Can two distinct irrational numbers be final segments of each other?

Question

Let $x$ and $y$ be real numbers. I define the relation "$y$ is a final segment of $x$" if the decimal expansion of $x$, considered as an infinite string, has the decimal expansion of $y$ as a suffix. We omit any leading zeroes in the decimal expansion. Let me give some examples. The real number $\pi - 3$ is a final segment of $\pi$. Also, the real numbers $0.23232323...$ and $0.32323232...$ are final segments of each other. My question is, are there two distinct irrational numbers which are final segments of each other?

Answer 1

Another way of putting it:

Let the digits of $x$ be $d_0,d_1,d_2,......$ and let the first digit be in the $m$th position-- that is to say so that $x = d_0\times 10^m + d_1\times 10^{m-1}+.... = \sum_{k=0}^{\infty}d_k \times 10^{m-k}$-- and if we let the digits of $y = c_0, c_1, c_2,.....$ and let the first digit be in the $n$th position so that $y = \sum_{k=0}^\infty c_k \times 10^{n-k}$.

$y$ is the final segment of $x$ if there is some value $w$ where $d_w= c_0$ and $d_{w+j} = c_j$ for all $j$.

And $x$ is the final segment of $y$ if there is some value $v$ where $c_v =d_0$ and $c_{v+j} =d_j$ for all $j$.

So if $x$ and $y$ are both final segments of each other then we'd have $d_{w+v+j} = c_{v+j} = d_j$ for all $j$.

But if $d_{w+v + j} = d_j$ for all $j$ then $x$ is a repeating decimal with period of $w+v$.... in other words $x$ must be rational. (Same for $y$.)

Answer 2

If $x$ is a final segment of $y$, then $10^ry=m+x$ for some positive integer $r$ and some integer $m$. If $y$ is a final segment of $x$, then $10^sx=n+y$ for some positive integer $s$ and some integer $n$. So we have two linear equations, with integer coefficients, for the unknowns $x,y$. So $x$ and $y$ are rational.