User blog comment:Eners49/The secret 0th hyper-operator?/@comment-2601:142:2:EC49:9D90:4852:1A34:1904-20180827170404

Hi, I'm the guy who figured out ? and ?? notation a few comments down using bounding. However, I just found a proof that could help eliminate the need for bounding x?y numbers at all!

Here are my findings:

x{y}z can be re-written as x{y-1}(x{y}z-1) (This can be useful for writing computer code for knuth arrows.)

Therefore, x+y = x?(x+y-1)

Therefore, 3+2 = 3?(3+1) = 5

Therefore, 3?4 = 5.

Mystery one of x?y solved! 3?4 = 5! (Not a factorial, just very excited.)

But wait, by that logic then 3+1 = 3?(3+0) = 4

And therefore: 3?3 = 4

But we can show that 3?3 = 3+2, using the x{y}2 = x{y-1}x rule that Eners49 started this rabbit hole with. so...

3+1 = 3+2

Which means...

4=5

And if we subtract 3 from both sides...

1=2

1=2

Huzzah! We've found a new way to rip math to pieces using these negative hyperoperator things that Eners's discovered.

The problem stems from the fact that x{y}1 equals x, allowing everything to stop and actually produce a number. No problem then, we'll just stop every time there's an x+1 in the equation, and change it to x before it can cause problems.

Let's write 3+2 in ? notation with my new rules.

3+2 = 3?(3+1) = 5

Oop, there's a +1 in there, let's get rid of it.

3+2 = 3?3 = 5

Hey, that lines up with the x{y}2 rule! We're off to a good start already! Let's try 3+3 with this new system.

3+3 = 3?(3+2) = 6

3+3 = 3?(3?(3+1)) = 6

And if we get rid of the +1...

3+3 = 3?(3?3)

We know from earlier that 3?3 = 5, so...

3+3 = 3?5 = 6

This lines up with Eners's numbers as well! We're on a ROLL! Let's try one more value before we get excited. How about 3+1?

3+1 = 4

Wait, hold the phone. We already have a +1 to deal with, so let's get rid of that.

3=4

And if we subtract 2 from both sides...

1=2

1=2

We did it again. What's wrong THIS time?

The answer is actually quite simple, it's that X DOESN'T EQUAL X+1!

Although x{y}1 = x for positive integer values of y, x{-1}1 (AKA addition AKA WHAT WE'VE BEEN USING THIS WHOLE TIME!) equals x+1. And our whole system I've just made over the course of this post assumes that x{-1}1 = x WHICH IS COMPLETELY WRONG.

We can solve this conundrum by just having x be equal to x+1, but this

A: Is a bit of a get-out-of-jail free card.

and B: Means that we can break math again.

If x = x+1 therefore

x+0 = x+1 (because x = x+0)

Subtracting x to both sides gets us

0=1

0=1

I said this could remove the need to bound numbers, not that it '''would. '''And it most definitely did not.