37 lines
787 B
Markdown
37 lines
787 B
Markdown
#Math #NT #Probability
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# Problem
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Calculate:
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$$
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P(x, y \in \mathbb{N}: gcd(x, y) = 1)
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$$
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# Solution
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Each number has a $\frac 1 p$ chance to be divisible by prime $p$, so the probability that two numbers do not share prime factor $p$ is
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$$
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1 - p^{-2}
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$$
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Therefore, the probability two numbers are coprime is:
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$$
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\prod _{p \in \mathbb{P}} 1 - p^{-2}
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$$
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Since $1 - x = (\frac 1 {1 - x})^{-1} = (\sum _{n = 0}^{\infty} x^n)^{-1}$, we can express the above as:
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$$
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(\prod _{p \in \mathbb{P}} \sum _{n = 0}^{\infty} p^{-2n})^{-1}
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$$
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We can choose any $n$ for $p^{2n}$ for each prime $p$, so by the Unique Factorization Theorem (any natural number can be prime factored one and only one way), we get:
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$$
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(\sum _{n = 0} n^{-2})^{-1} \\
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= (\frac {\pi^2} 6)^{-1} \\
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= \frac 6 {\pi^2}
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$$ |