Can we Construct an Exact Segment of Value \sqrt(n) for Any Natural Number n?

[tpm]

coud we find using rule and compass :Confused: an exact segment of value:

$$\sqrt (n)$$ for every natural number n

- the same but for $$n^{1/m}$$ where n and m are positive integers

- given a segment of length a known can we find an exact segment of length $$a 2^{-1/2}$$

for the case m=2 using Pythagorean theorem is easy to find but what about the other cases.

 

[robert Ihnot]

It is quite possible to take the square root of a number, and use the four arithmetic functions. However, everything by ruler and compass is equivalent to solving linear and quadratic equations.

In general we can not find for integers, N &M, N^(1/M).

 

[HallsofIvy]

More generally, "constructible" numbers are those that are algebraic of order a power of two- that is, that they satisfy an irreducible polynomial equation with integer coefficients of degree a power of 2 (1, 2, 4, 8, etc.).
$\sqrt{n}$ satisfies the equation x²- n= 0 and so is either algebraic of order 2 (if x²- n is irreducible- cannot be factored with integer coefficients) or algebraic of order 1: both of which are powers of two.
We can construct a segment of length $n^{1/m}$ if and only if m is a power of two.

That, by the way, is the reason we can't trisect all angles. cos(20 degrees) satisfies the irreducible polynomial equation 8x³- 6x- 1= 0: it is algebraic of order 3 and so is not "constructible". It's easy to construct a 60 degree angle: give an unit length, strike arcs of that length from each end point and draw the line from one end point to a point of intersection of those arcs. IF it were possible to trisect that 60 degree angle, it would be possible to construct a 20 degree angle. Now strike of a unit length on one leg of that angle and drop a perpendicular to the other. You would have constructed a segment of length cos(20) which is impossible.