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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. | ||
| + | |||
| + | |||
| + | ---- | ||
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". | 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 | + | 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, | a) checking what happens when at most one of 2n+1 and 2n+2 are chosen, | ||
b) what happens if both are chosen. | b) what happens if both are chosen. | ||
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if n+1 was not chosen, pretend you also get to chose k+1 and go from there. | 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. | ||
| + | <pre> | ||
| + | 1 3 5 7 9 11 13 15 | ||
| + | 2 6 10 14 | ||
| + | 4 12 | ||
| + | 8 | ||
| + | 16 | ||
| + | </pre> | ||
Revision as of 12:00, 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.
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
