Can you find an irrational number $x$ such that $x$, $1/x$ and $x^2$ all have exactly the same digits after the decimal point? Good luck!
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asked Sep 22 at 4:39
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3$\begingroup$ Good question, by the way. It's easy to forget the magic of this number. $\endgroup$– SlowMagicSep 22 at 14:39
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1$\begingroup$ Are there any non-irrational numbers which satisfy this other than 1? $\endgroup$– James HurleySep 22 at 18:47
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2$\begingroup$ You're not the first by a long way, it's a bad idea for multiple reasons @SlowMagic $\endgroup$– NijSep 22 at 21:19
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$\begingroup$ @JamesHurley I know one more rational number that satisfies this. Can you find it? $\endgroup$– Dmitry KamenetskySep 23 at 7:39
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4$\begingroup$ -1 (and this is not a downvote!) $\endgroup$– Florian FSep 23 at 19:25
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3 Answers
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The first one which leapt to mind satisfies $x^2=x+1$ and $1/x=x-1$. Upon closer examination, only the positive solution works: $x=(1+\sqrt5)/2$, which is commonly known as $\phi$, the 'golden ratio'.
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3$\begingroup$ I'm beginning to suspect not, but my proof had a loophole I'll need coffee to mend. $\endgroup$ Sep 22 at 6:03
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3$\begingroup$ $x= \frac{1-\sqrt(5)}{2}$ isn't a solution. $x = -0.618...$. $\frac{1}{x}=1.618...$. which is fine, but $x^2=0.381...$. $\endgroup$– tellSep 23 at 13:08
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1$\begingroup$ I figured Mr. Gibson's answer below covered for my undercaffeination, but figured I would at least share my lateral-thinking way of salvaging it. ^_^ If you distribute the negative sign across all the digits (so instead of -(6/10 + 1/100+ 8/1000...) we have -6/10 + -1/100 + -8/1000... then x^2 could (unconventionally) be written +1 + -6/10 + -1/100 + -8/1000... Seriously, though, good catch (we all need to be careful with sign flips sometimes ^_^) $\endgroup$ Sep 25 at 12:38
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1$\begingroup$ I try to make a point of not burying my mistakes, but hadn't considered the broader context of leaving the accepted answer unclear. Sorry for any resulting confusion. $\endgroup$ Sep 29 at 21:03
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1$\begingroup$ Here’s some coffee ☕️ so you’ll be ready for next time. Cheers! $\endgroup$ Sep 29 at 21:09
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Proof that the only positive solution is
$$x=\frac{1 + \sqrt{5}}{2}\text, \quad \text{ i.e. } \quad x=\phi $$
Let $y$ be a positive solution. Then
$y^2$ and $1/y$ are also positive and there exists some integers $k,l$ such that: $$y^2=y+k \quad \text{ and } \quad \frac{1}{y}=y+l$$ Multiplying the second equation by $y\neq0$ and substituting $y^2$ in the first equation leads to: $$1-ly=y+k\text, \quad \text{ i.e. } \quad 1-k=y(1+l)$$ In that last equation the LHS is an integer, but $y$ is irrational and $1+l$ is an integer, so the only possibility for the RHS to be an integer too is $l=-1$, and then $k=1$. That brings us back to $y^2=y+1$, and thus $y=\phi$.
Similarly, we can prove that
$y = \frac{- 1 - \sqrt{5}}{2} = -\phi$
is the only negative solution.
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Previous answers seem to say that there are two solutions:
$$x=\frac{1 \pm \sqrt{5}}{2}$$
but for $$x=\frac{1 - \sqrt{5}}{2}$$
the base 10 representations of $x$ and $x^2$ are:
$$-.618\dots$$ and $$.381\dots$$
which do not have the same digits after the decimal point.
answered Sep 22 at 22:44
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$\begingroup$ right, only as exponent of $e^(i*2pi*y)$, both $y=x$ and $y=x*x$ appear equal in digits after decimal point (both real and imaginary). $\endgroup$ Sep 25 at 1:12