Is there an easier way to prove the divisibility of integers?

  • Thread starter tronter
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In summary, the conversation discusses an easier way to prove a statement using a proof by cases approach. The speaker suggests considering the sets of consecutive integers and how they partition the integers. They also mention that it is impossible for a certain number to divide n exactly without overlapping with another set of integers.
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Well, consider the sets of integers (1,...,n) (n+1,...,2n) and so on (and the negatives likewise). Now, these partition the integers. Take n consecutive integers anywhere in the integers and for none of them to divide n exactly you would have to fit them inside one of these sets. That's impossible of course. For two to divide n, you would have to overlap mn and (m+1)n for some integer m. That's also impossible without at least n+1 integers. QED.
 

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