Timeline for About the proof of the proposition "there exists irrational numbers a, b such that a^b is rational"
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
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| Mar 24, 2015 at 12:33 | comment | added | user44143 | This proof can be made constructive easily enough. It is a constructive theorem (basically an exercise in Bishop and Bridges) that for any countable sequence, there is a positive real $x$ which differs from all its members. Using this, construct $x$ which differs from all rational $q$ and from all $c^q$ where $q$ is rational. Let $y$ be such that $x^y=c$. Then both $x$ and $y$ are irrational. | |
| Mar 24, 2015 at 12:14 | comment | added | jmc | @EmilJeřábek — Ah, of course! Thanks! I don't know in which way I was thinking, but somehow I wanted to map pairs $(a,b)$ or $(a,c)$ into the rationals, or something like that. But this is nice: simple and elegant, and it brilliantly does the job. | |
| Mar 24, 2015 at 11:06 | comment | added | Emil Jeřábek | @jmc: $a\mapsto\log_ac$. | |
| Mar 24, 2015 at 9:59 | comment | added | jmc | Wait, what is your injective map? | |
| Mar 24, 2015 at 9:48 | history | answered | Stefan Kohl♦ | CC BY-SA 3.0 |