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:

  • @Artisian@lemmy.worldEnglish
    60•2 years ago

    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 car student 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.

    • @jscari@lemmy.world
      2•2 years ago

      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.

    • Evirisu
      1•2 years ago

      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.

    • @kernelPanic@lemmy.ml
      1•2 years ago

      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.worldEnglish
        1•2 years ago

        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•2 years ago

          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.worldEnglish
            1•2 years ago

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