Let A ⊆ Z a subset of the integers such that satisfies the following properties.?
(1) If n, l ∈ A then n + l ∈ A.
(2) If n ∈ A and k ∈ Z then kn ∈ A.
a) Let A ⊆ Z nonempty that satisfies such properties (1) and (2) that contains more than one element, show that exists l_A ∈ N such that A = {jl_A: j ∈ Z}.
B) Let A and l_A the set and the element of a). Suppose l_A is not a prime number neither is 1, show that there is a set B which satisfies properties (1) and (2) such that A ⊊ B ⊊ Z (inclusion strict). Also show that if l_A is a prime number no exists a set B as mentioned above.
More Questions From This User See All
Answers & Comments
Verified answer
(a) Let a be the smallest positive element of A. (Since A contains more than one element, it must contain a non-zero element x. If x is negative, then by property (2) the positive number -x also lies in A.) This number a is just the number l_A:
From property (2) it follows that all multiples ja with j ∈ Z lie in A.
Suppose an non-zero integer b lies in A. Then we can divide b by a and get b = qa + r with integers q,r and 0 ≤ r < a. From properties (1) and (2) we know that r = b - qa must lie in A, and since there is no non-zero element of A that is less than a, we have r = 0.
So A only contains the multiples b = qa + 0 of a.
(b) Let a = l_A = pq with two integers p,q > 1 (which need not necessarily be prime). Then B = { jp : p ∈ Z } contains a = pq, and by property (2) we have B ⊇ A.
However, B ∋ p < a, and so p ∉ A, which means that B ≠ A.
Now suppose a is prime, and that B ⊇ A, but B ≠ A. Let's define b = l_B as before, then b must be either prime or divisible by a prime that is no factor of a. In either case, B contains two prime numbers, a and b (or a prime factor of b).
But any set of integers that contains two prime numbers and fulfils properties (1) and (2) contains all integers (*), and so we have B = Z. This means we can NOT find a suitable set B with A ⊊ B ⊊ Z.
(*) This is a consequence of the Chinese Remainder Theorem.