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本帖最后由 ╰☆ヾo.海x 于 2013-10-11 05:12 编辑 报告！我已经会了。。。不过跟你们的方法都有点小小小的区别 还是用跟证明$\sqrt{2}\notin\mbb Q$一样的方法不过要用到真命题“如果$q$能被3整除，那么$q^2$也能”
这里我就直接证明标题了。。
Assume $\exists x\in\mbb Q$ such that $x^2=3$, we can express as a simplified fraction $\dfrac{p}{q}$
now $x^2=\dfrac{p^2}{q^2}=3\riff p^2=3q^2\riff p^2$ is divisible by 3
Also, we've proved (其实这一问是在第二小题里的，第一小题就证明了以下真命题，在这里就不证了...) that "if $q$ is not divisible by 3, then neither is $q^2$", which is equivalent to "if $q^2$ is divisible by 3, then also is $q$", since the latter is the contrapositive of the former statement.
Thus, $p$ is divisible by 3 as well.
So, let $p=3k$ for some $k\in\mbb Z$, now $p^2=(3k)^2=9k^2=3q^2\riff q^2=3k^2\riff q^2$ is divisible by 3.
Again, similarly, we can know that $q$ is divisible by 3 as well.
Therefore, $p$ and $q$ are both divisible by 3, so $\dfrac{p}{q}$ is not a simplified fraction.
CONTRADICTION$\riff$ initial assumption is false$\riff \nexists x\in\mbb Q$ such that $x^2=3$ |
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kuing
发表于 2013-10-4 10:01
其妙
发表于 2013-10-4 15:27
