2
$\begingroup$

$0\neq a\in \mathbb Q, b\in \mathbb R \setminus\mathbb Q \text{ (b is irrational)}$ Prove that $\frac a b$ is irrational.

From defintion $a=\frac m n$ such that $m,n\in \mathbb Z, n\neq 0$.

Take the contrapositive: suppose $\frac m {nb}\in \mathbb Q$ prove $\frac m n\notin \mathbb Q$.

Immediate contradiction from defining $m,n\in \mathbb Z, n\neq 0$. Thus $\frac m {nb}$ is irrational.

Well I'm not sure I'm using the contrapositive right and I tried to combine contrapositive with proof by contradiction but I have a feeling I'm wrong...

$\endgroup$
  • $\begingroup$ From $\frac{m}{nb}$ rational, I would try to show that $b$ is rational, contradicting the fact it isn't. $\endgroup$ – André Nicolas Jan 26 '15 at 20:34
2
$\begingroup$

You are definitely mixing up contrapositive and contradiction. It is quite easy to do as a lot of proofs that can be done with one can be done with the other. Many students make this mistake early on in their careers. What you should say is that "Suppose $\dfrac{m}{nb}\in\Bbb Q$, then $b\in \Bbb Q$." The reason is that you want to prove "if $b\in \Bbb R\setminus \Bbb Q$, then $\dfrac{m}{nb}\in\Bbb R\setminus \Bbb Q$." The contrapositive of this statement is the statement I gave above because we want to negate each portion (and reverse the direction). The negation of $\dfrac{m}{nb}\in \Bbb R\setminus \Bbb Q$ is $\dfrac{m}{nb}\in \Bbb Q$, and likewise for the other. Philosophically, your argument is okay but you need to fix it up a little so that you're not mixing up contrapositive and contradiction.

$\endgroup$
  • $\begingroup$ Yeah I knew I was mixing things up... So to finish it we just add a contradiction since $b$ is irrational then the assumption by contrapositive is false and we're done. $\endgroup$ – shinzou Jan 26 '15 at 21:17
  • $\begingroup$ No. Not quite. Don't think about contradiction here. If we proceed by contrapositive, we have that $\dfrac{m}{nb} \in \Bbb Q$. That is, we can write $\dfrac{m}{nb} = \dfrac{p}{q}$ for some integers $p,q$ (obviously $q\neq 0$). Then by cross multiplying, we have that $b = \dfrac{mq}{np}$, i.e. $b\in \Bbb Q$. Hence if $\dfrac{m}{nb}\in \Bbb Q$, then $b\in\Bbb Q$ which is the contrapositive. This in turn says that if $b\not\in\Bbb Q$, then $\dfrac{m}{nb}\not\in\Bbb Q$. $\endgroup$ – Cameron Williams Jan 26 '15 at 21:21
  • $\begingroup$ In a proof by contrapositive, you are "pretending" that you don't know anything. We know that $b$ is irrational but with the way the original statement is worded, we kind of have to "forget" that knowledge. Think of it like this: If it is the case that $\frac{m}{nb}$ is rational, what can I say about $b$? Well the obvious thing is that $b$ has to be rational (by our above argument). So... by contrapositive, if $b$ is not rational, then there is no way that $\frac{m}{nb}$ can be rational. $\endgroup$ – Cameron Williams Jan 26 '15 at 21:24
  • $\begingroup$ Oh right right. Thanks a lot. $\endgroup$ – shinzou Jan 26 '15 at 21:24
  • $\begingroup$ No problem :) Think on it. It takes a while to train yourself to separate contrapositive and contradiction since they are very similar a lot of the time. $\endgroup$ – Cameron Williams Jan 26 '15 at 21:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.