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.)
Some Thoughts {P, r, o, o, f} | Comment 1 of 13
For any set with one element, that element is the least element

For any set with two elements, the smaller element is the least element.

Consider set S {a, b, n1, n2, ... nx} with x+2 (x >= 1)elements. Let c be the smaller of a and b. A new and smaller set T with x+1 elements can be defined as {c, n1, n2, ...nx}. Since c is the smaller of a and b, the least element was not removed which implies the sets S and T have the same least element.

For any set R of k elements with k > 2, this argument can be applied k-2 times to create a set of two elements with the same least element as R.
  Posted by Brian Smith on 2003-05-08 06:55:32
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (1)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (18)
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