Write theorem's "scratch work" in Velleman's How to prove it 'given–goal diagrams' style

Question

I want to apply the proof strategies and overall scratch work diagram based framework introduced in the book "How to Prove It" in order to rewrite the following theorem:

Theorem Any positive rational number has an expression as a fraction in lowest form.

Proof. First write a given positive rational number as a quotient of positive integers $\frac{m}{n}$. We know that 1 is a common divisor of $m$ and $n$. Furthermore, any common divisor is at most equal to $m$ or $n$. Thus among all common divisors there is a greatest one, which we denote by $d$. Thus we can write $$ m = dr \qquad \text{and} \qquad n = ds $$ with positive integers $r$ and $s$. Our rational number is equal to $$ \frac{m}{n} = \frac{dr}{ds} = \frac{r}{s} $$

All we have to do now is to show that the only common divisor of $r$ and $s$ is $1$. Suppose that $e$ is a common divisor which is greater than 1. Then we can write $$ r = ex \qquad \text{and} \qquad s = ey $$ with positive integers $x$ and $y$. Hence $$ m = dr = dex \qquad \text{and} \qquad n = ds = dey $$ Therefore $de$ is a common divisor for $m$ and $n$, and is greater than $d$ since $e$ is greater than $1$. This is impossible because we assumed that $d$ was the greatest common divisor of $m$ and $n$. Therefore $1$ is the only common divisor of $r$ and $s$, and our theorem is proved. $\blacksquare$

The first thing that I did was to rewrite the statement using logical connectives and quantifiers $$ \forall a: (a \in \mathbb{Q}_+) \rightarrow \exists \ r,s : (r,s \in \mathbb{Z}_+) \ \land (a = r/s) \ \land (\operatorname{gcd}(r,s) = 1). $$ making it clear that we are trying to prove a conditional statement.

So we should assume that $(a \in \mathbb{Q}_+)$ and our scratch work diagram becomes \begin{array}{c c c c c} \text{Givens} & & \qquad \qquad \qquad \qquad \qquad & & \text{Goal} \\ \hline (a \in \mathbb{Q}_+) & & & & \exists \ r,s : (r,s \in \mathbb{Z}_+) \ \land (a = r/s) \ \land (\operatorname{gcd}(r,s) = 1) \\ \end{array}

How should I proceed from here? This seems like an existence proof using a proof by contradiction at some point, and while the original proof makes sense to me, I'm having trouble factoring it in small steps using this framework.

Answer 1

Our goal is the universal conditional statement

$$ \forall a \; \biggl[\; a \in \mathbb{Q}_+ \; \rightarrow \; \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr)\; \biggr]$$ with no additional assumptions.

Scratch work

The starting givens-goals diagram is \begin{array}{c c c} \text{Givens} & \qquad \qquad & \text{Goal} \\ \hline & & \forall a \; \biggl[\; a \in \mathbb{Q}_+ \; \rightarrow \; \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr)\; \biggr] \end{array}

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We introduce the new variable $a$ into the proof to stand for an arbitrary object

\begin{array}{c c c} \text{Givens} & \qquad \quad \qquad \qquad & \text{Goal} \\ \hline & & a \in \mathbb{Q}_+ \; \rightarrow \; \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr) \end{array}

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Since our goal is a conditional statement we begin by assuming that the antecedent is true

\begin{array}{c c c} \text{Givens} & \qquad \qquad \qquad \qquad \qquad \qquad & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr) \end{array}

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Assuming that $a \in \mathbb{Q}_+$ allows us to use the definition of positive rational number

\begin{array}{c c c} \text{Givens} & & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr) \\ \exists \ m \ \exists \ n \; \Bigl( \; m,n \in \mathbb{Z}_+ \; \land \; a = m/n \;\Bigr) & & \end{array}

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We let $m_0, n_0$ stand for positive integers such that $a = m_0 /n_0$ (existential instantiation)

\begin{array}{c c c} \text{Givens} & \qquad \qquad \qquad \qquad & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr) \\ m_0, n_0 \in \mathbb{Z}_+ & &\\ a = m_0/n_0 & & \\ \end{array}

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We let $d$ stand for $\operatorname{gcd}(m_0,n_0)$

\begin{array}{c c c} \text{Givens} & \qquad \qquad \qquad & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr) \\ m_0, n_0 \in \mathbb{Z}_+ & &\\ a = m_0/n_0 & & \\ d = \operatorname{gcd}(m_0,n_0) && \end{array}

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Let produce positive integers $r,s$

\begin{array}{c c c} \text{Givens} & \qquad \qquad \qquad \qquad \qquad \qquad & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \\ m_0, n_0 \in \mathbb{Z}_+ & &\\ a = m_0/n_0 & & \\ d = \operatorname{gcd}(m_0,n_0) && \\ r = m_0/d && \\ s = n_0/d && \end{array}

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We treat the conjunction as separate goals

\begin{array}{c c c} \text{Givens} & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & r,s \in \mathbb{Z}_+\\ m_0, n_0 \in \mathbb{Z}_+ & & a = r/s\\ a = m_0/n_0 & & \operatorname{gcd}(r,s) = 1\\ d = \operatorname{gcd}(m_0,n_0) && \\ r = m_0/d && \\ s = n_0/d && \end{array}

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The try to prove $\operatorname{gcd}(r,s) = 1$ by contradiction

\begin{array}{c c c} \text{Givens} & \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad & \text{Goal} \\ \hline a \in \mathbb{Q}_+ & & r,s \in \mathbb{Z}_+\\ m_0, n_0 \in \mathbb{Z}_+ & & a = r/s\\ a = m_0/n_0 & & \textrm{contradiction}\\ d = \operatorname{gcd}(m_0,n_0) && \\ r = m_0/d && \\ s = n_0/d && \\ \operatorname{gcd}(r,s) \neq 1 && \end{array}

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Form of final proof:

Let $a$ be arbitrary

$\quad$ Suppose $a \in \mathbb{Q}_+$. Let $m_0, n_0 \in \mathbb{Z}_+$ and $a = m_0 / n_0$. Let $d = \operatorname{gcd}(m_0,n_0)$ .

$\quad$$\quad$ Let $r = m_0/d$ and $s = n_0/d$.

$\quad$$\quad$$\quad$ [Proof of $r,s \in \mathbb{Z}_+$ goes here.]

$\quad$$\quad$$\quad$ [Proof of $a = r/s$ goes here.]

$\quad$$\quad$$\quad$ Suppose $\operatorname{gcd}(r,s) \neq 1$

$\quad$$\quad$$\quad$$\quad$ [Proof of contradiction goes here.]

$\quad$$\quad$$\quad$ Therefore $\operatorname{gcd}(r,s) = 1$

$\quad$$\quad$ Thus, $\exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr)$.

$\quad$ Therefore $a \in \mathbb{Q}_+ \; \rightarrow \; \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr)$.

Since $a$ was arbitrary, we can conclude that $\forall a \; \biggl[\; a \in \mathbb{Q}_+ \; \rightarrow \; \exists \ r \ \exists \ s \; \Bigl( \; r,s \in \mathbb{Z}_+ \; \land \; a = r/s \; \land \; \operatorname{gcd}(r,s) = 1 \;\Bigr)\; \biggr].$