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:

  • @Ethalis@jlai.lu
    12•1 year ago

    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

    • @Elderos
      5•1 year ago

      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!

    • Kogasa
      5•1 year ago

      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.

      • Case 1: You picked Alice. Monty eliminates Bob. Switching wins. (1/3)

      • Case 2: You picked Bob. Monty eliminates Alice. Switching wins. (1/3)

      • 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.

    • @Num10ck@lemmy.worldEnglish
      3•1 year ago

      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.

      • Kogasa
        1•1 year ago

        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.

      • @Marvin42@feddit.nl
        -3•1 year ago

        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.

        • @Aosih@lemm.eeEnglish
          3•1 year ago

          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.