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| − | All I have so far is the base case. If you set n = 1 then you have a set with 2 (or n+1 = 1+1) positive integers where both integers have to be less than or equal to 2 (or 2*n = 2*1) so the only option is that the set contains the | + | All I have so far is the base case. If you set n = 1 then you have a set with 2 (or n+1 = 1+1) positive integers where both integers have to be less than or equal to 2 (or 2*n = 2*1) so the only option is that the set contains the numbers 1 and 2. For this set it is true that at least one integer in the set divides another integer in the set since 2 is divisible by 1. |
Does this sound right to anyone else? | Does this sound right to anyone else? | ||
I'm not sure how to complete the inductive step. | I'm not sure how to complete the inductive step. | ||
-Rachel | -Rachel | ||
Revision as of 13:56, 21 January 2009
Does anyone know how to do this problem, because i have no idea on this one
All I have so far is the base case. If you set n = 1 then you have a set with 2 (or n+1 = 1+1) positive integers where both integers have to be less than or equal to 2 (or 2*n = 2*1) so the only option is that the set contains the numbers 1 and 2. For this set it is true that at least one integer in the set divides another integer in the set since 2 is divisible by 1. Does this sound right to anyone else? I'm not sure how to complete the inductive step.
-Rachel
