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From Dirichlet's theorem on arithmetic progressions, if $\text{gcd}(a,b)=1$ we know $\{ak+b\}_{k\ge 0}$ contains infinitely many primes. Let those primes be $p_1,p_2,\cdots$. Then the real $$\alpha=0.p_1p_2p_3\cdots\tag{1}$$ is it irrational? Here the primes are placed side by side, as in $p_1=13,p_2=53,\cdots$ then the expression would be like $\alpha=0.1353\cdots$. So to rephrase my question, is $\alpha$ irrational, as defined in $(1)$? Thanks for answering.

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  • $\begingroup$ If you like my answer, please accept it officially (so that it turns green). Thanks in advance! $\endgroup$
    – GH from MO
    Nov 20, 2023 at 15:39

2 Answers 2

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It is well-known that not only does the arithmetic progression $\{ak+b\}_{k \in \mathbb{Z}^{+}}$ contain infinitely many prime numbers, but also that the series of the reciprocals of those primes diverges. The answer to the OP's question can be obtained now from the following general result:

If $\{a_{i}\}_{i \in \mathbb{N}}$ is a strictly increasing sequence of natural numbers such that the series $\sum_{i=1}^{\infty} \frac{1}{a_{i}}$ diverges, then the unending decimal fraction $\alpha$ formed by juxtaposing the successive terms of the sequence $\{a_{i}\}_{i \in \mathbb{N}}$ represents an irrational number.

For a proof of this theorem, see D. J. Newman, R. Breusch, and F. Herzog. Solution to problem 4494. Amer. Math. Monthly 9 (60), Nov. 1953, pp. 632-633. or N. Hegyvári. On some irrational decimal fractions. Amer. Math. Monthly 8 (100), Oct. 1993, pp. 779-780.

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  • $\begingroup$ Hi, could you post the solution? I couldn't find it on the web. Thanks a lot. $\endgroup$
    – shadow10
    Jun 28, 2014 at 8:19
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    $\begingroup$ Apply reductio ad absurdum. Suppose that the number $\alpha$ is a rational number. We may even assume that $\alpha$ has a periodic decimal representation and that its period starts right after the decimal comma. Let $s$ denote the period length of $\alpha$. It is not difficult to establish, that for every $k\in \mathbb{N}$, the sequence $\{a_{i}\}_{i \in \mathbb{N}}$ has no more that $s$ terms of $k$ digits. Hence, $\sum_{i=1}^{\infty} \frac{1}{a_{i}} \leq \sum_{i=1}^{\infty} \frac{s}{10^{i-1}}$, which contradicts the divergence of the series $\sum_{i=1}^{\infty}\frac{1}{a_{i}}$... $\endgroup$ Jun 28, 2014 at 8:53
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Yes, it is irrational. This is because any finite digit sequence occurs as the initial digits of a prime in your sequence (in fact you can prescribe 41% of all the digits in the beginning), hence the concatenated sequence is not periodic.

Added. The OP asked for more details so here they are. Assume that $\alpha$ is rational. Then the decimal expansion of $\alpha$ is periodic after a certain digit. This implies that there is a digit $d\in\{0,\dots,9\}$ and a positive integer $n\in\mathbb{N}$ such that among any $n$ consecutive digits of $\alpha$, the digit $d$ occurs. On the other hand, by known results on the distribution of primes in arithmetic progressions, there is a $p_i$ which starts with $n+1$ digits distinct from $d$. This is a contradiction, which shows that $\alpha$ is irrational.

In fact, by a result of Huxley and Iwaniec (Mathematika 22 (1975), 188-194), the interval $(x,x+x^{0.584})$ contains a $p_i$ for any sufficiently large $x>0$. So if $x$ is sufficiently large, we can find a $p_i$ whose decimal expansion agrees with that of $x$ on the initial 41% of the digits, since the error $x^{0.584}$ only affects about the last 58.4% of the digits.

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  • $\begingroup$ Could you elaborate a bit? I don't find your second sentence that obvious. Thanks. $\endgroup$
    – shadow10
    Jun 28, 2014 at 8:22
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    $\begingroup$ @shadow10: I added more details. Of course the standard analytic form of Dirichlet's theorem on arithmetic progression suffices for the irrationality of $\alpha$, along the above lines. The result of Huxley-Iwaniec is deeper: I only displayed it because I mentioned that a sizable proportion of digits can be prescribed for infinitely many $p_i$'s. $\endgroup$
    – GH from MO
    Jun 28, 2014 at 9:04

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