this post was submitted on 03 Aug 2023
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No Stupid Questions

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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
236 comments fedilink hide all child comments
 

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:

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[–] kernelPanic@lemmy.ml 1 points 1 year ago* (last edited 1 year ago) (1 children)

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.

[–] Artisian@lemmy.world 2 points 1 year ago (1 children)

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.

[–] kernelPanic@lemmy.ml 1 points 1 year ago (1 children)

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

[–] Artisian@lemmy.world 2 points 1 year ago (1 children)

Ah, so it's the probability you win by playing randomly. Gotcha. That makes sense, it becomes a choice between 2 doors

[–] kernelPanic@lemmy.ml 1 points 1 year ago

Without condition would be more technically correct term but yes