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The excersise asks us to prove with an Indirect Proof that if $abc$ is an irrational number then at most two of $a,b,c$ are rational numbers.

So we have,

Let $P$ denote "$abc$ is irrational" and let $Q$ denote "at most two of $a,b,c$ are rational numbers". We want to prove: If $P$ then $Q$, but indirectly, i.e. $\neg P => \neg Q$. Formulating the negation of $P$ we have

$\neg P := $ "$abc$ is rational"

But i don't know how to formulate the negation of $Q$. I figured that it should be "at least three of $a,b,c$ are rational numbers", but then $\neg P $ doesn't imply $ \neg Q$ because we could have $a=b=\sqrt2$ and $c=1$ and so

$abc=1\sqrt2 \sqrt2 =2$

which is a rational number even though $a,b$ are irrational and clearly, this disproves $\neg P => \neg Q$. How should I formulate $\neg Q$? And in general, if we have more than one conditions in a statement, which conditions become negatives?

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2 Answers 2

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First of all $P$ implies $Q$ is not the same as $\neg P $ implies $\neg Q$. It is $\neg Q $ implies $\neg P$.

The negation of $Q$ is 'all three of $a,b,c$ are rational'.

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The negation of $P\implies Q$ is $P$ and not $Q$ Thus you need to assume that $abc$ is irrational and all three of them are rational and derive a contradiction out of that.

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