Is the following number algebraic or transcendental? $$P:=\prod_{i=0}^{\infty}\left(1+\frac{1}{2^{3^i}}\right)$$ We could also define it as follows: let A be the set of natural numbers which contain only 0's and 1's in base 3. Then our number is $$\sum_{a\in A}2^{-a}.$$ I have found a proof that $P$ is irrational (which is a partial result on this question): for $n\geq1$, let $P_n:=\prod_{i=0}^n(1+\frac{1}{2^{3^i}})$. Then $P_n$ is a rational number with denominator $D_n=2^{3^0+3^1+\dots+3^n}$. Now we have $$0<|P_n-P|=\sum_{i\in A\mid i\geq3^{n+1}}2^{-i}<\sum_{i=3^{n+1}}^{\infty}2^{-i}=\frac{1}{2^{3^{n+1}-1}}=\frac{1}{D_n^2}.$$ This holds for all $n$, and we have $D_n\rightarrow\infty$ as $n\rightarrow\infty$. Therefore, $P$ has irrationality measure at least $2$, which implies that $P$ is irrational.

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    $\begingroup$ Every real number has irrationality measure at least $2$, I think. What you need to show is that the irrationality measure is greater than $2$ (this would also show that $P$ is transcendent, btw). And the decomposition above isn’t going to work. But here, you can see that the base two expansion of $P$ is nonperiodic – thus the number is irrational. $\endgroup$
    – Aphelli
    Jun 29, 2021 at 8:48
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    $\begingroup$ According to Wikipedia, rational numbers have irrationality measure equal to 1: en.wikipedia.org/wiki/Liouville_number#Irrationality_measure $\endgroup$
    – Riemann
    Jun 29, 2021 at 8:53
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    $\begingroup$ There was a mistake in my proof; I fixed it. You are right, you can immediately show that the base two expansion is nonperiodic, so now we have two proofs that P is irrational. $\endgroup$
    – Riemann
    Jun 29, 2021 at 8:57
  • $\begingroup$ @Mindlack. "The binary expansion of $P$ is non-periodic" does it. We can also show that if $P\not\in\Bbb Q$ then $P$ is a Liouville number...BTW, using elementary properties of Farey sequences we can show that if $x\in\Bbb R\setminus \Bbb Q$ there exist infinitely many $(a,b)\in \Bbb Z\times \Bbb Z^+$ such that $|x-a/b|<1/(b^2\sqrt 5)$. We can't replace the $\sqrt 5$ with anything larger if $x$ is the Golden Ratio. A much deeper result, for which Roth won a Fields Medal, is that if an irrational real $x$ has irrationality measure $>2$ then $x$ is transcendental.(As you said). $\endgroup$ Jun 29, 2021 at 10:08
  • $\begingroup$ $P_n$ has denominator $2^{3^n}=LCM ( \{2^{3^i}: 0\le i\le n\})$ $\endgroup$ Jun 29, 2021 at 10:14

1 Answer 1


Using a technique called Mahler's transcendence method, we can prove that $\prod_{i=0}^{\infty}\left(1+\frac{1}{2^{3^i}}\right)$ is transcendental. We shall use the following statement from the beginning of this article:

Let $p\in\mathbb Z_{\geq2}$. Let $f(z)$ be a transcendental function which is holomorphic on the open unit disk and which satisfies the functional equation $f(z) = a(z) f(z^p) + b(z)$, where $a(z)$ and $b(z)$ are polynomials with algebraic coefficients, and for which $f(0)$ is algebraic. If $\alpha$ is an algebraic number, $0<|\alpha| < 1$, and if $a(\alpha^{p^k})\neq0$ for all $k\in\mathbb Z_{\geq0}$, then $f(\alpha)$ is transcendental.

We let $p:=3$ and define $f:D\rightarrow\mathbb C$ by $f(z):=\prod_{i=0}^{\infty}\left(1+z^{3^i}\right)$ for all $z$ in the open unit disk $D$. Then $f$ is a holomorphic function which satisfies the functional equation $f(z)=(1+z)f(z^3)$. Also, $f(0)=1$ is algebraic. However, before we can apply the theorem, we need to show that $f(z)$ is a transcendental function. For this we follow the strategy described in Wadim Zudilin's answer to this MathOverflow question.

We shall prove by contradiction that $f(z)$ is transcendental over $\mathbb C(z)$. This is a stronger statement than we need, but I don't know a shorter proof which proofs the weaker statement that $f(z)$ is a transcendental function. So suppose that $f(z)$ is algebraic over $\mathbb C(z)$. Then there is a unique irreducible equation \begin{equation}f(z)^n+a_{n-1}(z)f(z)^{n-1}+a_{n-2}(z)f(z)^{n-2}+\dots+a_1(z)f(z)^1+a_0(z)=0\end{equation} with coefficients $a_0,\dots,a_{n-1}\in \mathbb C(z)$. Substituting $z^3$ for $z$ we obtain that $$f(z^3)^n+a_{n-1}(z^3)f(z^3)^{n-1}+a_{n-2}(z^3)f(z^3)^{n-2}+\dots+a_1(z^3)f(z^3)^1+a_0(z^3)=0,$$ and using our functional equation $f(z^3)=\frac{1}{z+1}f(z)$ we obtain that $$\frac{1}{(z+1)^n}f(z)^n+\frac{1}{(z+1)^{n-1}}a_{n-1}(z^3)f(z)^{n-1}+\dots+\frac{1}{z+1}a_1(z^3)f(z)^1+a_0(z^3)=0.$$ This gives us that \begin{equation}f(z)^n+(z+1)a_{n-1}(z^3)f(z)^{n-1}+\dots+(z+1)^{n-1}a_1(z^3)f(z)^1+(z+1)^na_0(z^3)=0.\end{equation}

By irreducibility, the coefficients of our first and last equations for $f(z)$ must coincide. Therefore, we have $a_i(z)=(z+1)^{n-i}a_i(z^3)$ for $i=0,1,\dots,n-1$. For such an $i$, write $a_i(z)=c(z)/d(z)$, where $c(z),d(z)\in\mathbb C[z]$, $d(z)$ is not the zero polynomial, and $c(z)$ and $d(z)$ are relatively prime. Then it follows that $$c(z)d(z^3)=(z+1)^{n-i}c(z^3)d(z).$$ Suppose that $c(z)$ is not the zero polynomial. For all polynomials $q\in\mathbb C[z]$ which are not the zero polynomial, the multiplicity of the root $-1$ of $q(z)$ equals the multiplicity of the zero $-1$ of $q(z^3)$. Applying this to $c(z)$ and $d(z)$, we see that the multiplicity of the root $-1$ is the same in $c(z)d(z^3)$ as in $c(z^3)d(z)$. But this is a contradiction since $(z+1)^{n-i}$ also has at least one root $-1$. This proves that $c(z)$ is the zero polynomial, hence $a_i(z)$ is the zero polynomial.

Since $a_i(z)=0$ for $i=0,\dots,n-1$, we see that $(f(z))^n+0+\dots+0+0=0$. Therefore, $f(z)=0$, which is a contradiction. Therefore, $f(z)$ is transcendental over $\mathbb C(z)$. As a corollary, $f(z)$ is a transcendental function.

We have proved that $f(z)$ fulfills all the requirements, so we can apply the theorem on $f(z)$. Take $\alpha:=\frac12$. Then for all $k\in\mathbb Z_{\geq0}$, we have $a(\alpha^{p^k})=1+(\frac12)^{3^k}\neq0$. Therefore, $f(\frac12)$ is transcendental, and $$f\left(\frac12\right)=\prod_{i=0}^{\infty}\left(1+\left(\frac12\right)^{3^i}\right)=\prod_{i=0}^{\infty}\left(1+\frac{1}{2^{3^i}}\right).$$


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