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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. | 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. | ||
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+ | 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. |
Revision as of 09:32, 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.