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Suppose I have a set of rational numbers where elements have denominators are odd and numerators and denominators are co-prime. I need to show that the set is closed under addition. It is clear that the denominator will be odd since the denominator of the sum of the two fractions is the product of the denominators. But now I need to show that the numerator and denominator are co-prime themselves. So let $(m_1,n_1)=(m_2,n_2)=1$ and let $n_i=2k_i+1$. Then $$\frac{m_1}{n_1}+\frac{m_2}{n_2}=\frac{m_1}{2k_1+1}+\frac{m_2}{2k_2+1}=\frac{m_1(2k_2+1)+m_2(2k_1+1)}{2(2k_1k_2+k_1+k_2)+1}$$

So it is clearly closed in regards to the denominator. But how do I procede with showing that $$\gcd{(m_1(2k_2+1)+m_2(2k_1+1),2(2k_1k_2+k_1+k_2)+1)}=1$$

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    $\begingroup$ I think this needs clarification. I take it you aren't reducing your fractions? So $\frac ab +\frac cd = \frac {ad+bc}{bd}$, no? But then we have things like $$\frac 23 +\frac 13=\frac 99$$ $\endgroup$ – lulu Sep 4 '15 at 1:33
  • $\begingroup$ Ok, but if they are in lowest terms, then of course the numerator and denominator are coprime. That's what "lowest terms" means. $\endgroup$ – lulu Sep 4 '15 at 1:37
  • $\begingroup$ So why the clarification? I think it's fine as is? $\endgroup$ – Iceman Sep 4 '15 at 3:15
  • $\begingroup$ I think the point is that the gcd you are asking about is not necessarily 1, but if it isn't 1, then you reduce to lowest terms and you still have an odd denominator. $\endgroup$ – Gerry Myerson Sep 4 '15 at 7:21
  • $\begingroup$ The way I read the question, you are asking "Suppose $\frac ab$ and $\frac cd$ are rational numbers in lowest terms, such that both $b$ and $d$ are odd integers. Show that the unreduced sum, $\frac {ad+bc}{bd}$ is also in lowest terms." That statement, however is false. There are counterexamples. I gave one already, a simpler one is $\frac 12+\frac 12=\frac 22$. $\endgroup$ – lulu Sep 4 '15 at 11:04

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