Least But Not The Last (Posted on 2003-05-08)

	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.

	Subject	Author	Date
	re: Proof by Contradiction	Caleb	2019-04-02 23:23:02
	vote	Cory Taylor	2003-05-13 11:40:47
	proof	val	2003-05-13 05:15:48
	re(4): {P, r, o, o, f}	levik	2003-05-12 17:21:46
	Proof by Contradiction	friedlinguini	2003-05-11 07:03:35
	re(3): {P, r, o, o, f}	friedlinguini	2003-05-09 01:55:35
	re(4): {P, r, o, o, f}	Ravi Raja	2003-05-08 21:00:54
	re(3): {P, r, o, o, f}	Ravi Raja	2003-05-08 20:55:03
	re(2): {P, r, o, o, f}	Ravi Raja	2003-05-08 20:50:56
	re(3): {P, r, o, o, f}	friedlinguini	2003-05-08 14:14:54
	re(2): {P, r, o, o, f}	Gamer	2003-05-08 12:09:07
	re: {P, r, o, o, f}	friedlinguini	2003-05-08 11:54:36
	{P, r, o, o, f}	Brian Smith	2003-05-08 06:55:32