In the decimal expansion of $\sqrt{2}=1.4142\dots$, Isabelle finds a sequence of $k$ successive zeroes where $k$ is a positive integer.

Show that the first zero of this sequence can occur no earlier than at the $k$-th position after the decimal point.
1. Let \( g \) be the decimal expansion of \( \sqrt{2} \) to \( m \) decimal places, where the \( m \)-th decimal place is not zero. This means we can write:
   \[
   g = 1.4142\ldots \text{ (up to } m \text{ decimal places)}
   \]
2. Consider \( g^2 \). Since \( g \) is an approximation of \( \sqrt{2} \), we have:
   \[
   g^2 \approx 2
   \]
   However, because \( g \) is truncated at \( m \) decimal places, \( g^2 \) will be slightly less than 2. Specifically, the last non-zero digit of \( g^2 \) will be in the \( 2m \)-th decimal place. Let this digit be \( d \), so:
   \[
   g^2 = 2 - d \cdot 10^{-2m}
   \]
   where \( d \) is a non-zero digit.

3. Since \( \sqrt{2} \) is irrational, its decimal expansion is non-repeating and non-terminating. Therefore, there are digits subsequent to \( d \).

4. We can write:
   \[
   g^2 < 2 - 10^{-2m}
   \]
   Adding \( 10^{-2m} \) to both sides, we get:
   \[
   g^2 + 10^{-2m} < 2
   \]

5. Let \( \sqrt{2} = g + e \), where \( e \) is the error term due to truncation. Then:
   \[
   (g + e)^2 = 2
   \]
   Expanding the left-hand side, we get:
   \[
   g^2 + 2ge + e^2 = 2
   \]
   Since \( g^2 < 2 \), we have:
   \[
   g^2 + 2ge + e^2 > g^2 + 10^{-2m}
   \]
   Therefore:
   \[
   2ge + e^2 > 10^{-2m}
   \]

6. Since \( e \) is very small compared to \( g \), we can ignore the \( e^2 \) term. Thus:
   \[
   2ge > 10^{-2m}
   \]
   Solving for \( e \), we get:
   \[
   e > \frac{10^{-2m}}{2g}
   \]
   Given that \( g \approx 1.4142 \), we can approximate:
   \[
   g \approx \sqrt{2} \approx 1.4142 \implies 2g \approx 2.8284
   \]
   Therefore:
   \[
   e > \frac{10^{-2m}}{2.8284}
   \]

7. This value of \( e \) begins in the \( 2m+1 \)-th decimal place. Hence, the sequence of \( k \) successive zeroes can start no earlier than the \( k \)-th position after the decimal point.

The final answer is \( \boxed{ k } \)-th position after the decimal point.