Passage: Let u be a positive integer and f be a real number lying...

Given $\left(\sqrt{2}+1\right)^{10} = u+f and \left(\sqrt{2}-1\right)^{10} = v$.

Multiplying these two equations, we get:

$(u+f)v = \left(\sqrt{2}+1\right)^{10}\left(\sqrt{2}-1\right)^{10}$

$(u+f)v = \left((\sqrt{2})^2 - 1^2\right)^{10} = 1^{10} = 1$

$uv + fv = 1 \Rightarrow uv = 1 - fv$

Since $f$ is the fractional part, $0 \lt f \lt 1$.

Also, $v = \left(\sqrt{2}-1\right)^{10}$, and since $0 \lt \sqrt{2}-1 \lt 1$, we have $0 \lt v \lt 1$.

Because both $f and v$ lie strictly between $0 and 1$, their product $fv$ also lies strictly between $0 and 1$, i.e., $0 \lt fv \lt 1$.

Therefore, $uv = 1 - fv$ must satisfy:

$0 \lt uv \lt 1$

Answer: $0\lt uv\lt 1$