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.