Fix a real number $x$. We can consider its binary expansion, for instance $x = (0.01101001100101101001011\ldots)_2$. Now we consider the real number $y = (0.01101001100101101001011\ldots)_{10}$ : we just changed from binary expansion to decimal expansion.
If $x$ is irrational (or transcendental), is it true that $y$ is also irrational (or transcendental)?
In other words, does the irrationality/transcendence of $\displaystyle \sum_{n=0}^{\infty} a_n2^{-n}$ (with $a_n \in \{0,1\}$) give the irrationality/transcendence of $\displaystyle \sum_{n=0}^{\infty} a_n 10 ^{-n}$ ?
Of course this question can be generalized with other basis than $2$ and $10$ (provided that the $a_n$'s are smaller than the second basis).
To clarify the situation: the binary expansion of $y$ is not the same as the one of $x$. The decimal expansion of $y$ is the same as the binary expansion of $x$.
I tried to think about the contrapositive, say if $y$ is rational (decimal expansion with only $0$'s and $1$'s), does it follow that $x$ is rational. This is trivial if the decimal expansion of $y$ is finite. I wasn't sure what to do with the infinite case. It doesn't seem obvious since replacing $2$ by $10$ in the series above doesn't appear as an easy algebraic operation.
Thank you for your help!
Answer
The question is:
Does interpreting in a different base preserve rational and irrational numbers?
Does interpreting in a different base preserve algebraic and transcendental numbers?
(I use interpreting in a different base rather than change of base to refer to the process you described.)
The answer to (1) is yes, shown below.
An affirmative answer to (2) contradicts several open conjectures and thus (2) is likely false.
Specifically, it is conjectured that every algebraic irrational is normal, or more weakly, every algebraic irrational has every digit in its base-$b$ expansion for any $b$, see mathoverflow.
Note that $x = \sqrt{2} = 1.414\ldots$ contains every digit in its base-$10$ expansion, but if we form
$$
y = (1.414\ldots)_{11}
$$
then $y$ does not contain the digit $[10]$ in its base-$11$ expansion. So according to the (quite plausible) conjecture, it cannot be algebraic irrational. And it is certainly not irrational, by (1). So it must be transcendental, contradicting (2).
Conversely, in order to show (2) is false one would want to use an easily-describable expansion, such as that of Liouville numbers. But Liouville numbers (as constructed in that link) are transcendental no matter the base we use.
So I expect showing (2) to be false is be difficult -- as showing things to be transcendental often is.
Claim:
Let $r,s \ge 2$ be integer bases.
Let $x = \sum_{i=1}^\infty a_i r^{-i}$, and $y = \sum_{i=1}^\infty a_i s^{-i}$, where $a_i$ are integers such that $\boldsymbol{0 \le a_i < \min(r,s)}$.
Then $x,y$ are both rational or both irrational.
Proof:
In words, the reason for this is that an eventually-periodic list of digits is rational in every base, and a non-periodic list of digits is irrational in every base.
We first cite the following useful result.
The result has been stated in the comments, but I am being a bit more careful by allowing distinct base-$b$ representations which are equivalent, such as $(0.0\overline{1})_2 = (0.1\overline{0})_2$.
Prop. For a real number $x \ge 0$, the following are equivalent:
$x$ is rational;
There exists a base $b$ such that $x$ has an eventually-periodic base $b$ representation;
For every base $b \ge 2$, every base-$b$ representation of $x$ is eventually periodic.
Now suppose $x$ is rational, so that by $(1) \implies (3)$, $a_n$ is eventually periodic; therefore by $(2) \implies (1)$, $y$ is rational. And vice versa, if $y$ is rational then $x$ is rational.
$\square$
The condition "$\boldsymbol{0 \le a_i < \min(r,s)}$" is critical both in the above proposition and in the main claim.
Without that condition, we may take a rational number, say $1$, and write it in base 2 as:
$$
1 = (1\;03\;11\;03\;111\;03\;1111\;03\;11111\;03\;\ldots)_2
$$
which is not eventually periodic.
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