Set closed under multiplication I'm struggling with some very basic logic, I guess it is in the simplest question that will probably exist on stack exchange.

Define a set $N_{2,3} = \left\{ n \in \mathbb{Z} : 2 \nmid n \text{ and } 3 \nmid n \right\}$. Then prove it is closed under multiplication.

I was trying to use the contrapositive, and had the following solution;

By considering the contrapositive, we seek elements $n \in \mathbb{Z} : 2\mid n \text{ or } 3\mid n$. So given two elements $a,b \in N_{2,3}$ we seek to prove that $ab\in N_{2,3}$. Suppose then w.l.o.g. that $2\mid a$, then we can say $a=2c$ for some $c\in\mathbb{Z}$. Now $ab = (2c)b = 2(cb)$, and clearly $2\mid ab$, hence $ab \in N_{2,3}$.

Something just doesn't feel right with all of this, but I can't put my finger on it. I was also thinking about just saying that taking two elements $a,b \in N_{2,3}$, neither $3$ nor $2$ divide $n$ or $m$, therefore their product $ab$ is also not divisible by $3$ or $2$, hence the set is closed under multiplication.

Thanks for any help!!
By the set construction: "If $2\nmid ab$ and $3\nmid ab$, then $ab\in N_{2,3}$ ".

The contraposition is that: "If $ab\notin N_{2,3}$, then $2\mid ab$ or $3\mid ab$".

However, as you have shown, for any $a,b$ that are both $\in N_{2,3}$ it will always be so that $2\nmid ab$ and $3\nmid ab$.  

So every such $ab$ must be in $N_{2,3}$.



tl;dr  When considering the contrapositive you do not want to find a case were the consequent holds. &nbsp; You seek to deny that it ever can (for any product of elements from the set).
This is actually so stupid of a logic problem youll kick yourself a little bit.

This is the list of all numbers that you can write as prime factorizations without using a 2 or 3.  All numbers can be written as a prime factorial.  Thus any number that is in the set written without 2 or 3 is in the set of multiples.

Lets to 35 as an easy example:
not divisible by 2, nor 3... Check
(1^1)(5^1)(7^1)=35.  35 not divisible by 3, nor is 1,5, or 7 check.

Now lets say we want to know if 385 is in our set.

well 385 cannot be divided by 3 nor 2 check
its also 35*11 =  (1^1)(5^1)(7^1)*(11^1)

do you see whats going on yet?

no matter what numbers come after you cannot write the numbers in the set without 2 or 3 as a prime factorization.  Thus the resulting manipulation is inconsequential.  The set is closed under multiplication because it must be for that prime factorization to exist