# 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!

• Good question, by the way. It's easy to forget the magic of this number. Sep 22 at 14:39
• Are there any non-irrational numbers which satisfy this other than 1? Sep 22 at 18:47
• – Nij
Sep 22 at 21:19
• @JamesHurley I know one more rational number that satisfies this. Can you find it? Sep 23 at 7:39
• -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