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
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asked 2020-12-25
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 \(\displaystyle{A}_{{{2}}}\cup{A}_{{{3}}}\cup{A}_{{{4}}}\cup\ldots\).

Answers (1)

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 \(\displaystyle{A}_{{{1}}}\cup{A}_{{{2}}}\cup{A}_{{{3}}}\cup\ldots\)
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