we know prior that $[a_1,...,a_n](\frac{p}{a_1},...,\frac{p}{a_n})=|a_1...a_n|$ that $p=a_1...a_n$,$a_i \in \Bbb N$ now

how to prove $\forall n\in \Bbb N , n\ge 2$ exist $n$ number as $a_1,...,a_n$ such that $\gcd(a_1,...,a_n)=1$ but all of proper subset of ${a_1,...,a_n}$ is not relatively prime.

for example: $\gcd(6,10,15)=1$ but $\gcd(10,15) \not = 1$,$\gcd(10,6) \not = 1$,$\gcd(6,15) \not = 1 $

  • $\begingroup$ Your 'statement' quantifies the same variable twice with different quantifiers: $(\forall n\in \Bbb N_{>2})(\exists n)\ldots$ There's something wrong. $\endgroup$ – Git Gud Mar 16 '13 at 21:11
  • $\begingroup$ @GitGud: It’s okay: the existential quantifier applies to $a_1,\dots,a_n$. $\forall n\ge 2\exists a_1,\dots,a_n$ $\endgroup$ – Brian M. Scott Mar 16 '13 at 21:14
  • $\begingroup$ @BrianM.Scott I don't get it. Is $(\forall n\in \Bbb N)(\exists n\in \Bbb N)\bigl(P(n)\bigr)$ a statement? $\endgroup$ – Git Gud Mar 16 '13 at 21:17
  • $\begingroup$ @GitGud: You’re not reading it correctly. It’s not $(\forall n\in\Bbb N)(\exists n\in\Bbb N)\big(P(n)\big)$; it’s $$(\forall n\in\Bbb N)\Big(n\ge 2\to\exists a_1,\dots,a_n\in\Bbb Z^+\big(P(n)\big)\Big)\;.$$ $\endgroup$ – Brian M. Scott Mar 16 '13 at 21:19
  • $\begingroup$ @BrianM.Scott Thanks. The singular number mislead me completly. $\endgroup$ – Git Gud Mar 16 '13 at 21:20

HINT: Let $p_1,\dots,p_n$ be distinct prime numbers. What happens if you look at products of $n-1$ of these primes? In your example, for instance, you’re looking at $2\cdot3$, $2\cdot5$, and $3\cdot5$.

  • $\begingroup$ it's true for $n-1$ element subset .for $i$ element ?($i=2,...,n-2$) how to work? $\endgroup$ – agustin Mar 16 '13 at 21:37
  • $\begingroup$ @agustin: You don’t need to look at any other size subset: $\{p_1,\dots,p_n\}$ has $n$ subsets of size $n-1$, which you can use to construct $a_1,\dots,a_n$. $\endgroup$ – Brian M. Scott Mar 16 '13 at 21:38

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