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**Brjakewa** Question Details

We introduce the number of divisors function d previously. For this function, d: N->N, where d(n) is the number of natural number divisors of n.

A function that is related to this function is the so-called set of divisors function. This can be defined as a function S that associates with each natural number the set of its distinct natural number factors. For example, S(6)={1,2,3,6}.

1.) Does there exist a natural number n such that the absolute value of S(n) = 1? Explain.

2.) Does there exist a natural number n such that the absolute value of S(n) =2? Explain?

3.) Write the output for the function d in terms of the output for the function S. That is, write d(n) in terms of S(n).

4.) Is the following statement true or false? Justify your conclusion.

For all natural numbers m and n, if m doesn't equal n, then S(m) = S(n). If it's true, the justification should be a proof...

5.) Is the following statement true or false? Justify your conclusion.

For all sets T that are subsets of N, there exists a natural number n such that S(n) = T. Again, if it's true the justification should be a proof...

Any help on this would be appreciated. This is a nongraded homework problem purely for my understanding and I have almost no idea how do do it, but I'll include my work below...

1.) I said that there does exist a natural number such that S(n)=1, it is 1 itself, I think, but I'm not sure what that means? 1 is in a set by itself because it's its only divisor?

2.) Along the same lines as part 1, would this S(n)=2,3,5,7,9,11, etc and all the other prime numbers, or is it just the set 2 by itself?

3.) All I got is d(n)=S(n) but I know there is more to it than that.

4.) and 5.) no idea