Constructing a Straggler from an Enumeration of N, {y ∊ N | 1 ≤ y}

Constructing a Straggler from an Enumeration of N, {y € N | 1 ≤ y}

 

Theorem 1: Given an enumeration the set N,  {y ∊ N | 1 ≤ y} there exists a natural number Y and Y is a straggler.

Proof:  Theorem 1 is proved by running Program 2 to construct a number Y  that is not part of the enumeration, which contradicts the assumption that all elements of N,  {y ∊ N | 1 ≤ y} can be enumerated with none missing. It is necessary to show that Y will not be included in the enumeration of N for y = 1 where 1 is the set {1},  y = n where n is the set {1, 2, 3, 4, 5, 6} and y = n + 1 where n + 1is the set {1, 2, 3, 4, 5, 6, 7}. We can then use mathematical induction to infer that Y will never appear in the enumeration of N which will prove the theorem.

Constructing the number Y:

Case 1: y = {1}

Output of Program 2

y	Y
1	12
Y = 12

Case 2: y, {1, 2, 3, 4, 5, 6}

Output of Program 2

y	Y
1	12
2	123
3	1234
4	12345
5	123456
6	1234567
Y = 1234567

Case 3: y, {1, 2, 3, 4, 5, 6, 7}

Output of Program 2

 

y	Y
1	12
2	123
3	1234
4	12345
5	123456
6	1234567
7	12345678
Y = 12345678

   

 

This proves the theorem for y = 1, y = n & y = (n + 1) and supports the conclusion that Y will never appear as part of the enumeration no matter how large the enumeration becomes. Thus, given an enumeration of elements from N,  there is an element Y of N for which there is no corresponding  y  in the enumeration. That contradicts the assertion that all elements of N can be enumerated. The contradiction invalidates the assumption which proves the theorem, Y is a straggler.