False

Explanation:

Definition: A subgroup H of a group G is a proper subgroup if H is not equal to G.

Definition: A subgroup H of a group G is cyclic if there exists an element a in G such that H = ⟨a⟩, the subgroup generated by a.

To show that all non-trivial proper subgroups of (R, +) are not cyclic, we need to find a counterexample, i.e., a non-trivial proper subgroup that is not cyclic.

Counterexample:

Consider the subgroup H = {0, 1}, where 0 and 1 are elements of the group (R, +).

1. H is a subgroup of (R, +) because it satisfies the subgroup criteria:
- Closure: For any two elements a, b in H, a + b = 0 + 1 = 1, which is also in H.
- Identity: 0 is the identity element of (R, +), and 0 is in H.
- Inverses: For any element a in H, -a is also in H since -a + a = 0, the identity element.

2. H is a proper subgroup of (R, +) because it is not equal to (R, +).

3. H is not cyclic because there is no element a in (R, +) such that H = ⟨a⟩. In fact, the subgroup H only contains two elements, 0 and 1, and it is not possible to generate all real numbers using just 0 and 1 under addition.

Therefore, the subgroup H = {0, 1} is a non-trivial proper subgroup of (R, +) that is not cyclic.

Conclusion:

Since we have found a counterexample, it shows that not all non-trivial proper subgroups of (R, +) are cyclic. Hence, the statement "All non-trivial proper subgroups of (R, +) are cyclic" is false.