/sci/ - Science & Math

>>11527194
>pt 1

The Axiom of Infinity defines infinity as the size of the set of all Natural numbers
[math]N: [1,2,3,4,5,\dots][/math]
Thus N is the set of Naturals that has ∞ individual elements in it, and every element is a finite natural number

[math]S: [0.9, 0.99, 0.999, \dots][/math]
Let S be a bijecting set of N that also solves the partial sums of [math]\sum_{k=1}^{\infty}\frac{9}{10^k} [/math] where k=N[n] (the n'th element of N)

[math]S^*: [0.9, 0.09, 0.009, \dots][/math]
Let S* be a bijecting set of N that also solves [math]lim_{k} \frac{9}{10^k}[/math] where k=N[n]

N, S, and S* each have ∞ elements and thus ∞ size to their sets, with full bijection between each set, indexing from 1.

[math]\sum_{m=1}^{p} S^*[m] = S[p][/math]
Shows that any cumulative m iterative summation of S* elements starting from the first element of S* is equal to the p'th element of S.
[math]p=3;(S^*[1]+S^*[2]+S^*[3])=(0.9+0.09+0.009)=S[3]=0.999[/math]

As S* has ∞ size, and all it's elements are bijective real numbers with N, and all it's elements occupy a unique decimal location, then [math]\sum_{m=1}^∞S^*[m]=0.999\dots[/math], a number with ∞ decimal places which also exists as an element in S, for ∞ is defined as the quality which covers ALL elements of the infinite set N. As ALL elements of N are bijective to S, this further covers ALL elements of S.
[math]\sum_{m=1}^∞S^*[m]=0.999\dots=S[∞][/math]
The index location of 0.999... within S is undefined, as is the largest real number in N also undefined; but 0.999... does indeed exist in S, for ALL of N is infinite.
However, since each element of S is singularly unique and unrepeated, there exists 1 unique case of an element in S equal to 0.999...

Tried to condense this as best i could but words.
>pt1

The Axiom of Infinity defines infinity as the size of the set of all Natural numbers
[math]N: [1,2,3,4,5,\dots][/math]
Thus N is the set of Naturals that has ∞ individual elements in it, and every element is a finite natural number

[math]S: [0.9, 0.99, 0.999, \dots][/math]
Let S be a bijecting set of N that also solves the partial sums of [math]\sum_{k=1}^∞\frac{9}{10^k}[/math] where k=N[n] (the n'th element of N)

[math]S^*: [0.9, 0.09, 0.009, \dots][/math]
Let S* be a bijecting set of N that also solves [math]lim_{k} \frac{9}{10^k}[/math] where k=N[n]

N, S, and S* each have ∞ elements and thus ∞ size to their sets, with full bijection between each set, indexing from 1.

[math]\sum_{m=1}^{p} S^*[m] = S[p][/math]
Shows that any cumulative m iterative summation of S* elements starting from the first element of S* is equal to the p'th element of S.
[math]p=3;(S^*[1]+S^*[2]+S^*[3])=(0.9+0.09+0.009)=S[3]=0.999[/math]

As S* has ∞ size, and all it's elements are bijective real numbers with N, and all it's elements occupy a unique decimal location, then [math]\sum_{m=1}^∞S^*[m]=0.999\dots[/math], a number with ∞ decimal places which also exists as an element in S, for ∞ is defined as the quality which covers ALL elements of the infinite set N. As ALL elements of N are bijective to S, this further covers ALL elements of S.
[math]\sum_{m=1}^∞S^*[m]=0.999\dots=S[∞][/math]
The index location of 0.999... within S is undefined, as is the largest real number in N also undefined; but 0.999... does indeed exist in S, for ALL of N is infinite.
However, since each element of S is singularly unique and unrepeated, there exists 1 unique case of an element in S equal to 0.999...

>continued