Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $

share|cite|improve this question
Your 'statement' quantifies the same variable twice with different quantifiers: $(\forall n\in \Bbb N_{>2})(\exists n)\ldots$ There's something wrong. – Git Gud Mar 16 '13 at 21:11
@GitGud: It’s okay: the existential quantifier applies to $a_1,\dots,a_n$. $\forall n\ge 2\exists a_1,\dots,a_n$ – Brian M. Scott Mar 16 '13 at 21:14
@BrianM.Scott I don't get it. Is $(\forall n\in \Bbb N)(\exists n\in \Bbb N)\bigl(P(n)\bigr)$ a statement? – Git Gud Mar 16 '13 at 21:17
@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 $$(\forall n\in\Bbb N)\Big(n\ge 2\to\exists a_1,\dots,a_n\in\Bbb Z^+\big(P(n)\big)\Big)\;.$$ – Brian M. Scott Mar 16 '13 at 21:19
@BrianM.Scott Thanks. The singular number mislead me completly. – Git Gud Mar 16 '13 at 21:20
up vote 1 down vote accepted

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 $3\cdot5$.

share|cite|improve this answer
it's true for $n-1$ element subset .for $i$ element ?($i=2,...,n-2$) how to work? – agustin Mar 16 '13 at 21:37
@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 $a_1,\dots,a_n$. – Brian M. Scott Mar 16 '13 at 21:38

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.