Thread: Q is not a free abelian group

I have to do the following exercise form the fraleigh:
Show that Q under addition is not a free abelian group (Hint: Show that no two distinct rational numbers n/m and r/s could be contained in a set satisfying condition 2 of theorem 38.1)

Condition 2: X generates G, and n_1*x_1 + n_2*x_2 + ... + n_r*x_r = 0 for n_i in Z if and only if n_1 = n_2 =...= n_r = 0

Originally Posted by Diego

I have to do the following exercise form the fraleigh:
Show that Q under addition is not a free abelian group (Hint: Show that no two distinct rational numbers n/m and r/s could be contained in a set satisfying condition 2 of theorem 38.1)

Condition 2: X generates G, and n_1*x_1 + n_2*x_2 + ... + n_r*x_r = 0 for n_i in Z if and only if n_1 = n_2 =...= n_r = 0

first show that Q cannot be cyclic. then suppose that it's free abelian and let x = n/m and y = r/s be two distinct elements of a basis of Q. see that mrx - sny = 0, which mean that x and y are

not linearly independent. contadiction!