# How many pairs of numbers are there so they are the inverse of each other and they have the same decimal part?

I was wondering... $1$, $\phi$ and $\frac{1}{\phi}$, they have something in common: they share the same decimal part with their inverse. And here it comes the question:

Are these numbers unique? How many other members are in the set if they exist? If there are more than three elements: is it finite or infinite? Is it a dense set? Is in countable? Are their members irrational numbers??

Note that $x$ and $\frac1x$ always have the same sign. Also if $\lvert x\rvert > 1,$ then $0 < \left\lvert\frac1x\right\rvert < 1$ and vice versa.

So the pairs will always be two positive numbers as solved in the other answers, or two negative numbers which you can get from one of those solutions just by changing the signs of both numbers; and the general form of two positive numbers $x$ and $\frac1x$ that have the same decimal part is that of the two numbers $$\frac12\left(n + \sqrt{n^2+4}\right) \quad \text{and} \quad \frac12\left(-n + \sqrt{n^2+4}\right)$$ where $n$ is any non-negative integer.

For $n = 0$ both forms come out to $1$; for $n=1$ they come out to $\phi$ and $\frac1\phi.$

There are exactly a countable number of such pairs, since from each non-negative integer we get at most four pairs. (The exact number of pairs for each value of $n$ depends on how you count them: do you consider "$\phi,\frac1\phi$" the same pair as "$\frac1\phi,\phi$" or different?)

Since $n^2 + 4$ is not a perfect square for any value of $n > 0,$ it follows that $\sqrt{n^2+4}$ is not an integer for $n > 0,$ and a further result is that $\sqrt{n^2+4}$ is irrational whenever $n>0.$ The only rational numbers whose multiplicative inverse have the same fractional part are therefore $1$ and $-1.$

• Your two numbers (in display math) have opposite signs; they are not $x$ and $1/x$. The latter should be negated I think. – R.. GitHub STOP HELPING ICE Feb 25 '17 at 23:11
• @R.. Thanks for pointing that out. I think I fixed it now. – David K Feb 26 '17 at 2:42

So we want values of $0<x<1$ such that $x+k= \frac{\large 1}{\large x}$ for positive integer $k$, meaning $x^2+kx-1 =0$. This has a positive solution in the range for every $k$.

• But only for $k>0$ the positive solution $x_+$ satisfies $0<x_+<1$. – egreg Feb 26 '17 at 10:14
• Yes, clarified. – Joffan Feb 26 '17 at 17:43

There is a pair for every $n \in N$ \begin{eqnarray*} x_{\pm} = \frac{n \pm \sqrt{n^2+4}}{2} \end{eqnarray*}

whence \begin{eqnarray*} 1/(x_{\pm}) = \frac{-n \pm \sqrt{n^2+4}}{2} = x_{\pm}-n \end{eqnarray*}

Eg $n=2$ ... $x_+=2.414 \cdots$ & $x_-=0.414 \cdots$.

If you want to find these numbers, you should search in the limit values of sequence numbers like Fibonacci numbers. For example, Consider the following sequence

\begin{equation} a_n= \left\{ \begin{array}{cc} a_{n-3} & n=1(mod \hspace{1mm}2)~,\\ \\ a_{n-3}+ a_{n-2} & n=0(mod \hspace{1mm}2)~. \end{array} \right. \end{equation} with the following initial values

$$a_0=0~, \hspace{5mm} a_1=1~, \hspace{5mm} a_2=0~.$$

Now, consider the limit values of the $a_n$ sequence are defined as follows

$$\lim_{n\rightarrow\infty}\frac{a_{2n}}{a_{2n+1}}=\alpha_1~,\quad \lim_{n\rightarrow\infty}\frac{a_{2n+1}}{a_{2n+2}}=\alpha_2~.$$

with calculation, you can see that

\begin{equation} \left\{ \begin{array}{ccc} \alpha_1 &=& 1.4655712318767680267\, , \\ &&\\ \alpha_2 &=& 0.4655712318767680267\, . \end{array} \right. \end{equation}

• does $\frac{1}{\alpha_1}=\alpha_2$ ? – Vincent Feb 25 '17 at 20:35
• @Vincent $\alpha_2=\frac{1}{\alpha_1^2}$ – Amin235 Feb 25 '17 at 20:39

The golden ratio is one of only two numbers that share certain properties, such as that the golden ratio is one less than its square. I have a complete description of these morphic numbers in my response to another post here: What real number is exactly one less than its cube?