To determine whether the number (\theta_k) (obtained by interpreting the decimal expansion of (\sqrt{2}) as a base-(k) expansion for (k > 10)) is transcendental, we can leverage results from transcendence theory, particularly those related to automatic sequences and generating functions.

Key Steps:

1. Decimal Digits of (\sqrt{2}): Let (\sqrt{2}) have decimal expansion (\sqrt{...

Write a_n for the base‑10 digits of √2, so that √2 = ∑{n≥1} a_n 10^{-n}, and set θ_k := ∑{n≥1} a_n k^{-n}. Equivalently, if A(z) := ∑_{n≥1} a_n z^n (a power series with integer coefficients bounded between 0 and 9), then A(1/10) = √2 and θ_k = A(1/k).

• What is known unconditionally: θ_k is irrational for every integer k ≥ 2. Indeed, if θ_k were rational, then its base‑k expansion would be even...

Let

  √2 = 1 . d1 d2 d3 …     (di ∈ {0 , 1 , … , 9})

be the ordinary decimal expansion of √2 and, for an integer b ≥ 11, put

  θb := Σn≥1 di b-n. (1)

In words: we keep the same infinite string of decimal digits, but now regard it as the base-b fractional expansion of a new real number θb.
The question is whether one can prove, for some (o...

Let

 α = √2 = 1.a1 a2 a3 … (decimal expansion, a n ∈ {0,…,9})

and put

 θk = a1 k-1 + a2 k-2 + a3 k-3 + … = Σn≥1 an k-n , k ≥ 2.

In particular θ10 = α – 1 and we are interested in θ11.
(The initial integral part “1.” plays no rôle, so we omit it and work with the fractional part f(z) defined below.)

Step 1. Introduce the power series
 f(z) = Σn≥1 an z^n (an bounded inte...

This is a fascinating question about base change and transcendence. Let me address what's known about this type of problem.

The General Setup

When we take the decimal expansion of , we're looking at: where , , , etc.

Your is defined as:

<think> The problem is about taking the decimal expansion of \(\sqrt{2}\) and interpreting it in a different base. Specifically, it says to take the decimal expansion for \(\sqrt{2}\), but think of it as a base 11 expansion, and call that number \(\theta_{11}\). Then, it asks if \(\theta_{11}\) is transcendental. Similarly for other bases greater than 10.

First, recall that (\sqrt{2}) is irrati...

Nothing non‐trivial is known. In fact one cannot even prove that the “same’’ digit‐string of √2, viewed in a larger base, cannot produce a new algebraic number. Here is why it seems out of reach:

1. If you take the ordinary base–10 expansion of √2
   √2 = 1.a₁a₂a₃…₁₀
   and simply re‐interpret the digits a₁,a₂,a₃,… as the base–11 (or base–k>10) expansion
   θ₁₁ = 0.a₁a₂a₃…₁₁
   ...

This is an excellent and notoriously difficult question. The short answer is:

This is a famous open problem. There is no known proof that (or any for ) is transcendental. It is, however, widely conjectured to be true.

Let's break down why this is so hard and what is known in this area.

1. The General Conjecture (Borel's Conjecture)

Your question is a ...

Transcendence of Base-Changed Algebraic Numbers

This question asks whether is transcendental, where is constructed by taking the decimal (base-10) expansion of and interpreting those same digits as a base-11 expansion.

To be explicit, if: in base 10, then: