# If $r_1,r_2$ are rational numbers prove that $G = \{n_1r_1+n_2r_2:n_1,n_2 \in \mathbb Z\}$ forms a cyclic group under addition

If $r_1,r_2$ are rational numbers prove that $G = \{n_1r_1+n_2r_2:n_1,n_2 \in \mathbb Z\}$ forms a cyclic group under addition

Solution Attempt:

If $r_1= \dfrac {a_1 }{b_ 1}$ and $r_2= \dfrac {a_ 2}{b_ 2}$, then an element of $G$ can be written as :

$g = N_1 \dfrac {a_1 }{b_ 1} + N_2 \dfrac {a_2 }{b_ 2} = \dfrac {N_1 a_1b_2 +N_2a_2b_1} {b_1b_2}$ for some $N_1,N_2 \in \mathbb Z$

I must express $g$ as $N_3r_3$ for some natural number $N_3$ and some rational number $r_3$ to prove our result.

EDIT:

I have an intuition that if we take $a_3 = \gcd ( a_1,a_2), b_3 = lcm~ (b_1,b_2),$ then $\dfrac {a_3}{b_3}$ generates $G$. Am I correct? Could anyone please tell me how do I proceed from here?

Thank you very much for your help in this regard.

Hint $\$ The proof for integers (below) works here because the set $\,G\,$ has a least positive element, being discrete, indeed $\,b_1 b_2 G \subset \Bbb Z\,\Rightarrow\, G\subset \frac{1}{b_1 b_2}\Bbb Z.$

Lemma $\ \$ Let $\,\rm S\ne\emptyset \,$ be a set of integers $>0$ closed under subtraction $> 0,\,$ i.e. for all $\rm\,n,m\in S, \,$ $\rm\ n > m\ \Rightarrow\ n-m\, \in\, S.\,$ Then every element of $\rm\,S\,$ is a multiple of the least element $\rm\:\ell = \min\, S.$

Proof $\$ If not there is a least nonmultiple $\rm\,n\in S,\,$ contra $\rm\,n-\ell \in S\,$ is a nonmultiple of $\rm\,\ell.$

Remark $\$ Alternatively $\, b_1 b_2 G\,$ is a nonempty subset of $\,\Bbb Z\,$ closed under subtraction so it is a subgroup of $\,\Bbb Z\,$ so cyclic $= n\,\Bbb Z.\,$ Thus $\,b_1 b_2 G = n\Bbb Z,\,$ so $\, G = (n/b_1 b_2)\,\Bbb Z\,$ is cyclic.

• Thank you for the answer :) . If we take $a_3 = \gcd ( a_1,a_2), b_3 = lcm ~(b_1,b_2),$ then does $\dfrac {a_3}{b_3}$ generates $G$ ? – Shreya Taneja Feb 7 '15 at 16:29
• @ShreyaTaneja See here for gcd and lcm of ratonals. – Bill Dubuque Feb 7 '15 at 16:30

We should look for group generator. I call it c.

Now we look for c:

$$g=n_1\frac{a_1}{b_1}+n_2\frac{a_2}{b_2}$$

I should be able to generate the lowest possible c:

$$n_1\frac{a_1}{b_1}+n_2\frac{a_2}{b_2}=c$$

or

$$n_1 a_1 b_2+n_2 a_2 b_1=c b_1 b_2$$

So:

$$c=\frac{gcd(a_1 b_2,a_2 b_1)}{b_1 b_2}$$

Now I should prove that c can generate $r_1$ and $r_2$

There must be N where:

$$N_1 \frac{gcd(a_1 b_2,a_2 b_1)}{b_1 b_2}=\frac{a_1}{b_1}$$

and

$$N_2 \frac{gcd(a_1 b_2,a_2 b_1)}{b_1 b_2}=\frac{a_2}{b_2}$$

Which are evidently integer:

$$N_1=\frac{a_1 b_2}{gcd(a_1 b_2,a_2 b_1)}$$

$$N_2=\frac{a_2 b_1}{gcd(a_1 b_2,a_2 b_1)}$$

• The last part is "as evident" as could the whole question be. I'd say the OP maybe struggling to see that. – Timbuc Feb 7 '15 at 16:45
• Well, now he surely is. Thanks. +1 – Timbuc Feb 7 '15 at 16:52