Ah, thank you! Seeing the exact line highlighted makes the potential misunderstanding much clearer.

The statement in the image is:
"The fractional part is $n^{k-i-1} / 2^{i+1} = r_i / 2^{i+1}$"

This is slightly imprecise notation. It should formally read:
"The fractional part of $n^{k-i-1} / 2^{i+1}$ is $r_i / 2^{i+1}$."
Or, using curly braces for the fractional part function:
${ n^{k-i-1} / 2^{i+1} } = r_i / 2^{i+1}$

Why this statement is correct:

1. We defined $n^{k-i-1} = q \cdot 2^{i+1} + r_i$, where $q$ is the integer quotient and $r_i$ is the remainder ($0 \le r_i < 2^{i+1}$).
2. Therefore, $n^{k-i-1} / 2^{i+1} = (q \cdot 2^{i+1} + r_i) / 2^{i+1} = q + r_i / 2^{i+1}$.
3. The fractional part of a number $x$ is ${x} = x - \lfloor x \rfloor$.
4. Here, $x = q + r_i / 2^{i+1}$. Since $q$ is an integer and we established $1 \le r_i < 2^{i+1}$ (meaning $0 < r_i / 2^{i+1} < 1$), the integer part is $\lfloor x \rfloor = q$.
5. So, the fractional part is ${x} = (q + r_i / 2^{i+1}) - q = r_i / 2^{i+1}$.

Addressing the Mathematician's Concern:

The mathematician seems to believe that the proof later incorrectly equates $r_i$ with $n^{k-i-1}$. This is not what happens.

• The highlighted line correctly identifies the fractional part using the remainder $r_i$.
• The subsequent calculation $c_i = \lfloor 2n(r_i / 2^{i+1}) \rfloor = \lfloor n r_i / 2^i \rfloor$ correctly uses this fractional part.
• The following step uses the inequality $r_i \ge 1$ (which we proved because $n^{k-i-1}$ is odd and $2^{i+1}$ is even) to establish a lower bound: $c_i = \lfloor n r_i / 2^i \rfloor \ge \lfloor n \cdot 1 / 2^i \rfloor = \lfloor n / 2^i \rfloor$.

The proof never claims $r_i = n^{k-i-1}$. It uses $r_i = n^{k-i-1} \pmod{2^{i+1}}$ and the derived property $r_i \ge 1$.

Conclusion: The confusion likely arises from either the slightly ambiguous notation in the highlighted sentence (which should explicitly state "fractional part of ... is ...") or a misreading of the later step where the lower bound $r_i \ge 1$ is applied. The mathematical logic itself appears sound.