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$\sqrt{2}^\sqrt{2}$的有理性?

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青青子衿 posted 2013-12-14 16:30 |Read mode
$\sqrt{2}^\sqrt{2}$的有理性?
既然$\sqrt{2}^\sqrt{2}$是有理数,那么$\sqrt{2}^\sqrt{2}$就可以表示成两个互质整数的商的形式。
请表示出来!

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icesheep posted 2013-12-14 17:19
谁告诉你它是有理数的

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original poster 青青子衿 posted 2013-12-14 17:29
回复 2# icesheep
亲,请构造$x=\sqrt{2}^\sqrt{2},y=\sqrt{2},x^y=?$c
两个无理数的幂乘应该不是有理数吧!

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icesheep posted 2013-12-14 17:43
谁告诉你两个无理数的幂乘不能是有理数的?

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kuing posted 2013-12-14 18:03

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睡神 posted 2013-12-15 08:17
这个我记得我之前问过更一般的情况了

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kuing posted 2013-12-15 14:30
回复 6# 睡神

嗯,我当时给的好像也是5#的链接……

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kuing posted 2016-9-18 15:05
顺便再贴一链:matrix67.com/blog/archives/4984

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isee posted 2016-9-18 15:07
Last edited by isee 2016-9-18 16:20这个确是经典且精彩

事实上,$\sqrt{2}^\sqrt{2}$是无理数。

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其妙 posted 2017-3-18 23:55
回复 9# isee
怎么证明?

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isee posted 2017-3-19 18:51
回复  isee
怎么证明?
其妙 发表于 2017-3-18 23:55
我仅知道结论,不会证明

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其妙 posted 2017-4-9 23:29
回复 11# isee
结论在哪?

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TSC999 posted 2017-5-16 09:34
回复 12# 其妙

结论在书里呢。嘿嘿。

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其妙 posted 2017-8-15 18:57
回复 13# TSC999
拍照后贴出来看看

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hbghlyj posted 2022-11-25 09:10

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