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March 7th, 2012, 08:24 PM   #1
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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?
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March 10th, 2012, 03:19 AM   #2
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Re: integer

erm...anyone can help me
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March 10th, 2012, 05:42 AM   #3
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Re: integer

Do you know about mathematical induction? If not, look it up in your text book or on Wikipedia.
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March 11th, 2012, 08:03 AM   #4
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Re: integer

erm... I have read it, but I don't know where to start. Would you mind to show it? ... erm...
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March 11th, 2012, 12:12 PM   #5
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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?
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March 11th, 2012, 12:40 PM   #6
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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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