Respuesta :
[tex]\begin{gathered} \text{For all }a,b\text{ in the set of integers} \\ a\times b\text{ is in the set of integers} \end{gathered}[/tex][tex]\text{ This means that the operation }\times\text{ is closed on the set of integers}[/tex]
Also,
[tex]\begin{gathered} \text{For all }x,y\text{ in the set of integers} \\ x+y\text{ is in the set of integers} \end{gathered}[/tex][tex]\text{ This means that the operation }+\text{ is closed on the set of integers}[/tex]We will now continue the proof and finish it.
From the image, we are at the point where we have
[tex]x+y=\frac{p_1q_2+p_2q_1}{q_1q_2}[/tex][tex]\begin{gathered} p_1,p_2,q_1,q_2\in Z \\ \text{ Then} \\ p_1q_2,p_2q_1\in Z \\ \end{gathered}[/tex]Hence,
[tex]p_1q_2+p_2q_1\in Z[/tex][tex]\begin{gathered} \text{ Since} \\ q_1\ne0_{} \\ \text{and} \\ q_2\ne0 \\ \text{ then} \\ q_1q_2\ne0 \\ \text{and} \\ q_1q_2\in Z \end{gathered}[/tex]Therefore, x + y is rational
