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| − | + | The problem I had is if 2n+1, 2n+2 are chosen 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. | Here's the way I ordered the numbers in my proof, noticing where n+1 and n+2 go. | ||
Revision as of 12:02, 2 September 2008
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.
The problem I had is if 2n+1, 2n+2 are chosen 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
