Timeline for Euler's constant: irrationality and proof theory

5 events

when toggle format	what		by	license	comment
Jan 4, 2016 at 23:40	comment	added	Pietro Majer		(Was it clear what I wrote? For any number $a$ there is $x$ so that $x$ and $a+x$ are both irrational, and we can choose $x$ to be either $\sqrt{2}$ or $2\sqrt{2}$. Indeed $a+\sqrt{2}$ and $a+2\sqrt{2}$ can't be both rational, while $x$ is in any case irrational).
Nov 30, 2015 at 8:42	comment	added	Pietro Majer		@Noam D. Elkies: it think here the problem is that $x$ is not constructive, not the diagonal argument. For instance, it's also true that there is $x\in \{ \sqrt2,\ 2\sqrt2\}$ such that $x$ and $a+x$ are both irrational.
Nov 29, 2015 at 15:32	comment	added	Noam D. Elkies		The usual diagonal argument is constructive. (e.g. use the number whose $n$-th digit is 7 if the $n$-th rational's $n$-th digit is 3, and 3 otherwise.)
Nov 29, 2015 at 13:24	comment	added	Fedor Petrov		But this $x$ found by cardinality argument is not specified, what is probably required by OP.
Nov 29, 2015 at 13:19	history	answered	Gerald Edgar	CC BY-SA 3.0