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Least But Not The Last (Posted on 2003-05-08) Difficulty: 4 of 5
Prove that every Non-Empty set of Positive Integers contains a "Least Element".

  Submitted by Ravi Raja    
Rating: 2.7500 (8 votes)
Solution: (Hide)
Let S be a non-empty subset of N (the set of Natural Numbers). Let k be an element of S. Then k is a natural number.
We define a subset T by T = {x belongs to S : x is less than or equal to k}. Then T is a non-empty subset of {1,2,3,4,....,k}.
Clearly, T is a finite subset of N and therefore it has a Least element, say m. Then, m is one of {1,2,3,4,....,k}.
We now show that m is the least element of S. Let s be any element of S.
If s > k, then the inequality m less than or equal to k implies m < s.
If s is less than or equal to k, then s belongs to T; and m being the Least element of T, we have m less than or equal to s.
Thus m is the least element of S.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: Proof by ContradictionCaleb2019-04-02 23:23:02
voteCory Taylor2003-05-13 11:40:47
Solutionproofval2003-05-13 05:15:48
re(4): {P, r, o, o, f}levik2003-05-12 17:21:46
SolutionProof by Contradictionfriedlinguini2003-05-11 07:03:35
re(3): {P, r, o, o, f}friedlinguini2003-05-09 01:55:35
re(4): {P, r, o, o, f}Ravi Raja2003-05-08 21:00:54
re(3): {P, r, o, o, f}Ravi Raja2003-05-08 20:55:03
re(2): {P, r, o, o, f}Ravi Raja2003-05-08 20:50:56
re(3): {P, r, o, o, f}friedlinguini2003-05-08 14:14:54
Solutionre(2): {P, r, o, o, f}Gamer2003-05-08 12:09:07
re: {P, r, o, o, f}friedlinguini2003-05-08 11:54:36
Some Thoughts{P, r, o, o, f}Brian Smith2003-05-08 06:55:32
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