Inverse pair equidistant from 1

Question

Say you pick a number $x$, like $\frac 43$. Its inverse is of course $\frac 34$. $x$ is a distance of $\frac 13$ away from 1, and its inverse is a distance of $\frac 14$ away from 1. Is there any number $x$ that is a distance $d$ away from 1, whose inverse $\frac 1x$ is also a distance $d$ away from 1? I came up with the following equation and found the solution $x = 1$, but I was hoping there was another solution.

$x - 1 = 1 - \frac 1x$

Answer 1

Yes. Your equation only has the solution $x=1$ but that doesn't mean that the only solutions to the problem are solutions to said equation. You can easily check that -1 also satisfies your problem. Maybe consider the following equation, since you're measuring distance: \begin{equation*} |x-1| = \left|1-\frac{1}{x}\right|. \end{equation*}

Edit in response to OPs comment: You can consider all possible cases. First rewrite the equation to \begin{equation*} |x-1| = \left|1-\frac{1}{x}\right| \iff |x-1| = \left|\frac{1}{x}-1\right|. \end{equation*} Case 1: $x>1.$ Then \begin{equation*} |x-1| = x-1 = 1-\frac{1}{x} = \left|\frac{1}{x}-1\right| \iff x+\frac{1}{x}=2. \end{equation*} Therefore, there is no real solution for $x>1$.

We've already considered the case $x=1$. So let's consider $0<x<1$. Then \begin{equation*} |x-1| = 1-x = \frac{1}{x}-1 = \left|\frac{1}{x}-1\right| \iff x+\frac{1}{x}=2, \end{equation*} but we're only considering $0<x<1$, so the equation has no solution for $0<x<1$.

We're not going to consider the case $x=0$. So let's consider $-1<x<0$. Then \begin{equation*} |x-1| = 1-x = 1-\frac{1}{x} = \left|\frac{1}{x}-1\right| \iff x-\frac{1}{x} = 0, \end{equation*} which has no solution for $0<x<1$.

We've alread covered the case $x=-1$, so let's consider $x<-1$ as our last case. Then \begin{equation*} |x-1| = 1-x = 1-\frac{1}{x} = \left|\frac{1}{x}-1\right| \iff x-\frac{1}{x} = 0, \end{equation*} which, again, has no solution. So, yes, $\pm 1$ are the only real solutions to your problem.

Answer 2

From $|x-1| = \left|1-\frac{1}{x}\right|. $, squaring both sides, $x^2-2x+1 =1/x^2-2/x+1 $ or $x^2-1/x^2-2x+2/x =0 $ or $2(x-1/x) =x^2-1/x^2 =(x-1/x)(x+1/x) $.

If $x=1/x$, then $x = \pm 1$. These both satisfy $|x-1| = \left|1-\frac{1}{x}\right| $.

If $x\ne 1/x$, then $x +1/x=2$. THis implies that $x > 0$. Therefore $0 =x -2+1/x =(\sqrt{x}-\frac1{\sqrt{x}})^2 $ so $\sqrt{x} =\frac1{\sqrt{x}} $ so $x = 1$.