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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?

Thank you for your time,

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    $\begingroup$ The product of two integers is an integer; the difference of two integers is an integer; a rational is defined as one integer divided by another non-zero integer. $\endgroup$ – Emily Jan 26 '15 at 21:07
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    $\begingroup$ @Arkamis Yes, I totally understand this. What I'm curious about does not concern what you've answered. Why is it that this proves the difference between two integers is an integer when indeed the proof is utilizing the very process I'm attempting to prove. Therefore, it doesn't prove anything. And also, I apologize for not making it clear that I was aware that the set of integers was closed under the operations of subtraction, addition and multiplication. Therefore, your comment is of littler to no use to me. But I should have been more clear. My apologies. $\endgroup$ – jm324354 Jan 27 '15 at 2:12
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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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    $\begingroup$ I totally understand everything here -- that the product of two integers is an integer and the subtraction and addition of two integers is an integer (hence this set is closed under those operations). But I must prove that the difference of two integers is an integer. Yet this difference requires that I use the difference of two integers? How does that prove anything? It simply utilizes the process I'm attempting to prove. $\endgroup$ – jm324354 Jan 27 '15 at 2:09
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    $\begingroup$ You're trying to prove that the difference between rational is a rational. Integers are not rational. They are a more fundamental case. $\endgroup$ – Emily Jan 27 '15 at 2:11
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    $\begingroup$ Ahh this is my error. Sorry I just got back from a class and did not re-read the question I had posed. My apologies if you have interpreted what I've said in a negative way. Just a bit distraught about a professor I have. Thank you for your time. +1 to everything you've written for your trouble. $\endgroup$ – jm324354 Jan 27 '15 at 2:13

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