Can I construct a line segment with the length $e$ or $\pi$?

Question

What I really mean is that without restriction(only circle and ruler),can we construct it with geometric method or something else. If we can or not,how or why?I am just interested in this question.Maybe your explanation will be beyond my ability.

Answer 1

Not with compass and straightedge, starting from a given interval of length $1$.

The line segments that are so constructible have lengths that are a (small) subset of the set of real algebraic numbers, and $e$ and $\pi$ are known to be transcendental (non-algebraic).

Answer 2

In short, no, neither $e$ nor $\pi$ can be drawn in the way you describe. The reason for this is because all constructible numbers (which are the class of numbers that can be drawn with a compass and straightedge) are algebraic numbers, and both $e$ and $\pi$ are transcendental.

Wikipedia has a decent article on the subject here

Answer 3

Mathematically speaking, no. Even if you had infinite precision, you cannot draw a line segment whose length is an irrational number (with the exception of those irrational numbers that involve square roots $\sqrt{\ \ }$). The set of numbers you can draw are constructables, which are a subset of the algebraic numbers. In short, we can construct any length by starting with the integers and performing the following operations with a compass and straightedge:

$$+,-,\times,\div,\sqrt{\ \ }$$

If your question is a physical one, though, then the answer is still no! But it gets more interesting: the modern physics interpretation would say that any exact length is impossible to draw due to the Heisenberg uncertainty principle. There is a lower bound on our physical ability to measure distances, upon other things, as says our current understanding of quantum theory.

Answer 4

A sub-field of the reals $R$ is a set $F\subset R$ such that $0,1\in F$ and $\forall x,y \in F\;\ (x\pm y\in F\;\land x y\in F\; \land [x\ne 0\to 1/x\in F] \;).$

By using Cartesian co-ordinates we can show that $E$, the set of constructible lengths and their negatives,and $0$, comprises the smallest sub-field of the reals that contains the square roots of all its positive members. Every $x\in E$ is algebraic (algebraic over $ Q$), which means $0=x^n+\sum_{j=1}^{n-1}q_j x^j$ for some $n\in N$ and some rationals $q_0,...,q_{n-1}.$

But $E$ is a proper subset of the algebraic reals.For example $y=\cos \pi/9$ satisfies $y^3-3 y/4-1/8=0$ but $y\not \in E.$ So a regular nonagon is not constructible.

This can all be shown by a fairly long series of elementary steps.

"Most" reals are not algebraic and are therefore not in $E.$ The non-elementary Hermite-Lindemann transcendence theorem implies that $e$ and $\pi$ are not algebraic and hence not in $E.$

Answer 5

Read an article on constructables but the really breezy intuitive wrap your brain around it explanation:

Basically the things we can do with a compass and straightedge that are not picking arbitrary points, angles, and values at random (in other words the things we can be sure of creating and recreating) but are extending from a frame of reference of just 1 reference unit distance and two endpoints are:

extending lines; making parallel/parallel lines, making circles, and intersecting lines and circles.

In other words anything to do with but only things pertaining to lines and circles.

Lines and circles conform to linear and quadratic equations. And if we can only bootstrap from previously constructed points and measurements we can: add, subtract by extending lines. We can divide m by n by making a line m units long, another n units long, and marking off the intersections of parallel lines. We can construct perpendicular lines and by the pythagorean theorem construct square roots.

But that's more or less it. Lines and circles conform to linear and quadratic equations and we can only identify measures in terms of solving linear and quadratic equations with coefficients of previously constructed measures.

So basically we can construct the rational numbers, square roots, and the algebraic numbers that are solutions to polynomials with only power of 2 exponents.

Odd roots and transcendental numbers simply are not possible. (At least not if you are limited to bootstrapping from simply 0 and 1.)

[pre-empting: When we draw a line from 2 to 4 we will, of course, include points that are pi and e distance. But we have no way of determining precisely which of the points they are. And given an arbitrary point, if it isn't such an algebraic root of an even power termed polynomial, we will have no way of determining the point's value.]