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## Main Question or Discussion Point

Every subset of n+1 from a 2n set has a pair of numbers with gcd=1. How can I prove this?

- Thread starter xax
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Every subset of n+1 from a 2n set has a pair of numbers with gcd=1. How can I prove this?

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Well, you can't because it's not true. For example let [tex]S = \left\{2, 4, 6, 8\right\} [/tex].

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I think what xax meant was this:

Given is the set of integers from 1 to 2n. If you choose n+1 numbers from this set there is always a pair of numbers (among these n+1 numbers) with gcd=1.

Given is the set of integers from 1 to 2n. If you choose n+1 numbers from this set there is always a pair of numbers (among these n+1 numbers) with gcd=1.

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Edgardo, you're right.

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Xax, given any natural number, b, what is the gcd(b, b+1)? In your subset of size n+1 must there be a pair of the form {b, b+1}?

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Thank you both, but I solved it using pigeonhole principle and the fact that gcd(n,n+1)=1

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but the first counter example of rodigee is correct since the property this is true if and only if the set elements can be ordered sucht that the diference between consective numbers is equal to 1

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cause n+1 numbers in 1..2n includes 2 consecutive ones, whose gcd is 1

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