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[数论] Show that there is no $x\in\mbb Q$ satisfying $x^2=3$

RT...本来以为会了结果还是有问题。。。
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回复 1# ╰☆ヾo.海x
I don't know what is“本来会了结果还是有问题”…
睡自己的觉,让别人说去!!!

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回复 2# kuing

贴子8楼他最后几步说 “p^2整除q^2,因p,q互质,故q^2=1,于是x=q^2这与正整数x是非平方数相矛盾”,这里看不懂。。。什么整除完了就q^2=1了?

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回复 4# ╰☆ヾo.海x
他那里打错了,是 $p^2=1$,11#已指出
$\href{https://kuingggg.github.io/}{\text{About Me}}$

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回复 5# kuing
还是不懂啊。。。为什么就等于1了?

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回复 3# 睡神
本来以为可以用跟"证明根号2是无理数"一样的方法证明根号3是无理数呢结果我就是写不下去了

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回复 7# ╰☆ヾo.海x
你把2的写出来看看

PS、这贴还是移去初等区了
$\href{https://kuingggg.github.io/}{\text{About Me}}$

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还可以用无穷递降法,重复一个有限过程,导致无穷与有限的矛盾,于是得到$\sqrt3$是无理数。

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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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回复 10# ╰☆ヾo.海x
牛笔!

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