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:

  • JonC@programming.devEnglish
    301·
    1 year ago

    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.

    • Shiitake@lemmy.world
      2·
      1 year ago

      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:

      1. Does it end with a 0, 2, 4, 6, 8, A, C, E? Yes means divisible by 2.
      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?