1. **Problem statement:** (i) Starting from a right triangle with sides 1, 1, and $\sqrt{2}$, construct a segment of length $\sqrt{3}$ using straightedge and compass. (ii) Use this method to construct $\sqrt{n}$ for $n=2,3,4,5,6,\ldots$. 2. **Key idea and formula:** We use the Pythagorean theorem: in a right triangle with legs $a$ and $b$, the hypotenuse $c$ satisfies $$c=\sqrt{a^2+b^2}.$$ Starting with a segment of length 1, we can build $\sqrt{2}$ by constructing an isosceles right triangle with legs 1 and 1. 3. **Step-by-step construction for $\sqrt{3}$:** - Start with a segment of length 1. - Construct a right triangle with legs 1 and $\sqrt{2}$ (the hypotenuse of the previous triangle). - By Pythagoras, the hypotenuse of this new triangle is $$\sqrt{1^2 + (\sqrt{2})^2} = \sqrt{1 + 2} = \sqrt{3}.$$ 4. **General construction for $\sqrt{n}$:** - Assume you have constructed a segment of length $\sqrt{n-1}$. - Construct a right triangle with legs 1 and $\sqrt{n-1}$. - The hypotenuse will be $$\sqrt{1^2 + (\sqrt{n-1})^2} = \sqrt{1 + (n-1)} = \sqrt{n}.$$ 5. **Explanation:** This method uses the fact that each new $\sqrt{n}$ can be obtained by adding a leg of length 1 to the previous $\sqrt{n-1}$ segment in a right triangle, then applying the Pythagorean theorem. 6. **Summary:** - Start with segment length 1. - Construct $\sqrt{2}$ from the isosceles right triangle. - For each $n \geq 3$, construct a right triangle with legs 1 and $\sqrt{n-1}$ to get $\sqrt{n}$. This provides a straightedge and compass construction for $\sqrt{n}$ for all positive integers $n$.