Prove that group $G=\{n_1r_1+n_2r_2 | n_1,n_2\in\mathbb{Z}\}$ is cyclic

Question

$r_1$ and $r_2$ are rational numbers. Prove that the group $G=\{n_1r_1+n_2r_2 | n_1,n_2\in\mathbb{Z}\}$ under addition is cyclic. Generalize to the case where you have $r_1, r_2,....,r_k$ rationals.

I know that for a group to be cyclic, there must exist a $a$ such that $\langle a \rangle =G$. But then I don't know how to proceed.

Answer 1

Let $r_1 = \dfrac{a}{b}$ and $r_2 = \dfrac{c}{d}$.

Then $G=\{n_1r_1+n_2r_2 \mid n_1,n_2\in\mathbb{Z}\} = \left\{\dfrac{n_1ad+n_2cb}{bd} \;\middle|\; n_1,n_2\in\mathbb{Z}\right\}$

By Euclid's Algorithm, for any integers $A$, $B$ there exist integers $m_1$, $m_2$ such that $m_1A + m_2B =\gcd(A, B)$ (where “gcd” means greatest common divisor).

So $G= \left\{n\dfrac{\gcd(ad, cb)}{bd} \;\middle|\; n\in\mathbb{Z}\right\} = \left\langle\dfrac{\gcd(ad, cb)}{bd}\right\rangle$