|September 4th, 2008, 09:28 AM||#1|
Joined: Sep 2008
Please help me figure out how to get started with this problem..
For every positive integer n, prove that a set with exactly n elements has exactly 2^n subsets (counting the empty set and entire set).
|September 4th, 2008, 10:24 AM||#2|
Joined: May 2008
From: York, UK
Re: abstract alg
Try induction. The case for n=1 is simple; for n>1, let the set S have n elements and subsets, and let then for each subset both and are subsets of the set which contains n+1 elements. Moreover, these subsets are all the possible subsets of (you will need to show this); since there are two for each , and there are possible subsets , there must be possible subsets of the set
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