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I was wondering idly this morning about real numbers $r$ with the property that the decimal representations of $r$ and $1/r$ both contain the same nonzero digits (not necessarily the same number of times or in the same order). Some obvious example are:

  • The only nonzero digit in the decimal representation of $1/3$ is $3$.
  • The only nonzero digit in the decimal representation of $10^n$ and $1/10^n$ is $1$.

Of course this property depends on the number base one uses to represent the real number. In binary, every number has this property (because there is only one nonzero digit available!). In any base $b$, the numbers $b^n$ all have this property. Somewhat more subtly, if the base $b$ has the form $n^2+1$, then the representation of $1/n$ will be $0.nnnnnn \dots$ (which generalizes the example of $1/3$ in base $10$).

But all of these examples involve whole number $r$ and seem in some sense trivial. Are there any nontrivial examples? Is it possible to prove whether there exist irrational numbers with this property?

mweiss
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1 Answers1

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Here's a nice class of non-trivial examples that will work in any base. In particular, suppose that $r$ satisfies $$ r = 1/r + n $$ for some integer $n$. Then $r$ and $1/r$ will have all the same digits after the exact same digits after the decimal place. Usually, both will contain all $10$ digits somewhere in there.

For example, with $n = 1$, we have the solution $$ r = \frac{1 + \sqrt{5}}2 $$ (aka the golden ratio), and we find that $$ r = 1.61803\dots\\ 1/r = 0.61803\dots $$

Ben Grossmann
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  • Very nice. This can be generalized, I suppose, to finding $r$ that satisfies $r = 1/(10^k r) + n$, as that would simply pad the digits of $r$ with some zeroes after the decimal point. – mweiss Jul 13 '16 at 21:33
  • (Or in base $b$, we would have $r = 1/(b^k r) + n$.) – mweiss Jul 13 '16 at 21:34
  • Can you add some precision to what "usually" means? – mweiss Jul 13 '16 at 21:35
  • @mweiss honestly, I'm not really sure. It is conjectured that $\sqrt{n}$ is irrational for any non-perfect square integer $n$. If this holds, then it would follow that any irrational solution $r$ to a quadratic of the type we described would also be normal, and thus would necessarily contain every digit. – Ben Grossmann Jul 13 '16 at 21:40
  • There might be a weaker condition that works here, but I am not aware of it. – Ben Grossmann Jul 13 '16 at 21:46
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    "It is conjectured that $\sqrt{n}$ is irrational for any non-perfect square integer $n$." I assume you meant "normal" rather than "irrational" -- otherwise I think I can prove that conjecture for you. :) – mweiss Jul 14 '16 at 02:12
  • Right... I meant normal, not irrational. – Ben Grossmann Jul 14 '16 at 03:47