Question

Im confused on this question for Discrete Mathematics. Let A_{2} be the set of all multiples of 2 except for 2. Let A_{3} be the set of all multiples

Discrete math
Im confused on this question for Discrete Mathematics.
Let $$\displaystyle{A}_{{{2}}}$$ be the set of all multiples of 2 except for 2. Let $$\displaystyle{A}_{{{3}}}$$ be the set of all multiples of 3 except for 3. And so on, so that $$\displaystyle{A}_{{{n}}}$$ is the set of all multiples of n except for n, for any $$\displaystyle{n}\geq{2}$$. Describe (in words) the set $$\displaystyle{A}_{{{2}}}\cup{A}_{{{3}}}\cup{A}_{{{4}}}\cup\ldots$$.

2020-12-26
Step 1
Given that $$\displaystyle{A}_{{{2}}}$$ is the set of all multiples of 2 except for 2. In this set all the even numbers except 2 are covered.
Similarly, $$\displaystyle{A}_{{{n}}}$$ is defined such that it is the set of all multiples of n except for n.
Now consider a composite number m.
Then $$\displaystyle{m}={a}{b}$$ with $$\displaystyle{\left({a},{b}\right)}={1}$$
Then the number m belongs to the set $$\displaystyle{A}_{{{a}}}$$ and $$\displaystyle{A}_{{{b}}}$$
Hence it belong to the set $$\displaystyle{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\ldots$$ And thus it cannot belong the set $$\displaystyle{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\ldots$$
Step 2
On the other hand, consider a prime number p.
Then it cannot be written as a multiple of any element other than p and 1.
By the definition of the set $$\displaystyle{A}_{{{p}}}$$, p does not belong to $$\displaystyle{A}_{{{p}}}$$.
which gives that p belongs to $$\displaystyle{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\ldots$$
Finally, the set of all odd primes will constitute the set $$\displaystyle{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\ldots$$