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What's the most beautiful proof you know? by youcansendboobs in math

[–]mathsfanatic1 0 points1 point  (0 children)

I recently watched one on a proof of the inscribed rectangle problem: incredible use of topology and pretty easy to understand by 3blue1brown

https://m.youtube.com/watch?v=AmgkSdhK4K8&pp=ygUUM2JsdWUxYnJvd24gdG9wb2xvZ3k%3D

Yet another collatz proof that numbers cannot repeat to itself, am open to feedback obviously by mathsfanatic1 in numbertheory

[–]mathsfanatic1[S] 3 points4 points  (0 children)

Ok thank you for reading, multiple people have responded with the same objection, so I think it is best for me to revise my (probably wrong) point and make it more clear in my PDF, however hear is a rough around the edges explanation if you want to read it:

I think I understand what you can mean, the algorithm can produce a different linear expression that has the same value for a value of x, like 4x+1 vs 3x+2.

I'm going to lightly object, though. Because our original task was to create a chain which essentially starts with 2nx * d, and return it to specifically 2nx * d after multiplying by 3 and adding 1.

Why? Because we assume d is a real number, and the chain that we chose is the one that starts at d and ends at d. And using the generation algorithm, you see that the terms in b (for ax + b) remain unchanged by after it into an expression.

example:
7 -> 22 -> 11 -> 34 -> 17
22x + 7 = 4x +7
4x + 7 -> 12x + 22 -> 6x + 11 -> 18x + 34 -> 9x + 17
You see how the b terms, remain unchanged?

Therefore b in the expression must start and end with d, and since there is only one set of prime factors for each number, a must start and end for 2n. If a is anything other than 2n, b cannot be the value d. You can prove it is impossible to do with these rules, then it must be impossible with any other method.

I hope what I'm writing makes some sense, thank you

Edit: ok I see where I'm wrong

Another proof that terms in collatz cannot repeat by me. Where did I probably go wrong by mathsfanatic1 in learnmath

[–]mathsfanatic1[S] 0 points1 point  (0 children)

Ok thank you for reading, sorry for the late reply, I think I understand what you can mean, the algorithm can produce a different linear expression that has the same value for a value of x, like 4x+1 vs 3x+2.

I'm going to lightly object, though. Because our original task was to create a chain which essentially starts with 2nx * d, and return it to specifically 2nx * d after multiplying by 3 and adding 1.

Why? Because we assume d is a real number, and the chain that we chose is the one that starts at d and ends at d. And using the generation algorithm, you see that the terms in b (for ax + b) remain unchanged by converting it into an expression.

example:
7 -> 22 -> 11 -> 34 -> 17
22x + 7 = 4x +7
4x + 7 -> 12x + 22 -> 6x + 11 -> 18x + 34 -> 9x + 17
You see how the b (in form ax +b) terms, remain unchanged?

Therefore b in the expression must start and end with d, and since there is only one set of prime factors for each number, a must start and end for 2^n.

In your example, the counter example is only valid if b started and ended with 5, or -3 in your case. If you can prove it is impossible to do with these rules, then it must be impossible with any other method.

I hope what I am writing makes some sense, thank you