0

I was showing someone an example of a proof where contrapositive is the way to go. Clearly, I cooked up the wrong implication. In any event, now I want to complete the proof regardless. And, do so by showing the contrapositive even though I know invoking a contradiction is easier in this case. I want to show, let $a \neq 0 \neq b$, $$ a \not\in \mathbb{Q} \wedge b \in \mathbb{Q} \Rightarrow ab \not\in \mathbb{Q} $$

Then the contrapositive will be, $$ ab \in \mathbb{Q} \Rightarrow a \in \mathbb{Q} \vee b \not\in \mathbb{Q} $$ We start by assuming $ab \in \mathbb{Q}$ then by definition of rationals, $$ ab = \frac{p}{q}, \quad \gcd(p,q)=1, \quad (p,q) \in \mathbb{Z}^2 $$ I am bit confused from here, I tried, $$ a = \frac{p}{bq}, \quad b = \frac{p}{aq} $$ But how does this help me show $a \in \mathbb{Q} \vee b \not\in \mathbb{Q}$?

scribe
  • 576
  • How is the contrapositive wrong? Could you explain a little more? Maybe write the original statement and then show me the correct contrapositive. – scribe Jan 31 '23 at 07:35
  • The contrapositive you have written is correct and $a=b=0$ is not a counterexample. – Anne Bauval Jan 31 '23 at 07:45
  • 2
    "The contrapositive you have written is not correct.", "The contrapositive you have written is correct", guys, I said no contradictions! :D – scribe Jan 31 '23 at 07:49

1 Answers1

1

You assume $ab\in\Bbb Q$ and $b\ne0,$ and want to prove $a \in \mathbb{Q} \vee b \not\in \mathbb{Q},$ i.e. $$b\in \mathbb{Q}\implies a\in \mathbb{Q}.$$ Simply assume moreover that $b\in \mathbb{Q}$ and write $a=\frac{ab}b.$

Anne Bauval
  • 49,005
  • How does $ab \in \mathbb{Q} \Rightarrow b \in \mathbb{Q}$? Why can I just assume it? – scribe Jan 31 '23 at 08:14
  • You want to show that "$a\in \mathbb Q\vee b\not\in \mathbb Q$" Annes answer suggests to substitute this by its classically equivalent "If not $b\not\in \mathbb Q$ then $a\in \mathbb Q$". – Evgeny Kuznetsov Jan 31 '23 at 08:23
  • Well, I am not sure how to get from $ab \in \mathbb{Q}$ to $b \in \mathbb{Q}$, since $ab \in \mathbb{Q}$ can be made true even when both $a$ and $b$ are irrational. Take $\sqrt{2}\times \sqrt{8}=4$. – scribe Jan 31 '23 at 08:33
  • I did nothing more than (very naively!) 1) consider $P\Rightarrow Q$ as equivalent by definition to $Q\lor\neg P$ and apply this to $P=(b\in\Bbb Q)$ and $Q=(a\in\Bbb Q)$; 2) In order to prove $P\Rightarrow Q$, assume $P$ and try to prove $Q$ (like you intended to do in order to prove $ab \in \mathbb{Q} \Rightarrow (a \in \mathbb{Q} \vee b \not\in \mathbb{Q})$). – Anne Bauval Jan 31 '23 at 09:00
  • I might be missing something subtle here. Our whole thing comes down to proving, $$ab \in \mathbb{Q} \Rightarrow b \in \mathbb{Q} \Rightarrow a \in \mathbb{Q}.$$ We assume $ab \in \mathbb{Q}$ and now have to show some intermediate steps via forward reasoning that $b \in \mathbb{Q}$ then having shown $ab \in \mathbb{Q} \Rightarrow b \in \mathbb{Q}$, the rest $a \in \mathbb{Q}$ follows. What are the explicit-intermediate steps that take us from $ab \in \mathbb{Q}$ to $b \in \mathbb{Q}$? – scribe Feb 01 '23 at 00:45
  • "Our whole thing comes down to proving"$$ab \in \mathbb{Q} \Rightarrow( b \in \mathbb{Q} \Rightarrow a \in \mathbb{Q}),$$not$$(ab \in \mathbb{Q} \Rightarrow b \in \mathbb{Q}) \Rightarrow a \in \mathbb{Q}.$$So, I assumed $ab \in \mathbb{Q} $ (like you) and moreover $b \in \mathbb{Q}$ (similarly), and deduced $a\in \mathbb{Q}$ from these 2 assumptions. – Anne Bauval Feb 01 '23 at 06:14