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 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: 20,386 Thanks: 2012 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 t-1 ? a, and so t-1 is in X. If t-1 is in S, so is t-1+1 = t, contrary to the supposition that t is not in S. Hence t-1 is not in S. However, t-1 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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