621
What are the most mindblowing things in mathematics?
(lemmy.world)
submitted
1 year ago* (last edited 1 year ago)
by
cll7793@lemmy.world
to
c/nostupidquestions@lemmy.world
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) > 10^10^10^10^10^10^10^10^10^10^10^10^10^10^10 (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
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 2D~16~ 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:
So if 1 and 2 are "yes", it's divisible by 10.
E.g.
Is this actually true? Have I found a new party trick for myself? How would I even know if this is correct?
You're gonna have to go to a lot of parties