March 7th, 2012, 08:24 PM  #1 
Member Joined: Dec 2011 Posts: 61 Thanks: 0  integer
Let a be an integer, and let X consist of all integers with x is greater and equal to a. Let S be a subset of X such that a is in S and whenever an integer k is in S, so is k+1. Then S=X. Prove it. Can anyone help me to prove it?

March 10th, 2012, 03:19 AM  #2 
Member Joined: Dec 2011 Posts: 61 Thanks: 0  Re: integer
erm...anyone can help me

March 10th, 2012, 05:42 AM  #3 
Senior Member Joined: Nov 2010 From: Berkeley, CA Posts: 174 Thanks: 35 Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT  Re: integer
Do you know about mathematical induction? If not, look it up in your text book or on Wikipedia.

March 11th, 2012, 08:03 AM  #4 
Member Joined: Dec 2011 Posts: 61 Thanks: 0  Re: integer
erm... I have read it, but I don't know where to start. Would you mind to show it? ... erm...

March 11th, 2012, 12:12 PM  #5 
Senior Member Joined: Nov 2010 From: Berkeley, CA Posts: 174 Thanks: 35 Math Focus: Elementary Number Theory, Algebraic NT, Analytic NT  Re: integer
The wikipedia article on mathematical induction may be found here. Especially look at the section titled Starting at some other number. Does that help? If not, where are you stuck?

March 11th, 2012, 12:40 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,833 Thanks: 2161 
If there is an integer u in X that is not in S, let t be the smallest such integer. (There are u  a + 1 integers from a to u inclusive, so there would have to be such a smallest.) Since t isn't in S, t is not a, and so t > a. It follows that t1 ? a, and so t1 is in X. If t1 is in S, so is t1+1 = t, contrary to the supposition that t is not in S. Hence t1 is not in S. However, t1 is in X and is smaller than t, which is contrary to the definition of t. It follows that there is no integer in X that is not in S, i.e., X is a subset of S. Since S is a subset of X, S and X must be equal. 

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