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Proposition : $\nexists{x}\in\mathbb{Q}\;|\;x^2=12$.

Hypothesis : We can rewrite $x$ like $\left(\frac{p}{q}\right)^2$ where $\forall{p}\in\mathbb{Z}$ and $\forall{q}\in\mathbb{N}$ and $\left(\frac{p}{q}\right)$ can't be reduced ($p$ and $q$ can't be both even).

Demonstration : We have to reject the hypothesis. $$\left(\frac{p}{q}\right)^2=12 \rightarrow \frac{p^2}{q^2}=12 \rightarrow p^2=12q^2.$$ That means that $p^2$ must be even. That can be possible only if $p$ is even too. Therefore, we can rewrite $p$ like $2k$ where $\forall{k}\in\mathbb{Z}$. $$p^2=(2k)^2=4k^2=12q^2\\k^2=3q^2$$

And there's where I stuck!

Can your help me?

hlapointe
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  • Hint: you can conclude that $k$ must be a multiple of 3, so $k=3m$ for some $m\in\mathbb{Z}$. – symplectomorphic Jan 16 '15 at 18:35
  • $3|k^2\Rightarrow 3|k$. – Janko Bracic Jan 16 '15 at 18:35
  • Here is a simple proof that if $n$ is not a perfect square, then $\sqrt{n}$ is irrational: http://math.stackexchange.com/questions/471648/sqrt17-is-irrational-the-well-ordering-principle/472040#472040 – marty cohen Jan 16 '15 at 18:39
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    You've got issues with your quantifiers and with your whole hypothesis. Your hypothesis should be : "$x=\frac{p}{q}$ for some $p\in\mathbb{Z}$ and some $q\in\mathbb{N}$ such that the fraction is reduced (if you don't suppose this you won't get the contradiction you are seeking)". – Guest Jan 16 '15 at 18:42

2 Answers2

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Now continue, with 3 instead of two.

Since $k^2 = 3q^2$, $k$ must be divisible by $3$. Writing $k = 3j$, $9j^2 = 3q^2$, or $3j^2 = q^2$, so $q$ must also be divisible by $3$.

marty cohen
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Even/odd won't help here. Try division by $3$: $$p^2=12q^2\implies 3\mid p^2\implies 3\mid p\implies (3k)^2=12q^2\implies 3k^2=4q^2\implies 3\mid 4q^2\implies 3\mid q$$

There's also an alternative way which makes it very easy to see the problem:

If $\frac pq$ is not an integer, then $\bigl(\frac pq\bigr)^2$ also isn't, so the square root must be an integer if it's rational. But $$12=2^2\cdot 3=n^2$$ is clearly impossible since $3\mid n^2\implies 9\mid n^2$. This method works for any square root.

user2345215
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