RE: Proof of Math Challenge #1 [DE/EN]
(Edited)
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I would like to exclude myself from the prize but have a go for fun if that's ok ;)
lets take a diffrent approach, how do define distance between any two points (x1,y1) and (x2,y2) ? ((X2-x1)^2 + (y2-y1)^2))^0.5 = distance
Or distance ^2 = (X2-x1)^2 + (y2-y1)^2)
In a right angle triangle , distance is C, the difference between x points is length a and y points is length b
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You are doing the classic beginner mistake of using Pythagoras to prove itself. You can calculate the distance on a coordination field this way because you know Pythagoras is true not the other way around.
Do you understand what I am saying? It is a rough hurdle to take, dw.
Ok fine then, I'll use relatively next time ;p no I get what your saying just never relasied pythagoras lead to this !
Tbh I dont have a Pythagoras Proof in my head. I just know that there are dozens of different ways to prove it and it is generally seen as a proof for beginners. I always struggled with it myself because geometry and graphical proofs are not my forte.
Maybe I can give you one or two of my favorites Proofs so you know what I am talking about, they are my favorite because they are so easy:
There is an unlimited amount of prim numbers.
You can prove it by thinking about what would happen if you know all prim numbers p1,p2,p3...., pn.
Now you multiply all primnumbers to the Number P. If you look at P-1 it is not devidable by any prim number since P is dividable by all of them.--> (P-1) must be a primnumber itself but is not part of p1,p2,p3,...,pn.
Since you lead the assumption of knowing all Prim Numbers to a logical fallacy you now know that there has to be an unlimited amount of Primnumbers.
Makes sense assuming the set is infinite right?
you assume the set is finite and known, but you get a new prim that is not part of the set even though it should be. There is the contradiction which lets you know the assumption must be wrong.
Of course you could then argue that primnumbers might be therefore finite and unknown and there is just no way to know them all. But thats quite trivial.