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| + | I believe that is correct for the base case. I am also stuck on the inductive step. If we assume that k is true, then k+1 integers, each less than 2k, must have k or fewer prime numbers. I'm stuck on the next part. How do I prove that k+2 integers, each less than 2k+2, have fewer than k+1 prime numbers? | ||
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| + | --[[User:Jberlako|Jberlako]] 19:47, 21 January 2009 (UTC) | ||
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| + | http://batty.mullikin.org/uga_courses/math2610/spring03/induction.pdf This website gives a very detailed explanation on how to solve it, I just noticed it tonight, I hope everyone gets to see it in time. However I will note that at the bottom, the "Pigeon-hole Principle" I first thought this would be the easiest way to prove it, however it does not use induction. | ||
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| + | -Chris Ruderschmidt | ||
Latest revision as of 17:24, 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
I believe that is correct for the base case. I am also stuck on the inductive step. If we assume that k is true, then k+1 integers, each less than 2k, must have k or fewer prime numbers. I'm stuck on the next part. How do I prove that k+2 integers, each less than 2k+2, have fewer than k+1 prime numbers?
--Jberlako 19:47, 21 January 2009 (UTC)
http://batty.mullikin.org/uga_courses/math2610/spring03/induction.pdf This website gives a very detailed explanation on how to solve it, I just noticed it tonight, I hope everyone gets to see it in time. However I will note that at the bottom, the "Pigeon-hole Principle" I first thought this would be the easiest way to prove it, however it does not use induction.
-Chris Ruderschmidt
