# Interesting irrational number

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!

• 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 Sep 22 at 6:27
• 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. Sep 22 at 14:34
• Good question, by the way. It's easy to forget the magic of this number. Sep 22 at 14:39
• – Nij
Sep 22 at 21:19
• -1 (and this is not a downvote!) Sep 23 at 19:25

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'.

• I'm beginning to suspect not, but my proof had a loophole I'll need coffee to mend. Sep 22 at 6:03
• $x= \frac{1-\sqrt(5)}{2}$ isn't a solution. $x = -0.618...$. $\frac{1}{x}=1.618...$. which is fine, but $x^2=0.381...$.
– tell
Sep 23 at 13:08
• 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 ^_^) Sep 25 at 12:38
• 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. Sep 29 at 21:03
• Here’s some coffee ☕️ so you’ll be ready for next time. Cheers! 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.

• Very nice. Thank you. 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.

• 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). Sep 25 at 1:12