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A rational number of the form a~a with a irrational

机译:a〜a形式的无理数

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摘要

Here is a very well known elementary proof that there are irrational numbers a and b so that a~b is rational. We know 2~(1/2) is irrational. If 2~(1/2) is rational, then we are done, while if it is not rational, then(2~(1/2)~(2~(1/2)))~(2~(1/2))=2 and we are done. This elementary but non-constructive argument motivates, but does not quite answer, the question of whether there is a single irrational number a so that a~a is rational. The answer to that question is a resounding affirmative and is also elementary.
机译:这是一个众所周知的基本证明,其中存在无理数a和b,因此a〜b是有理数。我们知道2〜(1/2)是不合理的。如果2〜(1/2)是有理数,那么我们就完成了,而如果2〜(1/2)〜(2〜(1/2)))〜(2〜(1 / 2))= 2,我们完成了。这个基本但非建设性的论点激发了(但并不能完全回答)是否存在单个无理数a的问题,因此a〜a是合理的。这个问题的答案是肯定的,也是基本的。

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