Revision as of 11:00, 2 September 2008 by Norlow (Talk | contribs)

Does anyone know how to go about starting problem 50 for 4.1

Use mathematical induction to show that given a set of n + 1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set.



Here is an idea: P(n)="given a set of n + 1 positive integers, none exceeding 2n, there is at least one integer in this set that divides another integer in the set".

Then assume P(n) and prove P(n+1) by a) checking what happens when at most one of 2n+1 and 2n+2 are chosen, b) what happens if both are chosen.

b) is harder, investigate what happens if n+1 was chosen.

if n+1 was not chosen, pretend you also get to chose k+1 and go from there.


If one or none is chosen is easier than if both are chosen. The problem is if 2n+1, 2n+2 and n+1 is not chosen. It's a great start, but it took me a while to figure out where to go from there.

Here's the way I ordered the numbers in my proof, noticing where n+1 and n+2 go.

 1  3  5  7  9 11 13 15
 2  6 10 14
 4 12
 8
16

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