All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Logic
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".

See The Solution Submitted by Ravi Raja    
Rating: 2.7500 (8 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution proof | Comment 11 of 13 |
Let S be the set. Since S is non empty, it has at least one element, which we call x. Since S has positive integers only, then x is integer and x is larger than 0. Let T = S intersected with {1, 2, .., x}. T is finite (it has at most x elements) and is non-empty (it contains at least the element x); thus T has a least element. The least element of T is also the least element of S. qed.
  Posted by val on 2003-05-13 05:15:48
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (3)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information