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This is a terribly simple question I’m sure, but I can’t find a work-around in my proof. I must prove that the difference between two rational numbers is thus rational. Here is my attempt:

Let $a$ and $b$ be rational numbers. Therefore, \begin{align}

a=\frac{\lambda}{\beta},\:b=\frac{\xi}{\zeta},\:\ni\:\lambda,\beta,\xi,\zeta\in\mathbb{Z},\tag{1} \end{align} which gives us \begin{align}

a-b=\frac{\lambda}{\beta}-\frac{\xi}{\zeta}=\frac{\lambda\zeta-\xi\beta}{\beta\xi}.\tag{2}

\end{align}

So I have shown that $a$ and $b$ are rational numbers which, by definition, can be represented by the quotient of two integers. But now how do I tackle the problem of the difference? By definition the difference between two integers is an integer. Does that require that this difference is thus rational?

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Thank you for your time,

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$$\begin{align*}

\lambda &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\beta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\xi &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\zeta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\alpha\zeta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\xi\beta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\alpha\zeta – \xi\beta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\beta\zeta &\color{red}\leftarrow \color{red}{\textrm{integer}} \\

\frac{\alpha\zeta-\beta\xi}{\beta\zeta} &\color{red}\leftarrow \color{red}{\frac{\textrm{integer}}{\textrm{integer}}}

\end{align*}$$

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