A number of people are asking why this problem is “dangerous”. It’s described as dangerous because its difficulty has defeated the world’s greatest mathematical minds for generations. Paul Erdos, a famous mathematician, said, "Mathematics is not yet ripe enough for such questions." Jeffrey Lagarias called it "an extraordinarily difficult problem, completely out of reach of present day mathematics". The problem is so maddeningly difficult, mathematicians are warned to stay away from it. Watch the full video for more context: ruclips.net/video/094y1Z2wpJg/видео.html
So, even the greatest of mathematicians cannot exactly solve it. But something I don't get, is solving the problem. What even is the problem in the first place?
Maybe because it’s not a problem to solve… what or why would be a reason it would need a solution? Or more so what would actually be achieved if it were or any of the other unsolved math problems… are these problems preventing some sort of breakthrough??? Just seems like number puzzles without purpose.
@@Overkillutube you remember the time people think infinity is un-slove-able and questions the others why they research infinity? The anz is told by Newton's maths.
@@AlexanderBrown77 Jesus said he would return before the people who he was talking to all died. He didn't. He's not coming back. You Belive in a 2000 year old fairy tale and are so stupid you don't even Belive Jesus own words about when he will return.
@@ApoorvaS-yt4iq It's only dangerous in the sense that it's dangerous to your career as a mathmetician. Some of the greatest mathematical minds have tried to solve it for years and failed. It's basically a warning given to any budding mathmeticians - stay away from the Collatz Conjecture if you want to get anywhere. But it is a very alluring problem because it's utterly fascinating and surely there has to be a solution to it one way or another.
@@ApoorvaS-yt4iq cannot prove that this will be the case for any natural number, although repeated checks confirm this. They cannot prove it in general terms, although this problem is thousands of years old
@@hickoryst.6961 if someone is able to come up with an abstract formulation that can prove that the conjecture is right, this person has just come up with a new branch of the mathematics of the natural numbers.
He calls it dangerous because it's considered to currently be pretty much unsolvable and yet many mathematicians quickly find interest in this problem and start trying to find a solution which can result in them essentially wasting years of their careers.
"Michael?? Where did you come from?" "Am I here?.. Where is 'here' anyway? Or... Was I always here. Truth is, I've always been here... Because, wherever I am, that is... Here... But wat if-" "Oh boy, here we go." 🤦
@@6uis1948 Wait, you mean people keep making this joke? Over and over again? Even though everyone has seen it and it's pointless to post it? Man, if only there were a word that helped relate or suggest the labors of Sisyphus. Specifically the requiring continual and often ineffective effort. Sisyphean is the word. I'm glad to have explained this joke to you.
@@datruommi not the exact same thing, because this one is simply taking the inverse operation everytime (is like the "take the jacket" and "put the jacket" loop from karate kid), whereas 3x+1 and x÷2 are not that directly conected
@@datruommi no, because this one is easy. If the number is odd, it will be multiplied by 2, giving an even result and will therefore be divided back to the original number. If the number is even, it will be divided by 2 until it is an odd number, which will lead to the loop mentioned above. That will be true for all N belonging to the natural numbers. However, can you ensure that all the natural numbers in the 3x+1 problem lead to the 4 ➝ 2 ➝ 1 loop? Because obviously you can't just brute force an infinite amount of numbers in a finite period of time.
It's fascinating to me too. If it is true, or not true, then why? If a mathematician could prove it's always true or find its not always true, then the underlying why would be answered
The goal is to find a number which does not end up in a 4,2,1 loop. But the problem is that there's no such number found yet even after calculating millions of numbers
And also we can't, it unsolvable, maybe absurd, purely because most numbers are bunch of coprimes [some are prime] , which, all the time gives us 421 loop, over and over. Only p will not give us a loop if you consider that as well..
You don't just need 1 number obviously. You need a whole infinite set of numbers or else if any one of those numbers goes to the loop then the whole set does. It's a stupid problem.
It's simple to understand why. 3x +1 will always result in an even number. You have a 50% chance to have to di idea by two after the jump, and a 50% chance to have to divide by two at least two times. This means that half the time you move up, half the time you move down. When you get an even number, you divide by two. This means that even numbers will always move down the number line. So on an odd number you have a half chance to move up, half to move down, On an even number you always move down. This means overall when you apply these rules, you have a 25% chance of moving up the number line and 75% chance to move down. The closet you get to 1, the closer the powers of 2 are. So take the biggest number you care to calculate, and watch it trend downwards until you step on a power of 2 which will cause a game end.
It’s called the collatz conjecture, the issue isn’t the loop it’s proving that extremely high numbers will eventually reach the loop (like googol type numbers)
Why do you have to prove every number when the formula is designed to set odd numbers to an even which will always result in more divisions than multiplations until the lowest point possible is reached. At any point hitting a power of 2 instantly sends you to the 4,2,1 loop. The +1 offsets the x3 from happening back to back so you can only ever have an even number after performing it. At best after dividing by 2 you get another x3.
@@Liv-0711 does 1+1=2? 2 is the answer, it is also the proof. Don't see that changing like powers of 2 disappearing after a few hundred of them just to make things complicated. It's accepting that seems to be the issue for people here.
It's conjecture until there's proof. Mathematicians have wasted entire careers attempting to find that proof and turn it into a formula. Until then, we won't be certain that every number will eventually resolve into 4.
@@technomatic6285 Hardly. That’s a tiny number in mathematics. The approximate lower bound at which this conjecture fails could be absurdly large, like other bounds in published proofs, such as Graham’s Number.
@@alexandermcclure6185 if that's supposed to be JavaScript, two things: 1. String are immutable, which means that at the end of it you will not have replaced the character. Also, you won't get an error. 2. The eval throws an error. 3. It's eval not evaluate.
@@Aki-ow9hdit's all arbitrary my man, it's not like a God came down and created the ten number system and decided how we're gonna do math. It's something we created to help us understand the world so when we don't understand our creation it doesn't matter, that's not what it was made for
You mean "convenient rules to a hypothetical problem" as with that addition it becomes unsolvable and a bait to mathematicians urge to solve random useless hypothetical problems.
@@prevailtm They're called hailstone numbers. The collatz conjecture is specifically the question of whether the hailstone numbers always reach 1 from any starting point.
The word "Problem" has seven letters The word "Seven" has Five letters The word "Five" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters The word "Four" has four letters....... No matter which word you start with it will always end up being four at last
Hippopotomonstrosesquipedaliophobia has 33 letters Thirty-tree has 10 letters Ten has 3 letters Three has five letters Five has .... Four... Gameover Try again Start Furthermore has eleven letters Eleven has six letters Six has three, Three has five... Five has four.... Gameover Try again Start Incompetence has twelve letters Twelve has six... Again Inconstituitionally has nineteen letters Nineteen has eight letters Eight has five... ..... ..... .. I tried
@@Skyluzz Also: While pneumonoultramicroscopicsilicovolcanoconiosis is the longest word in the English dictionary, an even longer word exists outside the dictionary. The extended term for “titin” has 189,819 letters, but the first 61 letters are methionylthreonylthreonylglutaminylarginyltyrosylglutamylsery So, going with titin's long form of 189,819 letters... One hundred eighty nine thousand eight hundred nineteen ... is 48 letters. Fourty Eight is 11 letters. Eleven is 6 letters. Six has 3 letters. Three... has four letters. You were never gonna win this one pal.
I will be using russian language for my display of proof against this statement the word aga meaning yeah has three letters which is tri letters and that will be the loop for russian that proves this statement wrong although there is another loop that does technically fall into your loop but differently in russian using the word Grud' has four letters Chetyre is 4 and has 6 letters Shest' is 6 and has 5 letters Pyat' is 5 and has 4 letters. . . IDK bleh I probably should be using the language's symbols
I learned about this problem in our class, just a few days back. Found it really interesting. Problems like this make me fall for mathematics and computer science.
Infinity is not a number for example if you try to add infinity and -infinity … it doesn’t exist, it’s just impossible … For example : A= 1+2+3+4+… B=1+1+… A-B= 0+1+2+… A-B =A Donc B=0 1+1+1+…=0 Where is my mistake ?
@@timonobel615 My mistake was at the begining A-B is not A, it doesn't exist even if we can write it the same way as A. There are a lot of dumb thing you can't do with infinity (that seem intuitive). For example : B = 1+1+1+1+... B= 1 + (1+1) + (1+1+1) + ... B=1+2+3+4+... B=A so A - B = 0 ? that doesn't work because A-B doesn't exist
@@anathos0369 your mistake is subtracting infinity from infinity, which is not allowed. it's like dividing by zero. you could get any number you want as a result, which is why it's just not possible. A - B ≠ A.
Yeah! And the whole reason of why it's 2 that remain is that we by the numbers choose to select odds and evens (by picking 2 as devisor) and then flipling the odds and evens with 1, while 3 preserves odds and evens but switches base number to include another prime number go dispose of (turn into a power of 2 by flipping it with a 1)
You can't, we don't have a proof that it would happen for all the numbers or we don't have a proof there isn't another sequence then 4-2-1 that loops into itself.
I have another dangerous one : take any integer, and remove one, if it's greater than 0, repeat, and eventually you will get to 0 no matter what number you pick 😅
You're yet another person who looks at this conjecture and thinks wrongly that it's simple and obvious. As of yet no way to prove that there are no sequences of entirely odd numbers where f(n+1) = (3 x f(n)+1)/2, never mind the other cases which might grow to infinity or stabilize into a loop other then 4-2-1. (edit: error correction.)
@@chakatfirepaw I get your point. However, there are more conjecture we can't prove than those that we can. For a math expert, this is delicious, but for those who are not blessed with that kind of mind, it just seems banale, unless there is a real life implication
@@tntg5 That it seems so simple and obvious is what makes it dangerous: It's a trap that generates a lot of wasted time as mathematicians keep looking over the same ground.
@@tntg5 "There are more conjectures that we can't prove than those that we can" I can't imagine how you would even begin to show that this statement is true.
The problem is in proving that you always end up in the 4-2-1 loop. Every number ever tried has done so, but no one has been able to prove that every number does. The problem is dangerous to young mathematicians. It's apparently very simple, but it defies analysis with every tool known to mathematics. It's enticing, almost impossible to avoid thinking about for a mathematician, and no real progress on it has ever been made. Mathematicians have wasted years of their careers on it, and young mathematicians are routinely warned not to work on it.
@@benkelly2024 To me it sounds similar to this phenomenon : if you take a solved rubik's cube and repeat the same sequence of moves over and over, no matter how long/chaotic that sequence is, it will cycle back to the solved state sooner or later. 3x+1 sounds like cycling over and over and over again until you end up on some power of 2 (the "solved state"). All roads lead to Rome, so you'll get there sooner or later (well, I can't prove it, but that's what it looks like).
@@MisterL777 A rubik's cube has only finitely many states it can be in, so ultimately it has to cycle through a loop as you describe. But there are an infinite number of integers - how can we know that there isn't some starting number that on average just keeps getting bigger?
@@benkelly2024 You can't I guess. But instictively that seems impossible to me. Because the numbers bringing the whole thing down are also infinite and can't be avoided forever. You'll bump into them at some point or another. A random even number has 50% chance to also give an even number once divided. An odd number through the 3X+1 machine will always give an even number. If you alternate the 2 it grows. But how long can you do that? Not forever I guess, statistically. One force is stronger than the other by design. Growth has to win forever. Anti growth just has to win once.
Proof. The “3x +1” step can only be applied once, never twice or more consecutively. However, the /2 step can be used more than once in a row, and 50% of even numbers are able to be divided by 2 at least twice in a row. Notice that 50% of all numbers are odd as well. As 4x > 3x+1 (4x because you are dividing by 2 twice) for most positive values of x, this means that after testing enough numbers, the number will generally decrease until it reaches 1. The reason why the loop goes like “4-2-1” is because 1-3 are the values where 4x < 3x+1, with the exception of 3 as since 3 is odd, the loop begins from 10.
Zero represents something that once existed but is no longer observable. You count it as both odd and even. When the rule set for even is applied it's still zero, but when the ruleset for odd is applied it becomes 1 and then starts the loop.
I recently wrote a thesis on this. I figured out that the amount of iterations ut takes for a number to get into this loop actually has an asymptote approximating the 4th root of 750000x plus 180. If someone who is actually smart would like to look into this further i would be happy to send you the report and code i used to get to this conclusion.
The thing is, it really doesn't matter what the number is... the reality is that they will end up there sooner or later, unavoidably. The number always gets smaller more than it gets bigger.
I would be interested to see a graphical representation of random number to which this algorithm is applied to. A web application would be wonderful. Hmm... maybe there is already a service like this. But I don't know.
Its not complex, it is incomplete. Two premisies without a conclusion isn't an arguement, which is what mathemaricians call a "problem". No solution = no conclusion. Not an arguement = not a Problem.
Smartest mathematician : "This is the biggest problem in math" Optimist : "Nah, it still can be solved" Businessman : "Nice problem to solve, but at what cost?" Non-mathematicians : "Wait, is that a problem?" Me : "What happened?"
I don't know how to express this correctly, but the conjecture is correct because A) every whole number must be ether odd OR even and the number line is (reasonably) divided evenly between odds and evens. B) 3n+1 ~ 150% of n(1) which always becomes even and is n(2)/2 which produces 75% of n(1), and n/2 = 50% of n. But that means for the sequence to reach a line never descending you would need odds = 2n, to balance the evens becoming 1/2n. And odds would have to > 2n for the values to have a chance of ascending. In other words, 3/4n < 1n and given the constraints (A), cannot sustain or ascend, but inevitably descends to the trivial loop (4,2,1).
@N0Xa880iUL true, but a power of two multiple of some number already known to end does still mean an eventual end is coming, though not necessarily a very fast one.
According to this question, I say If asked what are the chances of gerting a head, it is obvioisly 1/2 as it is a fair toss If asked chances of waking up on Monday with heads it 1/3, Chances of waking up on Monday with Tails is 1/3 and Chances of waking up on Tuesday with tales is 1/3 If asked chances of waking up on Monday, it is 2/3
I believe you are referring to “Living Death”which was first coined by famous warrior poet, Ken Shamrock. Props on spreading awareness on this tragically little known condition.
I mean, you can achieve the same result with just if odd +1 if even :2, u will end with a 2 1 2 1 2 1 loop. Simply because you divide only when the number is even and if it's not you make it even, so u will always end up with 1 which starts the loop
I looked up unsolved math equations in math yesterday and found this equation, today when I was scrolling I saw this and i’ve never ever seen a short to do with math
Okay, my excel sheet is currently still working, but it's big and filled with rows upon rows with pretty colours (set to recognize the pattern 4, 2 and 1) This feels solvable. So I see your point :)
The thing about this problem is it looks deceptively easy because it has been simplified to its most basic form. But if you try to prove the conjecture you need to use powers of 2 and 3 and numbers in different bases(base 2 is a good approach but some try with base 3 and base 6). You also need to overcome the problem of undecidability of the equation due to self-reference, try to predict how many times the next even number can be divided by 2 and somehow find a pattern that links 3x+1 and x/2 together to prove that this works for all numbers or not.
In binary, the problem is n+1+(bitshift left)n, guaranteeing at least 1 trailing 0. Div2 is the same as bitshift right or dropping trailing 0s. In this reference frame, the conjecture says the amount of significant figures approaches 1 instead of infinity
@@austinwoodall5423 You can effectively know how many times you can divide by 2 by counting the trailing 0s. You can also predict the number of times you will only be able to divide once after 3x+1 before being able to divide multiple times by counting the trailing 1s. But as of now I didn’t find a way to predict how many times you can divide by 2 when you can do it more than once. I think if we can find that we could change the self-referencing condition to an equation and from there prove that it applies to all numbers or not.
No! Not different bases, only base 2. Every even number can be written as b(2)^k, so you just divide by 2 until you reach an odd number. If you write out b(2)^k for, let’s say 5, {5, 10, 20, 40, 80…} the numbers that can get to this sequence form another sequence {3, 13, 53, 213…} which has a definition a(n) = 4*a(n-1) + 1. The crazy part is that this pattern holds true for ANY sequence b(2)^n
@@LAM1895 we know that any number that can be expressed as 2⁰+2²+2⁴... Is one step from a power of two. I forget how the maths work to build the tree, but my guess is we can say something about the sig figs, bit density, or maybe make it a geometry problem.
I actually have an intuitive sense that this can be generalized pretty easily (not a mathematician so this may be silly). For any odd number, multiplying it by 3 will always give you another odd number. Adding 1 will always then give you an even number, and half of that is always the whole integer you would get by rounding up from 1.5 times the odd number you started with. Half of all possible whole positive numbers are even, which means this process will immediately halve that number again, then the rules start again (and again). If this is generalized across all possible numbers, both of these possibilities (1. halving an even number or 2. multiplying an odd number by 1.5 and rounding up) seem to my brain to have a statistically 50/50 chance of resulting in a number that is itself even, and the downward change of *halving* an even number will be larger on average than the upward change of multiplying by ~1.5, assuming they're both happening about half of the time. This seems like it should lead to a downward trend overall, and landing on any power of 2 will immediately resolve into the 4-2-1 loop. Assuming this is all true and the pattern indeed tends downwards overall, this process gravitates towards lower numbers where there is a higher likelihood of landing on a power of two (compared with arbitrarily high numbers). And the only possible exception is if a power of 2 never occurs, but you will cycle through new numbers indefinitely until you get that power of 2 because multiplying by 3 then adding one and dividing by two will never be able to give you a recurring cycle of the same odd numbers looping forever, it will by definition constantly iterate into new numbers leaving it as a statistical impossibility to never encounter any power of 2. Am I thinking straight?
There might be a longer loop at values higher than have been explored or there might be a sequence of numbers where (3n+1)/2 is always odd and goes to infinity.
This is the intuitive approach to the problem and it isn’t necessarily wrong, it’s just really hard to prove that this conjecture is true for ALL natural numbers. Kind of like how hand sanitizer only kills 99.999% of germs
chakatfirepaw is right, even though you're showing that it will on average trend downwards, that doesn't prove that every sequence will, it just suggests that most will. If there were some 5 or 500 numbers way up there that managed to lead to each other, it would be a loop that never gets to 1, even though most numbers do trend downwards. Likewise, if there was a number built right to keep trending upwards, just saying "cycle through new numbers indefinitely until you get that power of 2" doesn't prove that you ever will get to a power of 2. It's a good start though! That's why this problem is (somewhat clickbaitingly) called "dangerous," because it's so easy to start playing around with it when you may never finish.
@@chakatfirepaw It's very easy to prove that you can't have an infinite sequence where (3n+1)/2 is always odd, so any starting number will always reach at least one number divisible by 4. But that doesn't help prove the overall conjecture.
For anyone struggling, rebase to binary and reframe as n+2n+1. Div2 is same as bit shift right. You're dropping trailing 0s in this reference frame. The conjecture now says the amount of binary significant figures approaches 1, not infinity.
yeah... but it's super easy to imagine a power of two so huge that if you did one division by two (i.e. binary shift right) every fempto-second for the entire length of the universe's existence, you would be nowhere close to getting the result to divide down close to 1. And the vast majority of counting numbers are bigger than that number... So this problem seems utterly foolish for the majority of counting numbers.
“3n+1” Don't waste your time tryna solve it guys, one comment said he tried it, but when he looked at his window in the car he found himself off the highway.
I mean, that does make sense. Multiplying an odd number by 3 will usually result in an odd number, adding one will make it even. And multiplying by three and then dividing by 2 kind of “shuffles” the number. Putting it into a new set of numbers divisible by 2. Idk how to describe it but you get the point
@@stevengordon3271 yeah, I meant like multiplying by three makes it a new set of numbers divisible by 3. If you divide any even number by 2 enough times, it’ll become an odd number. Multiplying by 3 shuffles the “chain” of numbers that you’re on. Like I said, I can’t explain very well
@@joshuaohuka7719 I mean, not quite. They’re trying to understand how and why specifically it works. I’m struggling to describe how it works generally speaking
Of course it's true. Every single odd number multipied by three is odd, plus one will always make it even, meaning there can never been more than a single odd number consecutively. Whereas half of all even numbers, when divided by two, will remain even. No matter how large a number is, it's only a matter of time before the number of divisions out-weighs the number of multiplications by enough to reduce it to a power of 2. Any power of 2 will then divide all the way down to 1, starting the loop.
Yes, half of even numbers will be even. But this is just a probability. So there is a chance for an alternating sequence "odd even odd even" for example 11 -> 34 -> 17 -> 52. This alternating sequence breaks in the next step ( 52 -> 26 ) 1) Can you prove that any alternating sequence will definitely break at some point? 2) If yes: Can you also prove that any breaking alternating sequence will not result in a number which was in the sequence before? Because that would create a loop other than 4 -> 2 -> 1
@@Desam1000 Can you prove there isn't a red sock hidden in the amazon rainforest somewhere? The question to the question is why should you care to even attempt to prove it? You can just type this into a computer and let it run for a couple of years until we get a number so high that we can reasonably tell it won't or that it will. I don't see a reason to care either way. I've seen others say it could lead us to other ways of thinking but so can playing with feces, it doesn't mean it's worth hearing about.
The loop is only a distraction of the underlying end of divisbility , or to sum it up: • This process always terminates because any number divisible by 2 will eventually reduce to either a prime number or a product with odd factors. The plus one gets it through the divisibility barrier of the odd divisors or prime divisors and therefore it will end up in the mechanics of the loop • A prime number will not be “safe” because of the multiplication plus one • A composite number with an odd divisor Or to phrase it differently the key insight here is: The 3n + 1 rule, combined with division by 2, effectively forces all numbers into a deterministic reduction process. The “+1” ensures no odd number, whether prime or composite, can escape being converted into an even number, which then repeatedly divides down until it enters the loop (4 → 2 → 1). Thus, the loop is not the mystery-it’s the endpoint of a system designed to eliminate divisibility barriers. Another helpful approach to understand the eliminating of Divisors is the point that 3n+1 can never become a prime number , ergo all the exorbitant high numbers will finally end up getting broken down
A number of people are asking why this problem is “dangerous”.
It’s described as dangerous because its difficulty has defeated the world’s greatest mathematical minds for generations.
Paul Erdos, a famous mathematician, said, "Mathematics is not yet ripe enough for such questions." Jeffrey Lagarias called it "an extraordinarily difficult problem, completely out of reach of present day mathematics".
The problem is so maddeningly difficult, mathematicians are warned to stay away from it.
Watch the full video for more context: ruclips.net/video/094y1Z2wpJg/видео.html
I'll never look at this problem the same after reading this!
So, even the greatest of mathematicians cannot exactly solve it.
But something I don't get, is solving the problem. What even is the problem in the first place?
@@uauausuuahshauaiausuuaususuis there number that doesn't end up in 4 2 1 loop??
Maybe because it’s not a problem to solve… what or why would be a reason it would need a solution? Or more so what would actually be achieved if it were or any of the other unsolved math problems… are these problems preventing some sort of breakthrough??? Just seems like number puzzles without purpose.
@@Overkillutube you remember the time people think infinity is un-slove-able and questions the others why they research infinity? The anz is told by Newton's maths.
Mathamaticians : "why cant i sleep "
3x + 1
Ok so basically what you do is this: 3x + 1 = 90.
Now this is solvable your welcome, go to bed and stop being a mathematical lunatic.
@@IWntGhost2YT did you watch the video or not?
@@IWntGhost2YT but 89/3 doesn't give any whole number
now I am in tension!!
@@IWntGhost2YT
Tell me you didn't watch the video without telling me you didn't watch the video.
@@IWntGhost2YTi think you dn even understand the context
I was wondering why this was the most dangerous problem. So got out my calculator and tried solving it, and my car ran off the highway. Point taken.
This turned into 0 to 200 real quick.
@@kur0nafc more like 100 to 0 too quickly
@@garlicbreadstick404 100 to 50 to 25 ... to 4 to 2 to 1
@@exusiai2336 OP crashed into the railing lmao
Agree
No Derek, I did not choose 7
Why does this read exactly as if you would have written “No, Karen, I did not choose 7” 😂
...promise!
i bet you picked 37
@@multiarray2320 what if he picked 69 or 42
The point is, it doesn’t matter what number you choose. If it’s a positive integer it will always end up in the 4,2,1 loop.
"Pick a number."
"One-"
"Seven! Good choice."
"Pick a number"
4
"7, good choice"
Oh ok
Well to be fair if you'd picked 4 it would have been a terrible video
That would be 1 actually
1 Corinthians 15 KJV ✝️🩸
1-4
🎺🌥️ Jesus is coming soon.
@@AlexanderBrown77 Jesus said he would return before the people who he was talking to all died. He didn't. He's not coming back. You Belive in a 2000 year old fairy tale and are so stupid you don't even Belive Jesus own words about when he will return.
@@AlexanderBrown77 nah u be trippin
I did this math problem and literally died guys its really dangerous
Wifi connections' gotta be pretty good in heaven
RIP 🙏
That sucks, but yada yada yada, Schrodinger’s cat, yada yada yada… and you can try again!
@@PurePain_1😂🤣🤣
well u proved that 3+1 = 4
u also proved that u can split 4 two times
xD
What normal people are afraid of:
Spiders, death, jumpscares, dark
What mathematicians are afraid of:
3x+1
And programmers.
No, they're afraid of x -> {mod(x,2) = 0: x/2, 3x+1}.
@@jagobabarron5501 Programmers are afraid of this;
Roses are red;
Violets are blue;
*Expected ";"*
*On line 2.*
@@alexandermcclure6185bro 💀💀😂
that thing? that thing scares me
Everyone needs this, so here it is :
Guys, this is called the 'Collatz Conjecture' or '3x+1 Problem' or
'Syracuse problem'
Ok why is this dangerous?
@@ApoorvaS-yt4iq It's only dangerous in the sense that it's dangerous to your career as a mathmetician. Some of the greatest mathematical minds have tried to solve it for years and failed.
It's basically a warning given to any budding mathmeticians - stay away from the Collatz Conjecture if you want to get anywhere. But it is a very alluring problem because it's utterly fascinating and surely there has to be a solution to it one way or another.
@@AxelSpinnet yeah I get it...but what is there to solve can't we just take it as a rule...like for example 4 is a blackhole number
@@ApoorvaS-yt4iq cannot prove that this will be the case for any natural number, although repeated checks confirm this. They cannot prove it in general terms, although this problem is thousands of years old
@@AxelSpinnet I don't see why it should have a solution..? It's just an effect surely.
I was still deciding to choose a number when he popped up saying
"7, good choice" 😭
yeah lol
I picked 1 and was immediately stuck in the loop
Same
Yeah
Same
It's no joke. I got stuck in the 4-2-1 loop for years once.
Story time?
What
Stay strong. The time to get out will be come
howd you get out?
damn... that's crazy
Crazy? I was crazy once
You try to impress someone at a party and they pick 1 right out the gate…
😂😂😂
"pick a number between 1 and 10 but not 1, 2 or 4."
You just made me laugh out loud.
900th like 🕶️✌️
"nahh, you win"
Nah bro how did he know that i choose 7😭
as a mathematician, that's one of the conjectures I was warned about not to waste any mind power on
What is the conjecture name
@@ClipSnacks collatz conjecture
@@hickoryst.6961 if someone is able to come up with an abstract formulation that can prove that the conjecture is right, this person has just come up with a new branch of the mathematics of the natural numbers.
@@hickoryst.6961prove it then mr genius
Collatz conjecture. Always ends in 1. No proof exists that it is true for all numbers, but we haven't found a number that doesn't.
Alternative title : What overthinking does to a mf
I'll assume mf stands for math fanatic
@@TwiceEvery14Days not wrong 👍
Mezzo Forte
@@Molds_s Master Fart
Mister Fancy
I never thought I’d hear someone say a math problem is dangerous 💀
Why does he say this is dangerous?
@@logixindie looks simple, so you spend a lot of time on it, only to make no progress
Yo you wanna know what’s more dangerous? My sanity for YT moderation deleting my good comments
@@trinityy-7That does not make something dangerous. You are illogical.
@@IWntGhost2YT one of your comments says "the problem is get a life". that is not a good comment
My question: What's the problem?
Right? Sounds like it's a done deal already.
Watch the full video.
The answer is obviously 37
In a row??
Rule 34
But that just becomes 7, which we already know from the video goes to 4-2-1
@@leftylizard9085 r/woosh
Veritasium has another video about how people see 37 as the most random number.
LOL
"Pick a number"
Huh
"7? Good choice"
Wait what
You are being rescued. Please, do not resist.
7 is the most common ”random” number people think of when asked to think of a random number between 1 and 10
الله اخبرنا بهذه المعادلة والى الان لم يستطيع العلماء حلها.
٣-٤/ ٥-٦ / ٧-٨ / +٩
لا يوجد حل
Thala for a reason
@@AritraMazumdar-th4fi lmao
He calls it dangerous because it's considered to currently be pretty much unsolvable and yet many mathematicians quickly find interest in this problem and start trying to find a solution which can result in them essentially wasting years of their careers.
A "solution" to what? What are we even trying to find?
"Solution?" Do you mean an int that boils down to something other than the 8421 loop?
@@hardware64 one number that this doesnt apply
@@hardware64it's explained in the linked full length video
@@hardware64trying to prove that we end up in a loop for any natural number
Zero: "Hold my beer"
Derek : This is the most dangerous prob....
VSauce : Wait !!!! OR IS IT ?
That's what I thought in the first place.
More like Vsauce2
"Michael?? Where did you come from?"
"Am I here?.. Where is 'here' anyway? Or... Was I always here. Truth is, I've always been here... Because, wherever I am, that is... Here... But wat if-"
"Oh boy, here we go." 🤦
This problem has been known to murder mathmeticians...
Derek vs Vsause would be World War III
"one must imagine sisyphus doing math"
@@6uis1948 And yet, like sisyphus, it persists.
Persistence strong enough to form a prime soul.
@@6uis1948 sisyphus is the embodiment of will power and you clearly have no will power
@@6uis1948 Wait, you mean people keep making this joke? Over and over again? Even though everyone has seen it and it's pointless to post it?
Man, if only there were a word that helped relate or suggest the labors of Sisyphus. Specifically the requiring continual and often ineffective effort.
Sisyphean is the word. I'm glad to have explained this joke to you.
@@6uis1948 I bet you get upset at the sun for being bright.
So dangerous. The numbers 4, 2, and 1 had almost killed me before...
I have a math problem too.
If the number is odd, I multiply by 2.
If the number is even, I divided by 2.
I will have infinity loop.
Mathematicians: Omg he is a genius please give your name to this amazing problem and we will waste 60 yrs of our lives trying to solve it.
My man is a genius
But seriously; isn't this the exact same thing, only less convoluted? Guess I'm too stupid to spot the difference.
@@datruommi not the exact same thing, because this one is simply taking the inverse operation everytime (is like the "take the jacket" and "put the jacket" loop from karate kid), whereas 3x+1 and x÷2 are not that directly conected
@@datruommi no, because this one is easy. If the number is odd, it will be multiplied by 2, giving an even result and will therefore be divided back to the original number. If the number is even, it will be divided by 2 until it is an odd number, which will lead to the loop mentioned above.
That will be true for all N belonging to the natural numbers. However, can you ensure that all the natural numbers in the 3x+1 problem lead to the 4 ➝ 2 ➝ 1 loop? Because obviously you can't just brute force an infinite amount of numbers in a finite period of time.
As a problem myself I find this really fascinating
💀
As a dangerous person, I find this problem very relatable
You expected something different?
It's fascinating to me too. If it is true, or not true, then why? If a mathematician could prove it's always true or find its not always true, then the underlying why would be answered
@@elijahknox4421brother is only dangerous to children
"Pick a number"
Let me thi-
"7? Good choice"
🗿
Exact
It could be any even or odd number. Just listen.
@@BillyBob-u5n No, it couldn't be "any odd or even number" when the only options listed were 1 to 9.
@@Fete_Fatale those werent the only options though, he just said "pick a number", not "pick a number between 1 and 9"
@@rarewubbox6413 You clearly didn't watch the video.
It's so dangerous that I am now scared to leave the house.
The goal is to find a number which does not end up in a 4,2,1 loop. But the problem is that there's no such number found yet even after calculating millions of numbers
And also we can't, it unsolvable, maybe absurd, purely because most numbers are bunch of coprimes [some are prime] , which, all the time gives us 421 loop, over and over. Only p will not give us a loop if you consider that as well..
Most likely it is an infinitely continuous fractal. Good luck calculating infinity.
You don't just need 1 number obviously. You need a whole infinite set of numbers or else if any one of those numbers goes to the loop then the whole set does. It's a stupid problem.
It's simple to understand why.
3x +1 will always result in an even number. You have a 50% chance to have to di idea by two after the jump, and a 50% chance to have to divide by two at least two times.
This means that half the time you move up, half the time you move down.
When you get an even number, you divide by two. This means that even numbers will always move down the number line.
So on an odd number you have a half chance to move up, half to move down,
On an even number you always move down. This means overall when you apply these rules, you have a 25% chance of moving up the number line and 75% chance to move down.
The closet you get to 1, the closer the powers of 2 are. So take the biggest number you care to calculate, and watch it trend downwards until you step on a power of 2 which will cause a game end.
3x+1 will not always be an even number…
It’s called the collatz conjecture, the issue isn’t the loop it’s proving that extremely high numbers will eventually reach the loop (like googol type numbers)
Just say it's true, who's gonna prove you wrong? Checkmate.
@@foxi13x You should see the knot video
Why do you have to prove every number when the formula is designed to set odd numbers to an even which will always result in more divisions than multiplations until the lowest point possible is reached.
At any point hitting a power of 2 instantly sends you to the 4,2,1 loop.
The +1 offsets the x3 from happening back to back so you can only ever have an even number after performing it. At best after dividing by 2 you get another x3.
@ajourneyb4destination558 The answer is quite intuitive. It is proving that all integers do this that is the hard part.
@@Liv-0711 does 1+1=2? 2 is the answer, it is also the proof. Don't see that changing like powers of 2 disappearing after a few hundred of them just to make things complicated.
It's accepting that seems to be the issue for people here.
Basically if it reaches a power of two, it's over.
The power of Twooooooo??!?!!
and it seems that the start to that loop can only be an even power of 2
the step before it being 5, 21, 85, 341 etc
It's conjecture until there's proof. Mathematicians have wasted entire careers attempting to find that proof and turn it into a formula. Until then, we won't be certain that every number will eventually resolve into 4.
@@bramweinreder2346why don't we just use AI or something to calculate till... let's say what... 20 quintillion numbers? isn't that sufficient proof
@@technomatic6285 Hardly. That’s a tiny number in mathematics. The approximate lower bound at which this conjecture fails could be absurdly large, like other bounds in published proofs, such as Graham’s Number.
this feels like info that's related to the prisoners having to find their own number problem.
Just pick 0, duh
I'll expect my Fields Medal in the mail shortly
it needs to be positive full number.
@@sonicwaveinfinitymiddwelle8555 then please prove zero is a negative number ;)
@@jorgefreitas5983 It is neither positive nor negative. it's nothing.
THEN IT WILL COME TO AT 1 AGAIN
@@jorgefreitas5983 zero is not negative nor positive, it is nothing.
"Doing math won't hurt you"
Math:
💀
Brain not braining
No, but it can waste your time…years of it.
let string = "Doing math won't hurt you";
for(i=0; i
@@alexandermcclure6185 if that's supposed to be JavaScript, two things:
1. String are immutable, which means that at the end of it you will not have replaced the character. Also, you won't get an error.
2. The eval throws an error.
3. It's eval not evaluate.
Mathematicians: oh my god we could never solve this hypothetical equation
Engineers: pi is 4
Actually is 3 😂
Erm
3.1415......
And an elephant is a sphere
Only moronic fake engineers say that.
No matter what number you choose,
High or low,
Even or not,
It will always come to 4, 2 ,1.
prove it! (just kidding)
Solution: Don't apply weird hypothetical rules to numbers.
Nuh uh, i will apply them
Calculus. It seems arbitrary for those new to it and would call the subject a bunch of "weird hypothetical rules."
@@Aki-ow9hdit's all arbitrary my man, it's not like a God came down and created the ten number system and decided how we're gonna do math. It's something we created to help us understand the world so when we don't understand our creation it doesn't matter, that's not what it was made for
Wheres the fun in that?
You mean "convenient rules to a hypothetical problem" as with that addition it becomes unsolvable and a bait to mathematicians urge to solve random useless hypothetical problems.
That problem is also called "Collatz Conjecture"
My school called it the hailstone series, i think it’s only a OSU specific thing tho, you can look it up
@@prevailtm They're called hailstone numbers. The collatz conjecture is specifically the question of whether the hailstone numbers always reach 1 from any starting point.
The word "Problem" has seven letters
The word "Seven" has Five letters
The word "Five" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters
The word "Four" has four letters.......
No matter which word you start with it will always end up being four at last
Hippopotomonstrosesquipedaliophobia has 33 letters
Thirty-tree has 10 letters
Ten has 3 letters
Three has five letters
Five has .... Four...
Gameover
Try again
Start
Furthermore has eleven letters
Eleven has six letters
Six has three,
Three has five...
Five has four....
Gameover
Try again
Start
Incompetence has twelve letters
Twelve has six...
Again
Inconstituitionally has nineteen letters
Nineteen has eight letters
Eight has five...
.....
.....
.. I tried
@@Skyluzz Superfragilisticipsyalidocious has 30 letters.
Thirty has 6 letters
Six has 3 letters
Three has four letters
@@Skyluzz Also:
While pneumonoultramicroscopicsilicovolcanoconiosis is the longest word in the English dictionary, an even longer word exists outside the dictionary. The extended term for “titin” has 189,819 letters, but the first 61 letters are methionylthreonylthreonylglutaminylarginyltyrosylglutamylsery
So, going with titin's long form of 189,819 letters...
One hundred eighty nine thousand eight hundred nineteen ... is 48 letters.
Fourty Eight is 11 letters.
Eleven is 6 letters.
Six has 3 letters.
Three... has four letters.
You were never gonna win this one pal.
I will be using russian language for my display of proof against this statement the word aga meaning yeah has three letters which is tri letters and that will be the loop for russian that proves this statement wrong although there is another loop that does technically fall into your loop but differently in russian using the word Grud' has four letters Chetyre is 4 and has 6 letters Shest' is 6 and has 5 letters Pyat' is 5 and has 4 letters. . . IDK bleh I probably should be using the language's symbols
@@Skyluzz you know that is the reason the no. Four is known as "A void number"
Seems simple and pretty easy tbh.
I'm going to start working on this tonight, assuming will have a solution by midnight.
It's called the collatz conjecture.
Indeed.
No, you are.
Back in my day we called it the cosmic sequence
I learned about this problem in our class, just a few days back. Found it really interesting. Problems like this make me fall for mathematics and computer science.
Why is it dangerous though?
@nykel3233 read the pinned comment
Its just binary.
Its not a comp-sci problem though?
@@chillout8185u can work on it with algorithm which is comp sci
"Pick A Number, 7? Good Choice."
Me : "??? I Didn't Choose Yet??"
0.5
It's like those times when Dora the Explorer would ask what my favorite part of the trip was and cut me off before I can say two words.
It’s like a infinite amount of roads all merging to one
"Pick a number"
"Infinity"
"Not that one"
"Ok zero"
... **sigh**
Infinity is not a number for example if you try to add infinity and -infinity … it doesn’t exist, it’s just impossible …
For example :
A= 1+2+3+4+…
B=1+1+…
A-B= 0+1+2+…
A-B =A
Donc B=0
1+1+1+…=0
Where is my mistake ?
@@anathos0369 you subracted the equation A from B instead of the solution to the equation
@@timonobel615 My mistake was at the begining A-B is not A, it doesn't exist even if we can write it the same way as A. There are a lot of dumb thing you can't do with infinity (that seem intuitive).
For example :
B = 1+1+1+1+...
B= 1 + (1+1) + (1+1+1) + ...
B=1+2+3+4+...
B=A
so A - B = 0 ?
that doesn't work because A-B doesn't exist
@@anathos0369 your mistake is subtracting infinity from infinity, which is not allowed. it's like dividing by zero. you could get any number you want as a result, which is why it's just not possible. A - B ≠ A.
infinity is a concept
It feels intuitive, because it's like you're weeding out all the other prime factors one by one until it's only 2s
Do you think it's that easy or do you think you're that smart?
@@JoanDarc1984who hurt you buddy
@@nizogos comments sections on maths videos
I'm afraid it's deeper than that
Yeah! And the whole reason of why it's 2 that remain is that we by the numbers choose to select odds and evens (by picking 2 as devisor) and then flipling the odds and evens with 1, while 3 preserves odds and evens but switches base number to include another prime number go dispose of (turn into a power of 2 by flipping it with a 1)
Didn’t even let me chose a number
“911 whats your emergency?”
“My brain isnt braining”
"Pick a number"
'Alright, two-'
"Seven? GoOD ChoICe!"
From Veritasium to Dora the explorer
the illusion of choice.
"Pick a number"
Proceeds to choose a number for us💀
Sounds completely harmless since you can predict it with 100% certainty.
solving the conjecture itself is whats difficult though, because you cant test ALL numbers, you have to find proof
You can't, we don't have a proof that it would happen for all the numbers or we don't have a proof there isn't another sequence then 4-2-1 that loops into itself.
@@Karak-_- not yet.
working on it.
don't hope I will find the answer but I'm bored at work.
@@DanielRossellSolanesyoure not him
@@tomf0olery A lot of discoveries are made by people just doing stuff for the heck of it, let him cook.
Now going to consult some medications to avoid such dangerous problem...
I have another dangerous one : take any integer, and remove one, if it's greater than 0, repeat, and eventually you will get to 0 no matter what number you pick 😅
-1
I win
You're yet another person who looks at this conjecture and thinks wrongly that it's simple and obvious. As of yet no way to prove that there are no sequences of entirely odd numbers where f(n+1) = (3 x f(n)+1)/2, never mind the other cases which might grow to infinity or stabilize into a loop other then 4-2-1.
(edit: error correction.)
@@chakatfirepaw I get your point. However, there are more conjecture we can't prove than those that we can. For a math expert, this is delicious, but for those who are not blessed with that kind of mind, it just seems banale, unless there is a real life implication
@@tntg5 That it seems so simple and obvious is what makes it dangerous: It's a trap that generates a lot of wasted time as mathematicians keep looking over the same ground.
@@tntg5 "There are more conjectures that we can't prove than those that we can"
I can't imagine how you would even begin to show that this statement is true.
"Pick a number"
8
"7, good choice"
:(
Aww lol :P
"Pick a number" se- "seven? Good choice"
MY TRAINING HAS FAILED ME.
I have a little theory... Number one is not a normal number, it is a special number. It has unique functions in math like this.
Regular people can't sleep: Count the sheep
Mathematician can't sleep: 3X + 1
its how the universe works. from start to finish it all seems random, yet theres still a complex pattern there and then it comes to an end/loop.
What is the problem exactly? And why is it dangerous. All you explained is a phenomenon
The problem is in proving that you always end up in the 4-2-1 loop. Every number ever tried has done so, but no one has been able to prove that every number does.
The problem is dangerous to young mathematicians. It's apparently very simple, but it defies analysis with every tool known to mathematics. It's enticing, almost impossible to avoid thinking about for a mathematician, and no real progress on it has ever been made. Mathematicians have wasted years of their careers on it, and young mathematicians are routinely warned not to work on it.
@@benkelly2024 To me it sounds similar to this phenomenon : if you take a solved rubik's cube and repeat the same sequence of moves over and over, no matter how long/chaotic that sequence is, it will cycle back to the solved state sooner or later.
3x+1 sounds like cycling over and over and over again until you end up on some power of 2 (the "solved state"). All roads lead to Rome, so you'll get there sooner or later (well, I can't prove it, but that's what it looks like).
@@MisterL777 A rubik's cube has only finitely many states it can be in, so ultimately it has to cycle through a loop as you describe. But there are an infinite number of integers - how can we know that there isn't some starting number that on average just keeps getting bigger?
@@benkelly2024 You can't I guess. But instictively that seems impossible to me. Because the numbers bringing the whole thing down are also infinite and can't be avoided forever. You'll bump into them at some point or another. A random even number has 50% chance to also give an even number once divided. An odd number through the 3X+1 machine will always give an even number. If you alternate the 2 it grows. But how long can you do that? Not forever I guess, statistically. One force is stronger than the other by design. Growth has to win forever. Anti growth just has to win once.
Oooooh, I like that. “Growth has to win forever. Anti Growth only has to win once.” Is that your saying originally?
I picked 2 and I INSTANTLY got looped
Proof. The “3x +1” step can only be applied once, never twice or more consecutively. However, the /2 step can be used more than once in a row, and 50% of even numbers are able to be divided by 2 at least twice in a row.
Notice that 50% of all numbers are odd as well. As 4x > 3x+1 (4x because you are dividing by 2 twice) for most positive values of x, this means that after testing enough numbers, the number will generally decrease until it reaches 1. The reason why the loop goes like “4-2-1” is because 1-3 are the values where 4x < 3x+1, with the exception of 3 as since 3 is odd, the loop begins from 10.
I think that makes perfect sense 🎉
That doesn't prove that there isnt a really large loop, especially since you're talking probability.
Get this guy a Nobel prize. He did it.
From wher does the inequality 4x > 3x+1 come from?
Really, why 4x, when we r supposed to divide twice by 2, not multiply?
@@akacaleb he made an impressive observation, but it didn't prove or disprove the conjecture.
"Pick a number"
Zero.
Positive integer.
Zero represents something that once existed but is no longer observable. You count it as both odd and even. When the rule set for even is applied it's still zero, but when the ruleset for odd is applied it becomes 1 and then starts the loop.
It was a joke guys.
It still start
@@steveh8724 no point it said had to be positive, first one i tried was -5 it loops and countless other negative numbers
It's not dangerous, it simply is what it is, the universe has no obligation to explain itself or make sense to anyone.
The real problem is:
Does this stay true for every number? Or does a starting number exist which goes to infinity?
@@Legendendear how is solving that gonna help me pick blueberries more efficiently?
@@linkbond08 if your main question is "what am I gonna use math for in real life", you're probably right, better stay away from it 😂
Universe your best universe, universe, that's what I always say.
@@Legendendear yes... This stays true for every number. That is why it is a conjecture.
The OG video is really cool! Love the channel.
Pick a number
Me: 7
Him: 7, good choice!
Me:
Most people pick 7 when asked for a random number. This has been studied.
👁👄👁
You've got 7 likes about now😂
@@jacksonwarugongo1203 life is a simulation, hence proved
You have 77 likes...
I recently wrote a thesis on this. I figured out that the amount of iterations ut takes for a number to get into this loop actually has an asymptote approximating the 4th root of 750000x plus 180. If someone who is actually smart would like to look into this further i would be happy to send you the report and code i used to get to this conclusion.
In theory, I'd be interested in it. In practice, I'd probably sit on it for years just like I've done with that differential geometry book.
Did you prove it for almost all natural numbers? Or did you run a code and extrapolated it?
@@abdillahahmad7025 no I ran a code up until like a billion
The thing is, it really doesn't matter what the number is... the reality is that they will end up there sooner or later, unavoidably. The number always gets smaller more than it gets bigger.
I would be interested to see a graphical representation of random number to which this algorithm is applied to. A web application would be wonderful. Hmm... maybe there is already a service like this. But I don't know.
"Choose a number. 7? Good choice."
Me: "But I wanted 4."
Really getting the jump on the “problem” aspect here, right to the point
Lol I picked 4 and got stuck in the “4, 2, 1” loop
Super funny how 3x + 1 seems so simple but the more you investigate the more realize how complex it is!
Its not complex, it is incomplete. Two premisies without a conclusion isn't an arguement, which is what mathemaricians call a "problem". No solution = no conclusion. Not an arguement = not a Problem.
wtf is bro yapping
Stop the yap
@@ellea3344 I love your way of viewing the problem. What I meant by complex was that it is much more than just an equation.
It would certainly be complex if x is in C
Smartest mathematician : "This is the biggest problem in math"
Optimist : "Nah, it still can be solved"
Businessman : "Nice problem to solve, but at what cost?"
Non-mathematicians : "Wait, is that a problem?"
Me : "What happened?"
Me: "How did I get gonorrhea!"
"Dormamu, I've come to arithmetize"
Lol🤣 dr. Strange upcoming movie be like:
I don't know how to express this correctly, but the conjecture is correct because A) every whole number must be ether odd OR even and the number line is (reasonably) divided evenly between odds and evens. B) 3n+1 ~ 150% of n(1) which always becomes even and is n(2)/2 which produces 75% of n(1), and n/2 = 50% of n. But that means for the sequence to reach a line never descending you would need odds = 2n, to balance the evens becoming 1/2n. And odds would have to > 2n for the values to have a chance of ascending. In other words, 3/4n < 1n and given the constraints (A), cannot sustain or ascend, but inevitably descends to the trivial loop (4,2,1).
I remember the long version of this video that you uploaded, that was your first video I ever saw…
Immediately subscribed 🎉
*no one
Mathematician*
"We gonna find angles of the circle "
?
This is calculable using formulas thought in the fourth year of high school what is your point here?
Once you get a power of 2 it's over.
Once you get a power of two multiple of a number that also goes to 4-2-1, it's also over then too
@@leftylizard9085 Not quite. A multiple of some power of 2 doesn't guarantee a swift end. But a pure power of 2 is game over.
@N0Xa880iUL true, but a power of two multiple of some number already known to end does still mean an eventual end is coming, though not necessarily a very fast one.
@@leftylizard9085 Exactly
@@leftylizard9085not necessarily, you just made that up.
According to this question, I say
If asked what are the chances of gerting a head, it is obvioisly 1/2 as it is a fair toss
If asked chances of waking up on Monday with heads it 1/3, Chances of waking up on Monday with Tails is 1/3 and Chances of waking up on Tuesday with tales is 1/3
If asked chances of waking up on Monday, it is 2/3
Ok we need full video on the "dangerous" part.
It's been since a while.
this is literally just a clip of the main video, look up "veritasium collatz conjecture"
The full video is called "The Simplest Math Problem No One Can Solve - Collatz Conjecture"
"we're stuck in a loop"
Programmers: story of my life!
break;
@@divinecomedian2 IMPOSSIBLE!
This math problem is so dangerous it killed me until I was dead. I am here to spread awareness.
I believe you are referring to “Living Death”which was first coined by famous warrior poet, Ken Shamrock.
Props on spreading awareness on this tragically little known condition.
I feel like this is part of a Thesis about how Percentage like going Lower rather than Higher.
"Pick a number"
Me:5
"Seven?Good choice!"
Me:😢
This must be how they calculate my bank account.
The final step: if it's 1, multiply by 0. And if it's 0, do nothing.
If I ever get to create a math problem for students, I'll find a way to add this somehow.
I mean, you can achieve the same result with just if odd +1 if even :2, u will end with a 2 1 2 1 2 1 loop. Simply because you divide only when the number is even and if it's not you make it even, so u will always end up with 1 which starts the loop
Except this is easy to prove, hence not "dangerous"
A problem in MATHEMATICS is already dangerous enough for me😅
I looked up unsolved math equations in math yesterday and found this equation, today when I was scrolling I saw this and i’ve never ever seen a short to do with math
@peni2326 They're listening 😂
They're always listening 😭
Okay, my excel sheet is currently still working, but it's big and filled with rows upon rows with pretty colours (set to recognize the pattern 4, 2 and 1)
This feels solvable. So I see your point :)
The thing about this problem is it looks deceptively easy because it has been simplified to its most basic form. But if you try to prove the conjecture you need to use powers of 2 and 3 and numbers in different bases(base 2 is a good approach but some try with base 3 and base 6). You also need to overcome the problem of undecidability of the equation due to self-reference, try to predict how many times the next even number can be divided by 2 and somehow find a pattern that links 3x+1 and x/2 together to prove that this works for all numbers or not.
In binary, the problem is n+1+(bitshift left)n, guaranteeing at least 1 trailing 0. Div2 is the same as bitshift right or dropping trailing 0s. In this reference frame, the conjecture says the amount of significant figures approaches 1 instead of infinity
@@austinwoodall5423 You can effectively know how many times you can divide by 2 by counting the trailing 0s. You can also predict the number of times you will only be able to divide once after 3x+1 before being able to divide multiple times by counting the trailing 1s. But as of now I didn’t find a way to predict how many times you can divide by 2 when you can do it more than once. I think if we can find that we could change the self-referencing condition to an equation and from there prove that it applies to all numbers or not.
No! Not different bases, only base 2. Every even number can be written as b(2)^k, so you just divide by 2 until you reach an odd number. If you write out b(2)^k for, let’s say 5, {5, 10, 20, 40, 80…} the numbers that can get to this sequence form another sequence {3, 13, 53, 213…} which has a definition a(n) = 4*a(n-1) + 1. The crazy part is that this pattern holds true for ANY sequence b(2)^n
@@LAM1895 we know that any number that can be expressed as 2⁰+2²+2⁴... Is one step from a power of two. I forget how the maths work to build the tree, but my guess is we can say something about the sig figs, bit density, or maybe make it a geometry problem.
I actually have an intuitive sense that this can be generalized pretty easily (not a mathematician so this may be silly). For any odd number, multiplying it by 3 will always give you another odd number. Adding 1 will always then give you an even number, and half of that is always the whole integer you would get by rounding up from 1.5 times the odd number you started with. Half of all possible whole positive numbers are even, which means this process will immediately halve that number again, then the rules start again (and again). If this is generalized across all possible numbers, both of these possibilities (1. halving an even number or 2. multiplying an odd number by 1.5 and rounding up) seem to my brain to have a statistically 50/50 chance of resulting in a number that is itself even, and the downward change of *halving* an even number will be larger on average than the upward change of multiplying by ~1.5, assuming they're both happening about half of the time. This seems like it should lead to a downward trend overall, and landing on any power of 2 will immediately resolve into the 4-2-1 loop. Assuming this is all true and the pattern indeed tends downwards overall, this process gravitates towards lower numbers where there is a higher likelihood of landing on a power of two (compared with arbitrarily high numbers). And the only possible exception is if a power of 2 never occurs, but you will cycle through new numbers indefinitely until you get that power of 2 because multiplying by 3 then adding one and dividing by two will never be able to give you a recurring cycle of the same odd numbers looping forever, it will by definition constantly iterate into new numbers leaving it as a statistical impossibility to never encounter any power of 2. Am I thinking straight?
There might be a longer loop at values higher than have been explored or there might be a sequence of numbers where (3n+1)/2 is always odd and goes to infinity.
This is the intuitive approach to the problem and it isn’t necessarily wrong, it’s just really hard to prove that this conjecture is true for ALL natural numbers. Kind of like how hand sanitizer only kills 99.999% of germs
chakatfirepaw
is right, even though you're showing that it will on average trend downwards, that doesn't prove that every sequence will, it just suggests that most will. If there were some 5 or 500 numbers way up there that managed to lead to each other, it would be a loop that never gets to 1, even though most numbers do trend downwards.
Likewise, if there was a number built right to keep trending upwards, just saying "cycle through new numbers indefinitely until you get that power of 2" doesn't prove that you ever will get to a power of 2. It's a good start though! That's why this problem is (somewhat clickbaitingly) called "dangerous," because it's so easy to start playing around with it when you may never finish.
@@chakatfirepaw It's very easy to prove that you can't have an infinite sequence where (3n+1)/2 is always odd, so any starting number will always reach at least one number divisible by 4. But that doesn't help prove the overall conjecture.
For anyone struggling, rebase to binary and reframe as n+2n+1. Div2 is same as bit shift right. You're dropping trailing 0s in this reference frame. The conjecture now says the amount of binary significant figures approaches 1, not infinity.
How would that help anyone struggling? If they struggled to understand 3x+1 and x/2, I don't think they understand binary or limits💀.
yeah... but it's super easy to imagine a power of two so huge that if you did one division by two (i.e. binary shift right) every fempto-second for the entire length of the universe's existence, you would be nowhere close to getting the result to divide down close to 1. And the vast majority of counting numbers are bigger than that number... So this problem seems utterly foolish for the majority of counting numbers.
@@andytraiger4079 Time has nothing to do with the problem, it's about understanding properties of numbers.
@@andytraiger4079wow😮In English? 😂
exactly, the only thing to demonstrate is rarefaction of "ones" in the number by the 3x+1 operation
"pick a number"
"∞"
Multiplying and dividing fractions by pi is a kicker
“3n+1”
Don't waste your time tryna solve it guys, one comment said he tried it, but when he looked at his window in the car he found himself off the highway.
I’m a little confused. What exactly is there to solve? What are we trying to find? Are we trying to model an equation for the situation?
@@musabawad7230 I guess a number which doesn't fit the code or something that proves all natural numbers fit this code
No Maths were harmed in the making of this video.
No mathematicians were harmed😅
I mean, that does make sense. Multiplying an odd number by 3 will usually result in an odd number, adding one will make it even. And multiplying by three and then dividing by 2 kind of “shuffles” the number. Putting it into a new set of numbers divisible by 2. Idk how to describe it but you get the point
Dividing by 2 does not make a number even.
@@stevengordon3271 yeah, I meant like multiplying by three makes it a new set of numbers divisible by 3. If you divide any even number by 2 enough times, it’ll become an odd number. Multiplying by 3 shuffles the “chain” of numbers that you’re on. Like I said, I can’t explain very well
That description is literally what mathematicians are trying to figure out...
@@joshuaohuka7719 I mean, not quite. They’re trying to understand how and why specifically it works. I’m struggling to describe how it works generally speaking
@@genericsidecharacter8915 The mathematical definition given in the video is already the best way to describe precisely how it works.
I like how he chose 7 for us, knowing that 7 is the least random, randomly chosen number.
You, my friend need to learn stats
@@M.K__ You need to learn stats. 7 is statistically the least random randomly chosen number. Followed closely by 3. Don't ☝🤓 if you aren't a 🤓.
@@infinitesalsa4422 you need to recheck your sources blud 🤓. 7 is the most randomly chosen number between 1 to 10
@@M.K__ That's literally what I said are you actually autistic????
@@M.K__ It's the opposite actually, 7 is the "least random" random number as it has the most bias to be chosen, hence the least randomness.
Of course it's true. Every single odd number multipied by three is odd, plus one will always make it even, meaning there can never been more than a single odd number consecutively. Whereas half of all even numbers, when divided by two, will remain even. No matter how large a number is, it's only a matter of time before the number of divisions out-weighs the number of multiplications by enough to reduce it to a power of 2. Any power of 2 will then divide all the way down to 1, starting the loop.
It’s true that on average the function should decrease, this doesn’t imply that there isn’t another cycle somewhere
Yes, half of even numbers will be even. But this is just a probability.
So there is a chance for an alternating sequence "odd even odd even" for example 11 -> 34 -> 17 -> 52.
This alternating sequence breaks in the next step ( 52 -> 26 )
1) Can you prove that any alternating sequence will definitely break at some point?
2) If yes: Can you also prove that any breaking alternating sequence will not result in a number which was in the sequence before? Because that would create a loop other than 4 -> 2 -> 1
@@Desam1000 Can you prove there isn't a red sock hidden in the amazon rainforest somewhere? The question to the question is why should you care to even attempt to prove it?
You can just type this into a computer and let it run for a couple of years until we get a number so high that we can reasonably tell it won't or that it will.
I don't see a reason to care either way.
I've seen others say it could lead us to other ways of thinking but so can playing with feces, it doesn't mean it's worth hearing about.
See how easy that was to solve? Why are mathematicians so freaked out by it, the theory is 100% correct so why do people still try to prove it wrong
@@deputyhobbs9683 Man didn't even solve the problem lol
Ok, but what's the problem? You never explained what the question being asked is...
Ur right. He didnt. The problem is to get a formal proof that al numbers will end on. 1
the question is does every number reach 1
It's a conjecture. The question is "can you prove it?"
Yeah, it's just a part of the longer video linked
The problem is, get a life
The loop is only a distraction of the underlying end of divisbility , or to sum it up:
• This process always terminates because any number divisible by 2 will eventually reduce to either a prime number or a product with odd factors. The plus one gets it through the divisibility barrier of the odd divisors or prime divisors and therefore it will end up in the mechanics of the loop
• A prime number will not be “safe” because of the multiplication plus one
• A composite number with an odd divisor
Or to phrase it differently the key insight here is:
The 3n + 1 rule, combined with division by 2, effectively forces all numbers into a deterministic reduction process. The “+1” ensures no odd number, whether prime or composite, can escape being converted into an even number, which then repeatedly divides down until it enters the loop (4 → 2 → 1).
Thus, the loop is not the mystery-it’s the endpoint of a system designed to eliminate divisibility barriers.
Another helpful approach to understand the eliminating of Divisors is the point that 3n+1 can never become a prime number , ergo all the exorbitant high numbers will finally end up getting broken down
Thank you! Your comment makes the most sense.