What does disjoint mean?

Disjoint sets mean that they have no elements in common.

easy u have $3\cdot4=12$ and done

Lionking212 wrote: easy u have $3\cdot4=12$ and done What if $3,4$ are in the same set while $12$ is in the other?

Oh i didn’t see that.

Assume the contrary that there exist a way to partition that doesn't satisfy the condition. Let the two disjoint sets be $A,B$. WLOG, let $3\in A$. $3\cdot 3=9$, so $9\in B$ $9\cdot 9=81$, so $81\in A$ $3\cdot 81=243$, so $243\in B$ $9\cdot 27=243$, so $27\in A$ But $3\cdot 27=81$, a contradiction.

Quidditch wrote: Assume the contrary that there exist a way to partition that doesn't satisfy the condition. Let the two disjoint sets be $A,B$. WLOG, let $3\in A$. $3\cdot 3=9$, so $9\in B$ $9\cdot 9=81$, so $81\in A$ $3\cdot 81=243$, so $243\in B$ $9\cdot 27=243$, so $27\in A$ But $3\cdot 27=81$, a contradiction. I thought $a,b,c$ aren't pairwise distinct.

jasperE3 wrote: Quidditch wrote: Assume the contrary that there exist a way to partition that doesn't satisfy the condition. Let the two disjoint sets be $A,B$. WLOG, let $3\in A$. $3\cdot 3=9$, so $9\in B$ $9\cdot 9=81$, so $81\in A$ $3\cdot 81=243$, so $243\in B$ $9\cdot 27=243$, so $27\in A$ But $3\cdot 27=81$, a contradiction. I thought $a,b,c$ aren't pairwise distinct. Sorry, I don’t quite understand what you’re trying to say…

Let's assume we partitioned the numbers into two disjoint set $A$ and $B$. For the sake of contradiction, we assume that the given condition is false. WLOG, $3 \in A$. Hence, $3^2 \not \in A \implies 3^2 \in B$. So, $3^2 \times 3^2 = 3^4 \not \in B \implies 3^4 \in A$. So, $3^5 \not \in A$. So, $3^5 \in B$. $3^3$ is not in any of the sets as we assumed the given condition is false. So, contradiction.