\begin{tabular}{l} Berween which two positive integers do the follawing irrational number lies? \ $$ \sqrt{11} $$ \ \hline\end{tabular}

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To find the two positive integers between which the irrational number $$ \sqrt{11} $$ lies, we can start by calculating the square roots of the integers surrounding 11.

1. **Identify the integers**: The integers we need to consider are 3 and 4, since $$ 3^2 = 9 $$ and $$ 4^2 = 16 $$. This means $$ 9 < 11 < 16 $$.

2. **Calculate the square roots**:
   - $$ \sqrt{9} = 3 $$
   - $$ \sqrt{16} = 4 $$

Since $$ 9 < 11 < 16 $$, we can conclude that:

$$
3 < \sqrt{11} < 4
$$

Thus, the two positive integers between which $$ \sqrt{11} $$ lies are **3 and 4**.
$$ \sqrt{11} $$ lies between 3 and 4.

\begin{tabular}{l} Berween which two positive integers do the follawing irrational number lies? \\ \( \sqrt{11} \) \\ \hline\end{tabular}

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