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Show N has Prime Numbers

 
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Jan8-13, 07:49 PM   #1
 

Show N has Prime Numbers


(For the following problem I don't just want a flat out answer, but steps and Ideas on how to solve it. The problem was given by my Universities newspaper and for solving it you get free Loot and stuff)
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The Problem:
Show that every set of n+1 positive integers, chosen from a set of 2n consecutive integers, contains at least one pair of relatively prime numbers.

relatively prime means: "Two integers are relatively prime if they share no common positive factors (divisors) except 1"

Example: 8 and 15 are relatively prime.

8 = 2 * 2 * 2 15 = 3 * 5

Example: 9 and 12 are NOT relatively prime

9 = 3 * 3 12 = 2 * 2 * 3
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My thinking thus far:
None really I've just started
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Jan8-13, 08:50 PM   #2
 
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Hint: If m is a positive integer, then m and m+1 are relatively prime.
Jan8-13, 09:25 PM   #3
 
Quote by Petek View Post
Hint: If m is a positive integer, then m and m+1 are relatively prime.
very true, but I think the problem is more difficult than this.

"Show that every set of n+1 positive integers, chosen from a set of 2n consecutive integers"

wouldn't this mean the set would be...
n= 0 1 2 3 4
Set=0,2,6,8

then you would do the n+1 one on that set?
Or something to that extent?
Jan8-13, 10:11 PM   #4
 
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Show N has Prime Numbers


Petek's hint is right. If you wanted to choose a subset without picking any pair that's relatively prime, what does that tell you about the option of picking consecutive numbers from the set?
Jan8-13, 10:19 PM   #5
 
Quote by C R P View Post
very true, but I think the problem is more difficult than this.

"Show that every set of n+1 positive integers, chosen from a set of 2n consecutive integers"
The statement means: I give you a set S = {a, a+1, a+2, ..., a+2n-1} and you pick a subset with n+1 elements.

The point of the problem is that no matter what you do, your subset always contains a pair which are relatively prime.
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college, loot, math, proof, relatively prime
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