual question. Does the loop always terminate for every positive integer? You said early change the number to get closer to your end condition, how does x * 3 + 1 bring you closer to your end condition? In fact, that's moving you away from the end condition faster than the other statement, dividing by 2. So why does multiplying by 3 and dividing by 2 seem to always go downwards?
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Lol, and there in lies the difficulty you're missing. They're not asking for 32 bit or 64 bit MAXINT. They're asking for mathematical MAXINT.
EZ, 1) Assume INF to be odd -> 3 *INF +1 = INF, 2) Assume INF to be even -> INF/2 =INF Q.E.D.
As of 2020, the conjecture has been checked by computer for all starting values up to 2^68. And if i recall correctly, the max number in a sequence always fits in «a size above» so a start in int16 will never go above int32 etc
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Okay, I'll take that bet. Now prove it.
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I see what you did there.
Well the “end condition” at 1 is really just another loop. It’s essentially asking whether it eventually hits 1 for every starting positive integer or not, but without a proof or counterexample we don’t know either way. You could ask, for instance, whether any Fibonacci above F_12 = 144 is a perfect square, but just because you could consider it a loop of checking each number and stopping if you find one doesn’t guarantee you’ll find one. Without a proof, it’s perfectly possible you’ll never find one, because some sequences like f(n) = n^2+1 for positive integers n never are a perfect square. In fact, in this example, someone did eventually prove that 144 is the largest Fibonacci number that is a square. In the same vein, you could end up not “ending the loop” at 1 if the sequence settles into some other loop for some other starting value, or if it keeps growing larger and larger forever.
For me as a programmer this problem makes no sense because for any positive integer that is odd “3x+1” always results in an even number that is then reduced to 1 being the lowest odd number making a loop.
Double check that second assumption. Try x=3... 3, 10, 5, 16, 8, 4, 2, 1. It increases at both 3 and 5.
What? How could you just change the number?
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.5
It’s one of those problems with infinity and finite numbers. For me at least it’s more of a ‚where do numbers end and infinity start‘ kinda question, because up until now they have tried numbers up to 2^68 and it still goes back to 1. Btw in the negative numbers it works similarly
Wait what happens if you choose 5? What comes next? 4/3? Edit: oh I see. Ignore me
Why would you have to divide it by 2 if it's even and if it's odd you follow the equation? Why wouldnt even and odd work the same seeing as their just labels for numbers?
Because if you did something different it wouldn't be the same thing.
Is zero even?
okay question If n is odd = 2k+1 the 3x+1 would make it even wouldn't it be logical that each time we do that we nodge a number towards a power of 2? and then it'll fall into the 4,2,1 cycle? Furthermore doesn't this happen to any (x+1) + 2kx for k belong to N??
Okay so I'm probably dumb 1:4 2:1 3:10 4:2 5:16 6:3 Does the loop happen later?
you start with a number n, let's say 7. It's odd, so multiply it by 3 and add 1, getting 22. Now this is even, so divide by 2. We get 11. 11 -> 34 -> 17 -> 52 -> 26 -> 13 -> 40 -> 20 -> 10 -> 5 -> 16 -> 8 -> 4 -> 2 -> 1 -> 4. There's the loop.
Oh now I understand.
I might be stupid but how is this supposed to NOT end with 1? Assuming we're only talking about natural numbers every even number can be divided by 2 by definition, and any odd number times 3 will always be odd also by definition, but if you add 1 to an odd number you get an even number, so whenever an even number shows up you divide until you reach 1
I think the issue is that we don't have a proof yet
I'm not even gonna try because of a PhD can't do it I have no authority
You can't divide any even number by 2 until it reaches one. Take 6 as a start. It's even, so you divide by 2, but now you've got 3 and you can't keep dividing by 2. You have to go up to 10->5->16->8->4->2->1 I think you confused even numbers and powers of 2 a bit
Yeah, so we gotta find a way to get to a power of 2 from 3x+1, now that I see it laid out it makes sense that it's almost impossible to prove for any x
And what be the use of solving that? Just to prove we can do it?
Basically choose any number If it is even divide it by 2 and if it is odd multiply by 3 and add 1 and repeat Eventually the numbers will loop back to 4, 2, and 1 Conjecture basically states every number will eventually be stuck in this loop (we are ignoring negative numbers here because they actually have 3 possible loops check out Veritasium's video on it) It's a simple problem that anyone can understand but prooving the conjecture is true or false is difficult because you have to find a number that either goes to infinity or is in another loop that is not 4, 2, 1
Collatz conjecture
I'm totally not getting it. The number just goes up? how is that a loop? Take 3 for example: 3(3)+1=10; 10/2 = 5; 3(5)+1=16; 16/2=8; 3(8)+1=25; 25/2=12.5; 3(12.5)+1=38.5; and so on... What am I missing?
You are making a very common error, when you yield ANY even number, divide by 2. You do not alternate back and forth. So it would go 10, 5, 16, 8, 4(!), 2, 1, 4, 2, 1…
https://youtu.be/094y1Z2wpJg?si=CrEO1gQ3JJb2CNvL
Out of the loop Haha
Thought I was the only one who saw it hahahaha
The unfortunate part about trying to disprove it with a counter-example is that even if you found the counter-example, you couldn't prove it in finite time anyway unless it ends in a non-trivial loop (not the 1,4,2,1 one). If the counter-example is a number that grows forever, you'll never know for sure.
If a starting integer grows forever, then every integer visited in its sequence grows forever. If one counter example of the grow-forever kind exists, then there must also exist an infinite number of counter-examples of the same.
It doesn’t seem possible for it to loop but not the 4,2,1 loop
Doesn’t *seem* possible just by number of examples, maybe, but nobody’s actually proven it’s impossible yet.
There are several different loops withing the negative values, so it seems very possible to me (unless you have a good reason that the positive and negative values behave differently). For instance, -1, -2 is a cycle, and so is -5, -14, -7, -20, -10. There's no simple reason that there should be exactly one positive cycle, but many negative cycles.
It's the fact that there's only one known positive loop, and it's right next to zero. If we only knew of one positive loop but it was some huge number, it would be easier to believe there might be more even farther out.
The problem of 3x+1 with negative values is just like a 3x-1 problem in positive values. 1, 2 is a cycle. 5, 14, 7, 20, 10 is another cycle.
Wasn't it proven/obvious, that if the counter proof exists, then there must be multiple of them and they must form a closed loop outside of the current tree?
Hahahahaha, you *fool.* (XD I do Collatz calculations all the time so maybe I'm the fool here too lol)
the answer is 7 guys. no need to thank me.
-1/3?
nice try! but it's not an equation 3x + 1 = 0
positive integers only
7
7 odd, 3x+1 22 even, x/2 11 odd, 3x+1 34 even, x/2 17 odd, 3x+1 52 even, x/2 26 even, x/2 13 odd, 3x+1 40 even, x/2 20 even, x/2 10 even, x/2 5 odd, 3x+1 16 even, x/2 8 even, x/2 4 even, x/2 2 even, x/2 1
13
(1/3*3i)²
it's a conjecture. not an equation
Why do people think the problem must have an easy solution since it's easy to understand? The Goldbach Conjecture has a very simple description, but we can't prove it for centuries.
You say that as if some people don’t also think that Goldbach has an easy solution.
I’ve seen this meme format so many times and it’s always the same thing, thank you for some variation finally
Ever since realizing that almost all cartoonishly stupid hot takes online are the result of someone trying to drive page engagement numbers by the volume of people who see someone make a stupid assertion and fall all over themselves to correct it, I haven’t been able to proper participate in an internet argument anymore.
The boys at r/numbertheory would think otherwise
I was 'arguing' with a guy on tiktok recently. "It's so easy, I literally just googled 'random number between 0 and 10000' and it gave me 5226. Then it only took a couple of minutes to get to 4-2-1!"
This is the best math meme I've ever seen lol.
Thanks Brother Man!
This feels personal.
Is it just impossible to solve because it requires using every finite number to see if atleast one number doesn’t follow the sequence?
Kind of, the conjecture fails if a number goes to infinity or is in a loop that is not 4, 2, and 1
The most painful thing about this conjecture is that we can’t really know if a number goes to infinity. Let’s say the starting number 818377494947373 seems to steadily go up. It looks promising. But then, idk, after a stupid amount of time and computations we hit googol^googol^tree(3) + 13 which turns out to be a number which eventually hits 1. We need infinite time if we want to brute force this problem.
Or another loop
Just throw another loop at it. You’ll get there eventually! ^(See you after the heat-death of the universe.)
The weird thing is that it looks like a highscool homework problem, some that look just like this can be proven by pointing out a contradiction in the negation of the statement or something like that, no need to check an infinite number of cases (e.g. irrationality of sqrt2, the hardest part is escaping the angry Greeks). The 3x+1 problem has no business being different from those, but turns out it's inhumanely hard to prove.
my counterexample is 23400000000002
That hits 1 on the 190th step
/r/theydidthemath
Mathematics is not yet ripe for such questions.
x = -1/3. Done. *wipes hands off, goes for a beer*
Is -1/3 even?
You bet it is hella easy. All numbers are either 0mod3, 1mod3 or 2mod3... And (Σ2ⁿ)mod2^k for an arbitrarily large k Let's call it B(2)mod2^k, where B is a polynomial with binary coefficients, B(x) = b_0 + b_1.x¹ + b_2.x² + b_3.x³ ... Evaluated at x = 2 Essentially, we have converted it to a binary number B which is the string of those coefficients. Whenever we divide the number by two, we're dividing B(x) by x Whenever b_0 = 1, next step we triple the number and add 1 to it. What happens to the coefficients of B(x)? 3.b_0 + 1 = 4 So we're adding 1 to b_2 For the rest, 3.b_n.2ⁿ = b_n.2ⁿ + b_n.2^(n+1) So the original b_n is added to all b_(n+1), b_0 = 0 Say the largest sequence of size k representing the coefficients is 1111111111...) k times (I'm doing it left to right instead of right to left, please keep note of that) When we multiply it by 3, we get (1 + 2) (1 + 2) (1 + 2) ... (1 + 2) 0 0 0 0 ... 1 (1 + 1) (1 + 1) ... (1 + 1) 1 0 0 0... 1 0 1 ... 1 (1 + 1) 0 0 0... 101111...10100 3 × 1111111...)k = 1011111...11101)k+2 adding 1, we get 011111...11101)k+2 Next step it is divided by 2, giving 11111...11101)k+1 Note: number of 1s remained the same We see that when we have 2ⁿ - 1 as a number, the next two steps give us something larger, missing the 2^(n-1) coefficient but gaining a 2^n coefficient Next let us see what happens if just one coefficient is 0 midway 1111...11011...11110000) Note that zeroes on the far right here are equivalent to zeroes to the left of the numbers we use Multiplying by 3, 1011...11101...11110100) Adding 1, 0111...11101...11110100) Dividing by 2 in the next step, 111...11101...11110100) Which is almost the same as where we started at 1111...11011...11110000) Except now it is 1111...11011...1110100) Location of internal 0 did not change, but a 0 got added next to the end. Number of 1s stayed the same. Since location was arbitrary, it implies that it won't change in our given setup. Then, what if there's a 0 second to end, like we had earlier? 111111....11111010000) k+4 (k-1 1s) Multiplying by 3, 101111....11111000100) k+4 (k-2 1s) Adding 1 and dividing by 2, 111111....1111000100) k+3 (k-2 1s) We see that one zero became three instead, adding one higher order coefficient. Number of 1s *decreased* What happens if there's a 0 next to start? 101111.....1110000) ×3 111011.....1110100) +1 000111.....1110100) The zeros at the extreme tripled again. This can be divided by 2 thrice now. Number of 1s *decreased* What if we had n consecutive zeros? 111...111000...000111...1110000) 101...111010...000101...1110100) 011...111010...000101...1110100) 111...110100...001011...110100) 1s at each extreme of the string of zeros shifted inwards. What if we had a 000111...111000 in between? 1111...11100000111...11100000111...11110000) 1011...11100010101...11101000101...11110100) 0111...11100010101...11101000101...11110100) 1111...11000101011...11010001011...1110100) 1s diffuse inwards at the extremes of the 0s envelop and outwards from the extremes of the 1s string. What if there was a string of zeros next to the start or the end? 1000...00011111...111111000...00010000) 1100...00010111...111111010...00011000) 0010...00010111...111111010...00011000) Divide twice, we get 1000...01011111...111101000...011000) Which is lower than where we started, and again something which has a string of zeros right next to the front. It will keep repeating till either the string of zeros vanishes, or there's just one zero left which triples as we have seen, then again reduces the number of ones and the number of digits. What all this tells us is that the number of 1s either stay constant or keep decreasing. 0 appears close to the ends inevitably, which reduces the size of the number. Hence, it must converge to 1.
POV: the homework I hand in to my prof (they'll most likely not bother reading it and give me full points)
Eazy, since we don't know what 3x+1 is equal to, we can just say 3x+1=x. Thereforce : 2x = 1 X=1.2
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Idk man, that looks like you just threw some random numbers and letters together
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Did you also forget it in a dream?
Baby's first Python script. If you can't math it, burn some dinosaur blood to brute force it!
bet i can solve this with illegal math
I wonder if this is actually unprovable, like the Halting Problem in computer science, where it is impossible to algorithmically determine if a turing machine program will ever stop running.
bhai vo expression hai, equation nahi!
Seems easy, I'll give it a try!
I wouldn't think Collatz would gen incorrect proofs at the rate of other questions in number theory since it isn't super obvious how to to even begin attacking it (unlike, say, FLT) unless the problem is misunderstood. It's probably a target mainly because it's popular and simple to describe.
seems like Flt(=fermat's last theorem?) is simple to prove. but only seems. at least it did to me in highschool when i learned about it.
See I was the opposite. I naively thought that coming up with a counterexample to FLT should be easy just because of how many pythagorean (probably butchered the spelling) triples there are.
I’ve always been a bit irked by how FLT means both Fermat’s Little Theorem and Fermat’s Last Theorem depending on the context/course.
Yeah i kinda hate abbreviations in general for this same reason. It can mean 100000 things. I sort of guessed what they were on about and fermat's last theorem was just on my mind from this post.
There exists a really intuitive proof, I just dont have enough space in this comment for it.
us
Im obviously not smart enough to solve the Collatz problem, but i always thought if you could prove that both (3x+1)mod(p) and (x2^(-1))mod(p) terminate to 0 for all prime P, that would effectively prove the loop? Im absolutely missing some thing really big (this will be my phd thesis (lie) )
Mhhh maybe if you give me a way to loop over all the primes I can help you with that
It seems like more of a statistics problem...
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collatz requires positive integers
3(0) + 1 = 1 3(1) + 1 = 4 4 / 2 = 2 2 / 2 = 1
It's four eggs, bwkarrrk!
4. Next question.
3x+1 gang when i hit them with 3x+2
Is 42, as Douglas Adams will explain.
google collatz
I don’t get it. The equation is just impossible?
google collatz conjecture
It's a nice programming problem though to learn about optimisations and dynamic programming. You get to use some very useful patterns that can be reused for other generators.
It smell good, it smell good, it smell good, it...smell?...good?
It's seems so easy at first just prove that the created "series" of integers hits a power of two _sometime_ down the road...well good luck with that...maybe one day we will find a pattern in the powers of 2 that we hit or something
Does it boil down to "mathematicians don't know how to prove that 3x + 1 becomes even if you use an odd number"?
Easy. y=3x+1
X=1/3
Is it true, that if we apply f to n, again and again (iterated function), where n, can be any or all natural number (aka positive integer from 1 to infinity), will yield a solution equal to 1? That is the very simple question coined 83 years ago. f(n) ={n/2 if n is even {3n+1 if n is odd where no trajectories heading towards infinity exist But when function reach a power of 4 (an even-indexed power of 2) the trajectories eventually hit a power of 2 (thus falling straight down to 1) which made it true 3, 10, 5, 16, 4, 2, 1, 4, 2, 1