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Working on Lang, Serge. "Basic Mathematics" (p. 39, ex. 4).

Let $a = m/n$ be a rational number expressed as a quotient of integers $m, n$ with $m \neq 0$ and $n \neq 0$. Show that there is a rational number $b$ such that $ab = ba = 1$.

Attempted Proof:

$ \def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \def\Ae#1{\qquad\mathbf{\forall E} \: #1 \\} \def\Ai#1{\qquad\mathbf{\forall I} \: #1 \\} \def\Ee#1{\qquad\mathbf{\exists E} \: #1 \\} \def\Ei#1{\qquad\mathbf{\exists I} \: #1 \\} \def\R#1{\qquad\mathbf{R} \: #1 \\} \def\ci#1{\qquad\mathbf{\land I} \: #1 \\} \def\ce#1{\qquad\mathbf{\land E} \: #1 \\} \def\ii#1{\qquad\mathbf{\to I} \: #1 \\} \def\ie#1{\qquad\mathbf{\to E} \: #1 \\} \def\be#1{\qquad\mathbf{\leftrightarrow E} \: #1 \\} \def\bi#1{\qquad\mathbf{\leftrightarrow I} \: #1 \\} \def\qi#1{\qquad\mathbf{=I}\\} \def\qe#1{\qquad\mathbf{=E} \: #1 \\} \def\ne#1{\qquad\mathbf{\neg E} \: #1 \\} \def\ni#1{\qquad\mathbf{\neg I} \: #1 \\} \def\IP#1{\qquad\mathbf{IP} \: #1 \\} \def\x#1{\qquad\mathbf{X} \: #1 \\} \def\DNE#1{\qquad\mathbf{DNE} \: #1 \\} $ $ \fitch{1.\, \forall x(x \in \mathbb{Q} \leftrightarrow \exists p\exists q(x = p/q \land q \neq 0)}{ 2.\,a \in \mathbb{Q} \leftrightarrow \exists p\exists q(a = p/q \land q \neq 0) \Ae{1} \fitch{3.\, a \in \mathbb{Q}}{ 4.\,\exists p\exists q(a = p/q \land q \neq 0) \be{2,3} \fitch{5.\, \exists q(a = m/q \land q \neq 0)}{ \fitch{6.\, a = m/n \land n \neq 0}{ \vdots\\ b \in \mathbb{Q} \leftrightarrow \exists p\exists q(b = p/q \land q \neq 0) \Ae{1} \fitch{b \in \mathbb{Q}}{ \exists p\exists q(b = p/q \land q \neq 0) \be{} \fitch{k.\, \exists q(b = n/q \land q \neq 0)}{ \vdots\\ }\\ \vdots\\ }\\ \vdots\\ }\\ \vdots\\ }\\ \vdots\\ }\\ \vdots\\ \exists x(x \in \mathbb{Q} \land ax=xa=1) } $

The solution given by the author, is:

Let $b = n/m$ (this is a rational number). Then $ab = m/n \cdot n/m = 1$; $ba = n/m \cdot m/n = 1$.

I have two questions:

  • Step k (attempted proof) is obviously incorrect since the instantiated variable, namely n, already appears in an undischarged assumption. How can I continue the proof and fix that error ?
  • In the author solution, what would be the logic rule for introducing the equality $b=n/m$ ?

P.S.: I am already aware of the question posted here (Basic Mathematics - Proofs - Proving rational numbers are equivalent to 1), but it does not answer my specific questions. I will appreciate any insights.

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I think you're starting way too far back. The assumptions you are given are $a\in \mathbb Q$, $m,n\in\mathbb Z$, $m\ne0$, $n\ne0$, and $a=m/n$. So everything you're doing before line 6 is unneeded instantiation. I think that answers your first question.

For your second question, you start by introducing $x=n/m$ as an assumption. Then you show that $x\in\mathbb Q\wedge ax=1\wedge xa=1$ inside that block using whatever facts of rational number existence and rational multiplication you're working with. Finally, by $\exists I$, you can end that block and infer that $\exists b (b\in\mathbb Q\wedge ab=1\wedge ba=1)$. Obviously, Serge is conflating $b$ and $x$ in his suggestion for the sake of human-readability.

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  • $\begingroup$ I appreciate your answer, @Matthew Daly. I reworked it based on what you suggest. My question is: how can I discharge the assumption $b=n/m$ in line 4 to fill line k ? $$ \def\fitch#1#2{\quad\begin{array}{|l}#1\\\hline#2\end{array}} \fitch{1.\, \forall x(x \in \mathbb{Q} \leftrightarrow \exists p\exists q(x = p/q \land q \neq 0)\\ 2.\,a \in \mathbb{Q} \\ 3.\, a = m/n \land m \neq 0 \land n \neq 0}{ \fitch{4.\, b = n/m}{ \vdots\\ \exists x(x \in Q \land ax=xa=1) }\\ k. ? } $$ $\endgroup$
    – F. Zer
    Commented May 10, 2020 at 13:00

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