WOP Summation Proof We proceed bycontradiation Thatis some natural number E 4 2n2 2n 24 which we SetC to contain all at leastone natural we know that I is a greater hereby will refer such counterexamples Since our suppose thatthere exists than 3 such thatthe theorem to as P n evaluates to False Define the assumption is thatthere is number greater than 3 for which this theorem does nothold non empty set Since I is itis a well ordered set This means thatthe well ordering principle applies and there mustbe a smallestelement of C Define natural number x to be this minimum non emptysubsetof the natural numbers a elementof C If natural example where the equation and PLX 1 mustbe true number x is the smallestelementof 3 meaning the smallestcounter smallestelementis x 1 does nothold then the next P x 1 14i 2 x 12 2 x 1 24 2 x 2x 1 2x 2 24 2x2 2x 24 We can also observe that when nix we have the following inequality P X 4 F2x2 2x 24 X i 4 Next equation preserves equality so the following equation musthold thatadding 4 X algebra we consider by to both sides ofan P X 1 84 2x2 2x 24 i 1 i 4 i 4 4X 4x 24i 242 2x 24 4x 2x2 2x 24 lies therein However cannotbe true atthe same time as the fact 2P X Thatis the above in conjunction with the statementthatXis the smallestcounter example can be the contradiction the equation thatwe derived because equation represented as follows 94i 2x2 2x 24 r 4i 2x2 2x 24 1 i 4 I derived from P x 1 I P X So far this means thatFnENISO 1 2 33 P n P n 1 However our proof currently does nottake into accountthatperhaps there is no x I that is smaller than X There is a natural number thatis greater this could occur and also exactly one case where than 3 When xis the smallestnatural number that is greater than 3 thatis X 4 However we can directly evaluate P 4 4 HS 4i 4 4 16 RHS 2 43 2 2 4 24 32 8 24 16 16 16v We know thatthe smallestpositive natural number thatis greater and P 4 is true since the explicitformula is also equal the summation evaluated when n 4 is equal to 16 than 3 is 4 to 16 and Therefore we have derived a contradiction we have proven thatacross all positive integers that if P n holds then P n 1 mustalso be true logic thatis AnEN58 1 2 33 P n zep n 1 1P 4 FneN158 1 2 33 P n the In
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