Post an erroneous Mathematical proof. The task is to identify it. The more outlandish the proof you make, the more points you get. Score = x•t Where x is the impact on Mathematics of your proof if true, and t is the time in minutes your proof goes without being challenged. You can also post a completely legitimate proof to screw with people's heads, but you won't get any points. The proof shouldn't require more than highschool Mathematics. (You can use single variable calculus). Combinatorics and Discrete Maths may be attempted if it's something a Highschool Mathematics student can wrap their head around, without needing to take a course on the subject. I'll start with an extension of a common well known proof. I'll try to fuck up Mathematics pretty badly. Spoiler Take two numbers a and b. Let a = b (1) a•a = b•b = a•b. (2) a•a - b•b = a•a - a•b (3) Factorise both sides. (a-b)(a+b) = a(a-b) (4) Divide through by (a-b). (a+b) = a (5) Substitute (1) into (5). (a+a) = a (6) 2a = a (7) Divide through by a. Therefore 2 = 1 (8) Multiply (8) through by K. 2K = K (9) All even numbers are numbers of the form 2m. From (9) 2m = m. Therefore the set of even numbers is equivalent to the set of odd numbers, among other conclusions that are trivial and left as exercise for the reader. Multiply (8) through by 'K+1'. 2(K+1) = (K+1) (10) 2K + 2 = K + 1 (11) Substitute (10) into (11). 2K + 2 = 2K + 1 (12) Subtract "2K + 1" from both sides. 2K + 2 - (2K + 1) = 2K + 1 - (2K + 1) 1 = 0 (13) Any number a = a•1 (14) Substitute (13) into (14). a = a•0 (15) From (15), (0 = a), and from (13), (1 = a) In light of the above and (9) and (11), every number is equal to every other number. The proof of this is trivial and left as an exercise for the reader. ( ͡° ͜ʖ ͡°) From (13). Taking the definition of the derivative. f'(x) = Lim{(f(x+delta(x)) - f(x))/delta(x)} as delta(x) tends to 0. But 0 = 1, Therefore the derivative of a function = f(x + delta(x)) - f(x). ( ͡° ͜ʖ ͡°) So with the conclusions above, you can mess up almost all of Mathematics. I'll define the impact of my proof on conventional Mathematics as X1, making it the yardstick. Estimate the impacts of your proof as a multiple/fraction of mine. ¯\_(ツ)_/¯ I'd tag fellow Math lovers, but I want to rack up as many points as possible XD.
(a-b)(a+b) = a(a-b) (4) You divided by (a-b) Since a = b, it's the same as dividing by (a-a) or (b-b) which equals zero. You cannot divide by zero.
Par terms in fibonacci (x_n is the n term of fibonacci sequence) x_n=x_n-1+x_n-2 x_n=2*x_n-2+x_n-3 x_n=3*x_n-3+2*x_n-4 .... x_n=x_p*x_n-p+x_p-1*x_n-p-1 x_2k=x_k^2x_k-1^2 So you can calculate par terms in fibonaci sequence without knowing the two previous term
I actually tried this XD; generating a general non recursive formula to describe the Fibonacci sequence. I expanded the decomposition, and you get. x_n = x_{n-1}•x_1 + x_{n-2}•x_2 x_2k = x_p•x_{2k-p} + x_{p-1}•x_{2k-p+1} Even by back induction to the base term, you end up with the terms you started with, this it's not possible to find par (what does this mean? Consecutive?) terms of the Fibonacci sequence without knowing the preceding two terms.
correct the formula is valid the notation of the terms is the incorrect (unless you are a beast an say that 1 1 are the terms 1 and second) is the formula for par terms n=2p