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
Goldbach’s Conjecture: Every even natural number > 2 is a sum of 2 prime numbers. Eg: 8=5+3, 20=13+7.
https://en.m.wikipedia.org/wiki/Goldbach’s_conjecture
Such a simple construct right? Notice the word “conjecture”. The above has been verified till 4x10^18 numbers BUT no one has been able to prove it mathematically till date! It’s one of the best known unsolved problems in mathematics.
Wtf !
How can you prove something in math when numbers are infinite? That number you gave if it works up to there we can call it proven no? I’m not sure I understand
There are many structures of proof. A simple one might be to prove a statement is true for all cases, by simply examining each case and demonstrating it, but as you point out this won’t be useful for proving statements about infinite cases.
Instead you could assume, for the sake of argument, that the statement is false, and show how this leads to a logical inconsistency, which is called proof by contradiction. For example, Georg Cantor used a proof by contradiction to demonstrate that the set of Natural Numbers (1,2,3,4…) are smaller than the set of Real Numbers (which includes the Naturals and all decimal numbers like pi and 69.6969696969…), and so there exist different “sizes” of infinity!
For a method explicitly concerned with proofs about infinite numbers of things, you can try Proof by Mathematical Induction. It’s a bit tricky to describe…
- First demonstrate that a statement is true in some 1st base case.
- Then demonstrate that if it holds true for the abstract Nth case, then it necessarily holds true for the (N+1)th case (by doing some clever rearranging of algebra terms or something)
- Therefore since it holds true for the 1th case, it must hold true for the (1+1)th case = the 2th case. And since it holds true for the 2th case it must hold true for the (2+1)=3th case. And so on ad infinitum.
Wikipedia says:
Mathematical induction can be informally illustrated by reference to the sequential effect of falling dominoes.
Bear in mind, in formal terms a “proof” is simply a list of true statements, that begin with axioms (which are true by default) and rules of inference that show how each line is derived from the line above.
Very cool and fascinating world of mathematics!
Just to add to this. Another way could be to find a specific construction. If you could for example find an algorithm that given any even integer returns two primes that add up to it and you showed this algorithm always works. Then that would be a proof of the Goldbach conjecture.
As you said, we have infinite numbers so the fact that something works till 4x10^18 doesn’t prove that it will work for all numbers. It will take only one counterexample to disprove this conjecture, even if it is found at 10^100. Because then we wouldn’t be able to say that “all” even numbers > 2 are a sum of 2 prime numbers.
So mathematicians strive for general proofs. You start with something like: Let n be any even number > 2. Now using the known axioms of mathematics, you need to prove that for every n, there always exists two prime numbers p,q such that n=p+q.
Would recommend watching the following short and simple video on the Pythagoras theorem, it’d make it perfectly clear how proofs work in mathematics. You know the theorem right? For any right angled triangle, the square of the hypotenuse is equal to the sum of squares of both the sides. Now we can verify this for billions of different right angled triangles but it wouldn’t make it a theorem. It is a theorem because we have proved it mathematically for the general case using other known axioms of mathematics.
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.
Same here! Great post but I’m out! lol
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.
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”.
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.
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.
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.
For me, personally, it’s the divisible-by-three check. You know, the little shortcut you can do where you add up the individual digits of a number and if the resulting sum is divisible by three, then so is the original number.
That, to me, is black magic fuckery. Much like everything else in this thread I have no idea how it works, but unlike everything else in this thread it’s actually a handy trick that I use semifrequently
That one’s actually really easy to prove numerically.
Not going to type out a full proof here, but here’s an example.
Let’s look at a two digit number for simplicity. You can write any two digit number as 10*a+b, where a and b are the first and second digits respectively.
E.g. 72 is 10 * 7 + 2. And 10 is just 9+1, so in this case it becomes 72=(9 * 7)+7+2
We know 9 * 7 is divisible by 3 as it’s just 3 * 3 * 7. Then if the number we add on (7 and 2) also sum to a multiple of 3, then we know the entire number is a multiple of 3.
You can then extend that to larger numbers as 100 is 99+1 and 99 is divisible by 3, and so on.
Waaaait a second.
Does that hold for every base, where the divisor is 1 less than the base?
Specifically hexidecimal - could it be that 5 and 3 have the same “sum digits, get divisibility” property, since 15 (=3*5) is one less than the base number 16?
Like 2D16 is16*2+13 = 45, which is divisible by 3 and 5.
Can I make this into a party trick?! “Give me a number in hexidecimal, and I’ll tell you if it’s divisible by 10.”
Am thinking it’s 2 steps:
- Does it end with a 0, 2, 4, 6, 8, A, C, E? Yes means divisible by 2.
- Do the digits add up to a multiple of 5 (ok to subtract 5 liberally while adding)? Skip A and F. For B add 1; C->2, D->3, E->4. If the sum is divisible by 5, then original number is too.
So if 1 and 2 are “yes”, it’s divisible by 10.
E.g.
- DEADBAE16 (=23349547010): (1) ends with E, ok. (2) 3+4+3+1+4=15, divisible by 5. Both are true so yes, divisible by 10.
- C4744416 (=1287482010): (1) ends with 4, ok. (2) 2+4+7+4+4+4=25, ok.
- BEEFFACE16 (=3203398350): (1) E, ok. (2) 1+4+4+2+4=15, ok.
Is this actually true? Have I found a new party trick for myself? How would I even know if this is correct?
How would I even know if this is correct?
You’re gonna have to go to a lot of parties
All numbers are dividible by 3. Just saying.
e^(pi i) = -1
like, what?
3Blue 1Brown actually explains that one in a way that makes it seem less coincidental and black magic. Totally worth a watch
If you think of complex numbers in their polar form, everything is much simpler. If you know basic calculus, it can be intuitive.
Instead of z = + iy, write z = (r, t) where r is the distance from the origin and t is the angle from the positive x-axis. Now addition is trickier to write, but multiplication is simple: (a,b) * (c,d) = (ab, b + d). That is, the lengths multiply and the angles add. Multiplication by a number (1, t) simply adds t to the angle. That is, multiplying a point by (1, t) is the same as rotating it counterclockwise about the origin by an angle t.
The function f(t) = (1, t) then parameterizes a circular motion with a constant radial velocity t. The tangential velocity of a circular motion is perpendicular to the current position, and so the derivative of our function is a constant 90 degree multiple of itself. In radians, that means f’(t) = (1, pi/2)f(t). And now we have one of the simplest differential equations whose solution can only be f(t) = k * e^(t* (1, pi/2)) = ke^(it) for some k. Given f(0) = 1, we have k = 1.
All that said, we now know that f(t) = e^(it) is a circular motion passing through f(0) = 1 with a rate of 1 radian per unit time, and e^(i pi) is halfway through a full rotation, which is -1.
If you don’t know calculus, then consider the relationship between exponentiation and multiplication. We learn that when you take an interest rate of a fixed annual percent r and compound it n times a year, as you compound more and more frequently (i.e. as n gets larger and larger), the formula turns from multiplication (P(1+r/n)^(nt)) to exponentiation (Pe^(rt)). Thus, exponentiation is like a continuous series of tiny multiplications. Since, geometrically speaking, multiplying by a complex number (z, z^(2), z^(3), …) causes us to rotate by a fixed amount each time, then complex exponentiation by a continuous real variable (z^t for t in [0,1]) causes us to rotate continuously over time. Now the precise nature of the numbers e and pi here might not be apparent, but that is the intuition behind why I say e^(it) draws a circular motion, and hopefully it’s believable that e^(i pi) = -1.
All explanations will tend to have an algebraic component (the exponential and the number e arise from an algebraic relationship in a fundamental geometric equation) and a geometric component (the number pi and its relationship to circles). The previous explanations are somewhat more geometric in nature. Here is a more algebraic one.
The real-valued function e^(x) arises naturally in many contexts. It’s natural to wonder if it can be extended to the complex plane, and how. To tackle this, we can fall back on a tool we often use to calculate values of smooth functions, which is the Taylor series. Knowing that the derivative of e^(x) is itself immediately tells us that e^(x) = 1 + x + x^(2)/2! + x^(3)/3! + …, and now can simply plug in a complex value for x and see what happens (although we don’t yet know if the result is even well-defined.)
Let x = iy be a purely imaginary number, where y is a real number. Then substitution gives e^x = e^(iy) = 1 + iy + i(2)y(2)/2! + i(3)y(3)/3! + …, and of course since i^(2) = -1, this can be simplified:
e^(iy) = 1 + iy - y^(2)/2! - iy^(3)/3! + y^(4)/4! + iy^(5)/5! - y^(6)/6! + …
So we’re alternating between real/imaginary and positive/negative. Let’s factor it into a real and imaginary component: e^(iy) = a + bi, where
a = 1 - y^(2)/2! + y^(4)/4! - y^(6)/6! + …
b = y - y^(3)/3! + y^(5)/5! - y^(7)/7! + …
And here’s the kicker: from our prolific experience with calculus of the real numbers, we instantly recognize these as the Taylor series a = cos(y) and b = sin(y), and thus conclude that if anything, e^(iy) = a + bi = cos(y) + i sin(y). Finally, we have e^(i pi) = cos(pi) + i sin(pi) = -1.
Damn that’s actually pretty simple way to use differential equations to show this.
Edit. The series one is also pretty simple.
If you write it as e^(i pi) - 1 = 0
Then it is a thing of beauty.
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.
Multiply 9 times any number and it always “reduces” back down to 9 (add up the individual numbers in the result)
For example: 9 x 872 = 7848, so you take 7848 and split it into 7 + 8 + 4 + 8 = 27, then do it again 2 + 7 = 9 and we’re back to 9
It can be a huge number and it still works:
9 x 987345734 = 8886111606
8+8+8+6+1+1+1+6+0+6 = 45
4+5 = 9
Also here’s a cool video about some more mind blowing math facts
I suspect this holds true to any base x numbering where you take the highest valued digit and multiply it by any number. Try it with base 2 (1), 4 (3), 16 (F) or whatever.
11 X 11 = 121
111 X 111 = 12321
1111 X 1111 = 1234321
11111 X 11111 = 123454321
111111 X 1111111 = 12345654321
You could include 1 x 1 = 1
But thats so cool. Maths is crazy.
Amazing in deed!
Just a small typo in the very last factor
1111111.holt shit
The Fourier series. Musicians may not know about it, but everything music related, even harmony, boils down to this.
Fourier transformed everything
Quickly a game of chess becomes a never ever played game of chess before.
Related: every time you shuffle a deck of cards you get a sequence that has never happened before. The chance of getting a sequence that has occurred is stupidly small.
Most of the time, but only as long as the shuffle is actually random. A perfect riffle shuffle on a brand new deck will get you the same result every time, and 8 perfect riffles on a row get you back to where you started.
I’m guessing this is more pronounced at lower levels. At high level chess, I often hear commentators comparing the moves to their database of games, and it often takes 20-30 moves before they declare that they have now reached a position which has never been reached in a professional game. The high level players have been grinding openings and their counters and the counters to the counters so deeply that a lot of the initial moves can be pretty common.
Also, high levels means that games are narrowing more towards the “perfect” moves, meaning that repetition from existing games are more likely.
Oh yes I agree totally!
At those (insane) levels “everyone” knows the “best” move after the choice of opening and for a long time.
But for the millions of games played every day, I guess it’s a bit less :-)
The one I bumped into recently: the Coastline Paradox
“The coastline paradox is the counterintuitive observation that the coastline of a landmass does not have a well-defined length. This results from the fractal curve–like properties of coastlines; i.e., the fact that a coastline typically has a fractal dimension.”
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 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.
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.
This is my silly contribution: 70% of 30 is equal to 30% of 70. This applies to other numbers and can be really helpful when doing percentages in your head. 15% of 77 is equal to 77% of 15.
I’ve seen this one used in the news when they want to make one side of a statistic stand out.
A/100×B=A×B/100
Seeing mathematics visually.
I am a huge fan of 3blue1brown and his videos are just amazing. My favorite is linear algebra. It was like an out of body experience. All of a sudden the world made so much more sense.
His video about understanding multiple dimensions was what finally made it click for me
e^(I*pi) = -1
My favorite form, just slightly different,
e^(i*π)+1=0Euler’s identity directly relates all the really cool mathematical constants into one elegant formula:
e,i,π,1, and0I think that is the cooler form but I first saw Euler’s ID as =-1
The square of any prime number >3 is one greater than an exact multiple of 24.
For example, 7² = 49= (2 * 24) + 1
Does this really hold for higher values? It seems like a pretty good way of searching for primes esp when combined with other approaches.
Every prime larger than 3 is either of form 6k+1, or 6k+5; the other four possibilities are either divisible by 2 or by 3 (or by both). Now (6k+1)² − 1 = 6k(6k+2) = 12k(3k+1) and at least one of k and 3k+1 must be even. Also (6k+5)² − 1 = (6k+4)(6k+6) = 12(3k+2)(k+1) and at least one of 3k+2 and k+1 must be even.
I’ve read books about prime numbers and I did not know this.
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I don’t get it,
5² = 25 != (2 * 24) + 111² = 121 != (2 * 24) + 1Could you please help me understand, thanks!
5² = 25 = (1 * 24) + 111² = 121 = (5 * 24) + 1The key thing is that
p² = 24n + 1(forpgreater than 3).Not 2 * 24, but x * 24.
So 5^2 = 1 * 24 +1
11^2 = 5 * 24 + 1
