decimal expansion which consists of the concatenation of the powers of $2$ is irrational The real number in $(0,1)$ which has as its decimal expansion $$0.248163264128256512\cdots$$ must be irrational. I am trying to prove the assertion by contradiction. I assumed that this decimal expansion is eventually periodic. Thus after a finite number of decimal places a fixed finite sequence ${a_1}{a_2}\cdots{a_p}$ repeats forever. Am wondering if the fact that any finite sequence of positive integers must appear in some large enough power of $2$ will help me.
Is it possible to find an argument that proves that a real number in $(0,1)$ which has as its decimal expansion $$0.{x_1}{x_2}\cdots$$ where ${x_j} = p^j$ written out in decimal noptation where $p$ is a fixed prime number is irrational. The above question being the case $p=2$.

Here is the proof of the main fact : I'll leave a sketch of the rest.

If $N$ is a natural number, then there is a power of $2$ that, when written in base $10$,starts with the digits of $N$.

Proof : $N$ has $n$ digits so that $1 \leq N/10^{(n-1)}<10$   , then note that it is sufficient to prove that for large enough $m$, we have $N/10^{(n-1)} \times 10^m \leq 2^k < (N+1)/10^{(n-1)}\times 10^m$. Taking logs and dividing by $\ln 10$ we get the equivalent inequality $$
\frac{\ln(N/(10)^{n-1})}{\ln 10} + m \leq k\frac{\ln 2}{\ln 10} < \frac{\ln((N+1)/(10)^{n-1})}{\ln 10} + m 
$$
Since $ 1 \leq N/(10)^{n-1}<10$, the above inequality is true if and only if
$$
\frac{\ln(N/(10)^{n-1})}{\ln 10} \leq \left\{k \frac{\ln 2}{\ln 10}\right\} \leq \frac{\ln((N+1)/(10)^{n-1})}{\ln 10}
$$
The irrationality of $\frac{\ln 2}{\ln 10}$ implies that such a $k$ exists by density of the sequence $\left\{l \frac{\ln 2}{\ln 10}\right\}$ in $[0,1]$  : in fact, infinitely many such $k$ exist. This follow as a result of $\frac{\ln 2}{\ln 10}$ being irrational. Picking a large enough $k$ to obtain a large enough $m \geq n-1$ finishes the argument.

Now assume that $0.2481632\ldots$ is rational, say $$
0.2481632 = 0.b_1\ldots b_n\overline{a_1\ldots a_m}
$$
where $b_1\ldots b_n$ and $\overline{a_1\ldots a_m}$ denote the fixed and repeating parts respectively. We choose $\overline{a_1\ldots a_m}$ to be the smallest string that upon repeating generates the repeating part. For example, $0.\overline{7373}\ldots = 0.\overline{73}\ldots$ so we choose $a_1a_2 = 73$ in this case and not $a_1a_2a_3a_4 = 7373$.
Now, consider the number $\underbrace{111\ldots 111}_{n+m+1 \text{ times}}$.

*

*It definitely appears in some power of $2$, so it must be a part of the decimal expansion at some point.


*However, in that case,it must happen that $a_1\ldots a_m = \underbrace{111\ldots 111}_{m \text{ times}}$ (this requires a not very difficult proof)


*$a_1\ldots a_m$, by choice, cannot be $111\ldots111$ unless $m=1$, because we've chosen it to be the smallest repetend.
This shows that it is not possible for the decimal expansion to contain this particular number, contradictory to the main fact. Hence, we are done.

If you wanted to avoid the main fact for its somewhat distant idea, you may also use the following approach to prove this : for any $k$ , there exists a power of $2$ that contains $k$ consecutive $0$s in its decimal expansion.
The idea is simple : fix $n$ and consider the number $2^n$. It has $n \frac{\ln 2}{\ln 10} + 1$ digits, which is smaller than $n$ for large enough $n$.
We can prove the following : $2^{n+10^n} \equiv 2^n \pmod{10^n}$. To prove this, use the Chinese Remainder Theorem : suppose that $2^{n+10^n} \equiv x \pmod{10^n}$. It is obvious that $x \equiv 0 \pmod{2^n}$ , and modulo $5^n$, note that the Euler function value $\phi(5^n)$ is a divisor of $10^n$ so $2^{10^n} \equiv 1 \pmod{5^n}$ and therefore $x \equiv 2^n \pmod{5^n}$. Since $x$ must be unique modulo $10^n$ by the CRT, we know that $x\equiv 2^n \pmod{10^n}$.
However, this means that the last $n$ digits of $2^{n+10^n}$ contain at least $n(1-\frac{\ln 2}{\ln 10}) - 1$ consecutive zeros. This is a number that goes to infinity as $n \to \infty$. It follows that powers of $2$ can contain arbitrarily long sequences of zeros. Now, use the same argument as we did for the first proof, but replace $1$s by $0$s instead. You will quickly see the answer emerge.
