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Let $q\in \mathbb{Q}$ and $r\in \mathbb{R}$ \ $\mathbb{Q}$, then $r\cdot q$ is irrational.

$r\in \mathbb{R}$ \ $\mathbb{Q}$ $\Rightarrow r\in\mathbb{R}$ such that $r\notin\mathbb{Q} $

Thus, $r$ is irrational.

Proof through contradiction: let's assume $irrational\cdot rational=rational$

Let $q=\frac { a }{ b } $ such that $a,b\in\mathbb{Z}$ and $b\neq 0$ let $m,n\in\mathbb{Z}$ such that $n\neq0$

$$irrational\cdot rational=rational \Rightarrow r\cdot \frac { a }{ b } =\frac { m }{ n } $$

$$(\frac { b }{ a } )\frac { a }{ b } \cdot r=\frac { m }{ n } (\frac { b }{ a } )$$

$r=\frac { mb }{ na } \Rightarrow $ $r$ must be rational since $mb$ and $na$ are integers.

This leads to a contradiction in our assumption. This means our assumption must be false. So, an $irrational\cdot rational$ must be irrational.

Is my proof any good? If not, how can I fix it or improve it?

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  • $\begingroup$ Fine except you have to handle $q=0$ separately, otherwise you're forming by $\frac b 0$. $\endgroup$ – BrianO Oct 16 '15 at 17:59
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Your proof is $99\%$ correct.

One correction is that your proof does not cover the possibility that $a=0$. In fact, your original statement is FALSE if $q=0$, so I assume you meant to say that $q$ can be anything except $0$.

Just a minor cosmetic correction, I would not use as many repetitions of the word "rational".

I would say:

Let's assume that there is some rational number and some irrational number such that they multiply to a rational number. Then $$x\cdot q = p$$ where $x$ is irrational, and $p,q$ are rational. Because $p,q$ are rational, there exist $a,b,m,n\in\mathbb Z$ such that $b,n\neq 0$ and $p=\frac ab$ and $q=\frac mn$.

Then, continue with your proof as-is.


For most mathematitians, this will be easier to read since we don't like full words in equations. Also, it's nice to use a disticntly different symbol for an irrational number (i.e., call the rationals $p,q,r,\dots$, but call the irrationals $x,y,z,\dots$). This is just convention, but it improves readability.

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  • $\begingroup$ Hmm. I think I did too much work then. The directions state "Prove, or disprove using a counter example the following claim:". Obviously if $q$ is $0$, that is a valid counter example. Right? $\endgroup$ – Cherry_Developer Oct 16 '15 at 16:49
  • $\begingroup$ @Cherry_Developer Well, to provide a counter example, you also need to provide an irrational number. But that's easy, since any one will do. $\endgroup$ – 5xum Oct 16 '15 at 16:51
  • $\begingroup$ So my answer is wrong, since there can be a $q$ that disproves this claim. Correct? $\endgroup$ – Cherry_Developer Oct 16 '15 at 16:52
  • $\begingroup$ @Cherry_Developer Correct. $\endgroup$ – 5xum Oct 16 '15 at 16:53
  • $\begingroup$ If $q=0$ and $r=\sqrt { 2 } $, $r\cdot q=0$ which is rational. Is this good? $\endgroup$ – Cherry_Developer Oct 16 '15 at 17:32
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Almost, you should probably handle the possibility that $a=0$ separately.

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