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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    $\begingroup$ wow cool so many minuses! To be honest I didn't know this result and only found out by accident. So I thought it could be interesting for others $\endgroup$ Sep 22 at 6:27
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    $\begingroup$ It would be helpful if downvotes prompted a mandatory question, "Why are you downvoting?" and the replies could be seen by the querent, who would then be free to learn from them or disregard them. $\endgroup$
    – SlowMagic
    Sep 22 at 14:34
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    $\begingroup$ Good question, by the way. It's easy to forget the magic of this number. $\endgroup$
    – SlowMagic
    Sep 22 at 14:39
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    $\begingroup$ You're not the first by a long way, it's a bad idea for multiple reasons @SlowMagic $\endgroup$
    – Nij
    Sep 22 at 21:19
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    $\begingroup$ -1 (and this is not a downvote!) $\endgroup$
    – Florian F
    Sep 23 at 19:25

3 Answers 3


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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    $\begingroup$ 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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    $\begingroup$ $x= \frac{1-\sqrt(5)}{2}$ isn't a solution. $x = -0.618...$. $\frac{1}{x}=1.618...$. which is fine, but $x^2=0.381...$. $\endgroup$
    – tell
    Sep 23 at 13:08
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    $\begingroup$ 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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    $\begingroup$ 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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    $\begingroup$ Here’s some coffee ☕️ so you’ll be ready for next time. Cheers! $\endgroup$ Sep 29 at 21:09

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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    $\begingroup$ Very nice. Thank you. $\endgroup$ Sep 22 at 9:48

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.

  • $\begingroup$ 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

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