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Last entry, I made an organized display of JF's Theorem (henceforth on here "JFT"). I said he continued his proof, and I wanted to make a series of entries here so that the reader wasn't inundated with information. Next, I want to expand on what we have already learned before we get to theorem two in JF's livestream. The next thing JF says is you cannot infinitely iterate D_x's. This is true, since E is a finite number. He makes some claims next that I will not verify in this post. But I think I can use some simple logical rules to lay a groundwork for his next theorem he presents in the stream. This is the case for when: And here are the logical rules I can give you:
To prove no R_x = R_x+1, you would have to suppose D_x was coprime with E, but not prime. (JFT) Suppose also that D_x has factors, A and C. If an R_x+1 could be generated in such a case, then R_x+1 would have the same value as R_x. For a proof by contradiction, one of the prime factors would have to be R_x. To put it more generally: So let A = R_x, and we can solve for C: And how does such a result relate with what we already know? Well, by the constraints, we know that E and R_x are coprime. We also know that C is an integer. The only way the left side of the equation will yield an integer is when R_x = 1. You are probably saying the proof looks familiar, because it does. This is just JFT, applied with a prime factor of R_x.
Proof of 4: Consequence of 3. Since, if, besides 1, no D_x and D_x+1 share common prime factors, then D_x and D_x+1 cannot be equal. I think this can lay the groundwork for some of the claims JF makes next. -AF
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AuthorI'm just trying to learn about everything I can. Archives
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