What concepts or facts do you know from math that is mind blowing, awesome, or simply fascinating?
Here are some I would like to share:
- Gödel’s incompleteness theorems: There are some problems in math so difficult that it can never be solved no matter how much time you put into it.
- Halting problem: It is impossible to write a program that can figure out whether or not any input program loops forever or finishes running. (Undecidablity)
The Busy Beaver function
Now this is the mind blowing one. What is the largest non-infinite number you know? Graham’s Number? TREE(3)? TREE(TREE(3))? This one will beat it easily.
- The Busy Beaver function produces the fastest growing number that is theoretically possible. These numbers are so large we don’t even know if you can compute the function to get the value even with an infinitely powerful PC.
- In fact, just the mere act of being able to compute the value would mean solving the hardest problems in mathematics.
- Σ(1) = 1
- Σ(4) = 13
- Σ(6) > 101010101010101010101010101010 (10s are stacked on each other)
- Σ(17) > Graham’s Number
- Σ(27) If you can compute this function the Goldbach conjecture is false.
- Σ(744) If you can compute this function the Riemann hypothesis is false.
Sources:
- YouTube - The Busy Beaver function by Mutual Information
- YouTube - Gödel’s incompleteness Theorem by Veritasium
- YouTube - Halting Problem by Computerphile
- YouTube - Graham’s Number by Numberphile
- YouTube - TREE(3) by Numberphile
- Wikipedia - Gödel’s incompleteness theorems
- Wikipedia - Halting Problem
- Wikipedia - Busy Beaver
- Wikipedia - Riemann hypothesis
- Wikipedia - Goldbach’s conjecture
- Wikipedia - Millennium Prize Problems - $1,000,000 Reward for a solution
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I came here to find some cool, mind-blowing facts about math and have instead confirmed that I’m not smart enough to have my mind blown. I am familiar with some of the words used by others in this thread, but not enough of them to understand, lol.
Please feel free to ask any questions! Math is a wonderful field full of beauty but unfortunately almost all education systems fail to show this and instead makes it seem like raw robotic calculations instead of creativity.
Math is best learned visually and with context to more abstract terms. 3Blue1Brown is the best resource in my opinion for this!
Here’s a mindblowing fact for you along with a video from 3Blue1Brown. Imagine you are sliding a 1,000,000 kg box and slamming it into a 1 kg box on an ice surface with no friction. The 1 kg box hits a wall and bounces back to hit the 1,000,000 kg box again.
The number of bounces that appear is the digits of Pi. Crazy right? Why would pi appear here? If you want to learn more here’s a video from the best math teacher in the world.
Thanks! I appreciate the response. I’ve seen some videos on 3blue1brown and I’ve really enjoyed them. I think if I were to go back and fill in all the blank spots in my math experience/education I would enjoy math quite a bit.
I don’t know why it appears here or why I feel this way, but picturing the box bouncing off the wall and back, losing energy, feels intuitively round to me.
Same here! Great post but I’m out! lol
There was a response I left in the main comment thread but I’m not sure if you will get the notification. I wanted to post it again so you see it
Response below
Please feel free to ask any questions! Math is a wonderful field full of beauty but unfortunately almost all education systems fail to show this and instead makes it seem like raw robotic calculations instead of creativity.
Math is best learned visually and with context to more abstract terms. 3Blue1Brown is the best resource in my opinion for this!
Here’s a mindblowing fact for you along with a video from 3Blue1Brown. Imagine you are sliding a 1,000,000 kg box and slamming it into a 1 kg box on an ice surface with no friction. The 1 kg box hits a wall and bounces back to hit the 1,000,000 kg box again.
The number of bounces that appear is the digits of Pi. Crazy right? Why would pi appear here? If you want to learn more here’s a video from the best math teacher in the world.
That’s so cool! Thanks for the reply and the link.
Nonsense! I can blow both your minds without a single proof or mathematical symbol, observe!
There are different sizes of infinity.
Think of integers, or whole numbers; 1, 2, 3, 4, 5 and so on. How many are there? Infinite, you can always add one to your previous number.
Now take odd numbers; 1, 3, 5, 7, and so on. How many are there? Again, infinite because you just add 2 to the previous odd number and get a new odd number.
Both of these are infinite, but the set of numbers containing odd numbers is by definition smaller than the set of numbers containing all integers, because it doesn’t have the even numbers.
But they are both still infinite.
I may be wrong or have misunderstood what you said but the sets of natural numbers and odd numbers have the same size/cardinality. If there exists a bijection between the two sets then they have the same size.
f(x) = 2x + 1 is such a bijection
For the same reason, N, Z and Q have the same cardinality. The fact that each one is included in the next ones doesn’t mean their size is different.
Agree. Uncountable infinities are much more mind blowing. It was an interesting journey realising first that everything like time and distance are continuous when learning math the then realising they’re not when learning physics.
Both of these are infinite, but the set of numbers containing odd numbers is by definition smaller than the set of numbers containing all integers, because it doesn’t have the even numbers.
This is provably false - the two sets are the same size. If you take the set of all integers, and then double each number and subtract one, you get the set of odd numbers. Since you haven’t removed or added any elements to the initial set, the two sets have the same size.
The size of this set was named Aleph-zero by Cantor.
Your fact is correct, but the mind-blowing thing about infinite sets is that they go against intuition.
Even if one might think that the number of odd numbers is strictly less than the number of all natural numbers, these two sets are in fact of the same size. With the mapping n |-> 2*n - 1 you can map each natural number to a different odd number and you get every odd number with this (such a function is called a bijection), so the sets are per definition of the same size.
To get really different “infinities”, compare the natural numbers to the real numbers. Here you can’t create a map which gets you all real numbers, so there are “more of them”.
The Julia and Mandelbrot sets always get me. That such a complex structure could arise from such simple rules. Here’s a brilliant explanation I found years back: https://www.karlsims.com/julia.html
This one isn’t terribly mind-blowing, though it does have some really cool uses. I always remember it because of its name: Witch of Agnesi
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you need a space before the second pi
As an engineer, i dont know how to feel about this. On the one hand, 19.99999 = 20. But on the other hand, 3^3 - 3 = 24.
Oh fuck you. I’m going to insert this in place of 20 in some formula and see people have breakdowns.
A simple one: Let’s say you want to sum the numbers from 1 to 100. You could make the sum normally (1+2+3…) or you can rearrange the numbers in pairs: 1+100, 2+99, 3+98… until 50+51 (50 pairs). So you will have 50 pairs and all of them sum 101 -> 101*50= 5050. There’s a story who says that this method was discovered by Gauss when he was still a child in elementary school and their teacher asked their students to sum the numbers.
For the uninitiated, the monty Hall problem is a good one.
Start with 3 closed doors, and an announcer who knows what’s behind each. The announcer says that behind 2 of the doors is a goat, and behind the third door is
a carstudent debt relief, but doesn’t tell you which door leads to which. They then let you pick a door, and you will get what’s behind the door. Before you open it, they open a different door than your choice and reveal a goat. Then the announcer says you are allowed to change your choice.So should you switch?
The answer turns out to be yes. 2/3rds of the time you are better off switching. But even famous mathematicians didn’t believe it at first.
It took me a while to wrap my head around this, but here’s how I finally got it:
There are three doors and one prize, so the odds of the prize being behind any particular door are 1/3. So let’s say you choose door #1. There’s a 1/3 chance that the prize is behind door #1 and, therefore, a 2/3 chance that the prize is behind either door #2 OR door #3.
Now here’s the catch. Monty opens door #2 and reveals that it does not contain the prize. The odds are the same as before – a 1/3 chance that the prize is behind door #1, and a 2/3 chance that the prize is behind either door #2 or door #3 – but now you know definitively that the prize isn’t behind door #2, so you can rule it out. Therefore, there’s a 1/3 chance that the prize is behind door #1, and a 2/3 chance that the prize is behind door #3. So you’ll be twice as likely to win the prize if you switch your choice from door #1 to door #3.
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But even famous mathematicians didn’t believe it at first.
They emphatically did not believe it at first. Marilyn vos Savant was flooded with about 10,000 letters after publishing the famous 1990 article, and had to write two followup articles to clarify the logic involved.
Oh that’s cool - I had heard one or two examples only. Is there some popular writeup of the story from Savant’s view?
I couldn’t tell you - I used the Wikipedia article to reference the specifics and I’m not sure where I first heard about the story. I just remember that the mathematics community dogpiled on her hard for some time and has since completely turned around to accept her answer as correct.
Also relevant - she did not invent the problem, but her article is considered by some to have been what popularized it.
It’s a good one.
First, fuck you! I couldn’t sleep. The possibility to win the car when you change is the possibility of your first choice to be goat, which is 2/3, because you only win when your first choice is goat when you always change.
x1: you win
x2: you change
x3: you pick goat at first choice
P(x1|x2,x3)=1 P(x1)=1/2 P(x3)=2/3 P(x2)=1/2
P(x1|x2) =?
Chain theory of probability:
P(x1,x2,x3)=P(x3|x1,x2)P(x1|x2)P(x2)=P(x1|x2,x3)P(x2|x3)P(x3)
From Bayes theorem: P(x3|x1,x2)= P(x1|x2,x3)P(x2)/P(x1) =1
x2 and x3 are independent P(x2|x3)=P(x2)
P(x1| x2)=P(x3)=2/3 P(x2|x1)=P(x1|x2)P(x2)/P(X1)=P(x1|x2)
P(x1=1|x2=0) = 1- P(x1=1|x2=1) = 1\3 is the probability to win if u do not change.
Why do you have a P(x1) = 1/2 at the start? I’m not sure what x1 means if we don’t specify a strategy.
Just count the number of possibilities. If you change there there two possible first choices to win + if you do not change 1 possible choice to win = 3. If you change there is one possible first choice to lose + if you do not change there two possible first choices to lose=3 P(x1)=P(x1’) = 3/6
Ah, so it’s the probability you win by playing randomly. Gotcha. That makes sense, it becomes a choice between 2 doors
Without condition would be more technically correct term but yes
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I know it to be true, I’ve heard it dozens of times, but my dumb brain still refuses to accept the solution everytime. It’s kind of crazy really
Let’s name the goats Alice and Bob. You pick at random between Alice, Bob, and the Car, each with 1/3 chance. Let’s examine each case.
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Case 1: You picked Alice. Monty eliminates Bob. Switching wins. (1/3)
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Case 2: You picked Bob. Monty eliminates Alice. Switching wins. (1/3)
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Case 3: You picked the Car. Monty eliminates either Alice or Bob. You don’t know which, but it doesn’t matter-- switching loses. (1/3)
It comes down to the fact that Monty always eliminates a goat, which is why there is only one possibility in each of these (equally probable) cases.
From another point of view: Monty revealing a goat does not provide us any new information, because we know in advance that he must always do so. Hence our original odds of picking correctly (p=1/3) cannot change.
In the variant “Monty Fall” problem, where Monty opens a random door, we perform the same analysis:
- Case 1: You picked Alice. (1/3)
- Case 1a: Monty eliminates Bob. Switching wins. (1/2 of case 1, 1/6 overall)
- Case 1b: Monty eliminates the Car. Game over. (1/2 of case 1, 1/6 overall)
- Case 2: You picked Bob. (1/3)
- Case 2a: Monty eliminates Alice. Switching wins. (1/2 of case 2, 1/6 overall)
- Case 2b: Monty eliminates the Car. Game over. (1/2 of case 2, 1/6 overall)
- Case 3: You picked the Car. (1/3)
- Case 3a: Monty eliminates Alice. Switching loses. (1/2 of case 3, 1/6 overall)
- Case 3b: Monty eliminates Bob. Switching loses. (1/2 of case 3, 1/6 overall)
As you can see, there is now a chance that Monty reveals the car resulting in an instant game over-- a 1/3 chance, to be exact. If Monty just so happens to reveal a goat, we instantly know that cases 1b and 2b are impossible. (In this variant, Monty revealing a goat reveals new information!) Of the remaining (still equally probable!) cases, switching wins half the time.
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like on paper the odds on your original door was 1/3 and the option door is 1/2, but in reality with the original information both doors were 1/3 and now with the new information both doors are 1/2.
Your original odds were 1/3, and this never changes since you don’t get any new information.
The key is that Monty always reveals a goat. No matter what you choose, even before you make your choice, you know Monty will reveal a goat. Therefore, when he does so, you learn nothing you didn’t already know.
Yes, you don’t actually have to switch. You could also throw a coin to decide to stay at the current door or to switch. By throwing a coin, you actually improved your chances of winning the price.
This is incorrect. The way the Monty Hall problem is formulated means staying at the current door has 1/3 chance of winning, and switching gives you 2/3 chance. Flipping a coin doesn’t change anything. I’m not going to give a long explanation on why this is true since there are plenty other explanations in other comments already.
This is a common misconception that switching is better because it improves your chances from 1/3 to 1/2, whereas it actually increases to 2/3.
To me, it makes sense because there was initially 2 chances out of 3 for the prize to be in the doors you did not pick. Revealing a door, exclusively on doors you did not pick, does not reset the odds of the whole problem, it is still more likely that the prize is in one of the door you did not pick, and a door was removed from that pool.
Imo, the key element here is that your own door cannot be revealed early, or else changing your choice would not matter, so it is never “tested”, and this ultimately make the other door more “vouched” for, statistically, and since you know that the door was more likely to be in the other set to begin with, well, might as well switch!
This explanation really helped me make sense of it: Monty Hall Problem (best explanation) - Numberphile
I know the problem is easier to visualize if you increase the number of doors. Let’s say you start with 1000 doors, you choose one and the announcer opens 998 other doors with goats. In this way is evident you should switch because unless you were incredibly lucky to pick up the initial door with the prize between 1000, the other door will have it.
How do we even come up with such amazing problems right ? It’s fascinating.
This is fantastic, thank you.
This is so mind blowing to me, because I get what you’re saying logically, but my gut still tells me it’s a 50/50 chance.
But I think the reason it is true is because the other person didn’t choose the other 998 doors randomly. So if you chose any of the other 998 doors, it would still be between the door you chose and the winner, other than the 1/1000 chance that you chose right at the beginning.
I now recall there was a numberphile with exactly that visualisation! It’s a clever visual
It really is, it’s how my probability class finally got me to understand why this solution is true.
I don’t find this more intuitive. It’s still one or the other door.
I think the problem is worded specifically to hide the fact that you’re creating two set of doors by picking a door, and that shrinking a set actually make each individual door in that set more likely to have the prize.
Think of it this way : You have 4 doors, 2 blue doors and 2 red doors. I tell you that there is 50% chance of the prize to be in either a blue or a red door. Now I get to remove a red door that is confirmed to not have the prize. If you had to chose, would you pick a blue door or a red door? Seems obvious now that the remaining red door is somehow a safer pick. This is kind of what is happening in the initial problem, but since the second ensemble is bigger to begin with (the two doors you did not pick), it sort of trick you into ignoring the fact that the ensemble shrank and that it made the remaining door more “valuable”, since the two ensembles are now of equal size, but only one ensemble shrank, and it was always at 2/3 odds of containing the prize.
The point is, the odds don’t get recomputed after the other doors are opened. In effect you were offered two choices at the start: choose one door, or choose all of the other 999 doors.
This is the way to think about it.
Same here, even after reading other explanations I don’t see how the odds are anything other than 50/50.
read up on the law of total probability. prob(car is behind door #1) = 1/3. monty opens door #3, shows you a goat. prob(car behind door #1) = 1/3, unchanged from before. prob(car is behind door #2) + prob(car behind door #1) = 1. therefore, prob(car is behind door #2) = 2/3.
Following that cascade, didn’t you just change the probability of door 2? It was 1/3 like the other two. Then you opened door three. Why would door two be 2/3 now? Door 2 changes for no disclosed reason, but door 1 doesn’t? Why does door 1 have a fixed probability when door 2 doesn’t?
No, you didn’t change the prob of #2. Prob(car behind 2) + prob(car behind 3) = 2/3. Monty shows you that prob(car behind 3) = 0.
This can also be understood through conditional probabilities, if that’s easier for you.
The thing is, you pick the door totally randomly and since there are more goats, the chance to pick a goat is higher. That means there’s a 2/3 chance that the door you initially picked is a goat. The announcer picks the other goat with a 100% chance, which means the last remaining door most likely has the prize behind it
Edit: seems like this was already answered by someone else, but I didn’t see their comment due to federation delay. Sorry
Don’t be sorry, your comment was the first time I actually understood how it works. Like I understand the numbers, but I still didn’t get the problem, even when increasing the amount of doors. It was your explanation that made it actually click.
The odds you picked the correct door at the start is 1/1000, that means there’s a 999/1000 chance it’s in one of the other 999 doors. If the man opens 998 doors and leaves one left then that door has 999/1000 chance of having the prize.
11 X 11 = 121
111 X 111 = 12321
1111 X 1111 = 1234321
11111 X 11111 = 123454321
111111 X 1111111 = 12345654321
holt shit
Amazing in deed!
Just a small typo in the very last factor
1111111.You could include 1 x 1 = 1
But thats so cool. Maths is crazy.
The Monty hall problem makes me irrationally angry.
I found the easiest way to think about it as if there are 10 doors, you choose 1, then 8 other doors are opened. Do you stay with your first choice, or the other remaining door? Or scale up to 100. Then you really see the advantage of swapping doors. You have a higher probability when choosing the last remaining door than of having correctly choosen the correct door the first time.
Edit: More generically, it’s set theory, where the initial set of doors is 1 and (n-1). In the end you are shown n-2 doors out of the second set, but the probability of having selected the correct door initially is 1/n. You can think of it as switching your choice to all of the initial (n-1) doors for a probability of (n-1)/n.
Holy shit this finally got it to click in my head.
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I find the easiest way to understand Monty Hall is to think of it in a meta way:
Situation A - A person picks one of three doors, 1 n 3 chance of success.
Situation B - A person picks one of two doors, 1 in 2 chance of success.
If you were an observer of these two situations (not the person choosing doors) and you were gonna bet on which situation will more often succeed, clearly the second choice.
I don’t understand how this relates to the problem. Yes 50 percent is greater than 33 percent, but that’s not what the Monty hall problem is about. The point of the exercise is to show that when the game show host knowingly (and it is important to state that the host knows where the prize is) opens a door, he is giving the contestant 33 percent extra odds.
But the issue is that by switching doors, you have a 66% chance of winning, it doesn’t drop to 50% just because there are 2 doors, it’s still 33% on the first door, 66% on the other doors (as a whole), for which we know one is not correct and won’t choose.
on the whole
is the key words here
individually the door has that 1:2 chance, but the scenario has more context and information and thus better odds. Choosing scenario B over scenario A is a better wager
you aren’t talking about the Monty Hall problem then
Only to somebody who didn’t know about the choice being made!
Collatz conjecture or sometimes known as the 3x+1 problem.
The question is basically: Does the Collatz sequence eventually reach 1 for all positive integer initial values?
Here’s a Veritasium Video about it: https://youtu.be/094y1Z2wpJg
Basically:
You choose any positive integer, then apply 3x+1 to the number if it’s odd, and divide by 2 if it’s even. The Collatz conjecture says all positive integers eventually becomes a 4 --> 2 --> 1 loop.
So far, no person or machine has found a positive integer that doesn’t eventually results in the 4 --> 2 --> 1 loop. But we may never be able to prove the conjecture, since there could be a very large number that has a collatz sequence that doesn’t end in the 4-2-1 loop.
maybe this will make more sense when I watch the veritasium video, but I don’t have time to do that until the weekend. How is 3x+1 unprovable? won’t all odd numbers multiplied by 3 still be odd? and won’t adding 1 to an odd number always make it even? and aren’t all even numbers by definition divisible by 2? I’m struggling to see how there could be any uncertainty in this
The unproven part is that it eventually will reach 1, not that it’s not possible to do the computation. Someone may find a number loop that doesn’t eventually reach 1.
The number 26 reaches as high as 40 before falling back to 4-2-1 loop. The very next number, 27, goes up to 9232 before it stops going higher. For numbers 1 to 10,000 most of them reach a peak of less than 100,000, but somehow, the number 9663 goes up to 27,114,424 before trending downwards. The uncertainty is that what if there is a special number that doesn’t just stop at a peak, but goes on forever. I’m not really good at explaining things, so you’re gonna have to watch the video.
Just after going through a few examples in my head, the difficulty becomes somewhat more apparent. let’s start with 3. This is odd, so 3(3)+1 = 10. 10 is even so we have 10/2=5.
By this point my intuition tells me that we don’t have a very obvious pattern that we can use to decide whether the function will output 4, 2, or 1 by recursively applying the function to its own output, other than the fact that every other number that we try appears to result in this pattern. We could possibly reduce the problem to whether we can guess that the function will eventually output a power of 2, but that doesn’t sound to me like it makes things much easier.
If I had no idea whether a proof existed, I would guess that it may, but that it is non-trivial. Or at least my college math courses did not prepare me to find one. Since it looks like plenty of professional mathematicians have struggled with it, I have no doubt that if a proof exists it is non-trivial.
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Which one?
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Szemeredis regularity lemma is really cool. Basically if you desire a certain structure in your graph, you just have to make it really really (really) big and then you’re sure to find it. Or in other words you can find a really regular graph up to any positive error percentage as long as you make it really really (really really) big.
The utility of Laplace transforms in regards to differential systems.
In engineering school you learn to analyze passive DC circuits early on using not much more than ohms law and Thevenin’s Theoram. This shit can be taught to elementary schoolers.
Then a little while later, you learn how to do non-finear differential equations to help work complex systems, whether it’s electrical, mechanical, thermal, hydrolic, etc. This shit is no walk in the park.
Then Laplace transforms/identities come along and let you turn non-linear problems in time-based space, into much simpler problems in frequency-based space. Shit blows your mind.
THEN a mafacka comes along and teaches you that these tools can be used to turn complex differential system problems (electrical, mechanical, thermal, hydrolic, etc) into simple DC circuits you can analyze/solve in frequency-based space, then convert back into time-based space for the answers.
I know this is super applied calculus shit, but I always love that sweet spot where all the high-concept math finally hits the pavement.
And then they tell you that the fundamental equations for thermal, fluid, electrical and mechanical are all basically the same when you are looking at the whole Laplace thing. It’s all the same…
ABSOLUTELY. I just recently capped off the Diff Eq, Signals, and Controls courses for my undergrad, and truly by the end you feel like a wizard. It’s crazy how much problem-solving/system modeling power there is in such a (relatively) simple, easy to apply, and beautifully elegant mathematical tool.
That you can have 5 apples, divide them zero times, and somehow end up with math shitting itself inside-out at you even though the apples are still just sitting there.
You try having 5 apples and divide them into 0 equal groups and you’ll shit yourself too.
Except that by dividing the total number zero times means you’re not dividing them at all, and therefore by doing nothing you are still left with 5 apples.
Not dividing at all is dividing by 1.
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x^n + y^n = z^n has no solutions where n > 2 and x, y and z are all natural numbers. It’s hard to believe that, knowing that it has an infinite number of solutions where n = 2.
Pierre de Format, after whom this theorem was named, famously claimed to have had a proof by leaving the following remark in some book that he owned: “I have a proof of this theorem, but there is not enough space in this margin”. It took mathematicians several hundred years to actually find the proof.
