Prove that we can construct a line segment with the length $\sqrt{n}$.

Question

Question

If a line segment with the length $1$ is given then prove that we can construct a line segment with the length $\sqrt{n}$ for any $n \in \Bbb N$.

My approach:

I am looking for a solution using induction. For the base case it is obvious and like always the problem is the case $n+1$. I'm trying to use

$$n=(\sqrt{n+1}-1)(\sqrt{n+1}+1)$$

but I can't go further.

Any idea?

Answer 1

Hint: If you have line segments of length $1$ and $\sqrt n$, then constructing a certain right triangle will give you something of length $\sqrt{1+n}$.

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The classical solution -- without induction! -- is to construct $\sqrt n$ as the mean proportional between line segments of length $1$ and $n$.

Answer 2

You are perhaps looking for something like the Spiral of Theodorus, an incremental set of right triangles with the root of every integer as the hypotenuses.

Picture from Wikipedia: