How to show that LCM and GCD of any two elements of a set lies in the set?

Question

Let $A := \{x : x \in \mathbb{N}$ and $x$ divides $30 \}$

$A = \{1, 2, 3, 5, 6, 10, 15, 30\}$

How can I show that for any two elements $(a, b) \in A$, there LCM and GCD also lies in $A$. I have no idea where to begin

Answer 1

The set $A$ is defined with a property, namely, its elements are precisely the divisors of $30$.

Are the LCM and the GCD of two divisors of $30$ a divisor of $30$?

Answer 2

This is true iff $\mid A\mid=\tau (lcm(x:x\in A))$ where $\tau (n)$=Number of positive divisors of $n$

You can use this result which is more applicable for practical usefulness: $\mid A\mid=\tau ($maximum element of A$))$

Answer 3

Let $n$ be a positive integer and define $D_n = \{ x \in \mathbb ZX^+ : x \mid n \}$ be the set of all divisors of $n$.

Let $u, v \in D_n$ and let $g = \gcd(u,v)$. Since $g \mid u$ and $u \mid n$, then $g \mid n$. Hence $\gcd(u,v) \in D_n$.

Since $u, v \in D_n$, then $u \mid n$ and $v \mid n$. So $\operatorname{lcm}(u,v) \mid n$. Hence $\operatorname{lcm}(u,v) \in D_n$.