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 August 26th, 2012, 02:11 PM #1 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Line segment How I can divide a line segment in a given ratio, except ratios of the form $\frac{p}{q}$ where $p,q \in \mathbb{N}$ but in ratio $k$ where $k$ is irrational?
 August 26th, 2012, 03:25 PM #2 Senior Member   Joined: Feb 2012 Posts: 628 Thanks: 1 Re: Line segment Are you asking for a geometric construction that divides a line segment in an irrational ratio?
 August 26th, 2012, 04:39 PM #3 Math Team   Joined: Sep 2007 Posts: 2,409 Thanks: 6 Re: Line segment Strictly speaking a ratio is a fraction so "ratio $\frac{p}{q}$" is the same as "ratio of p to q". Give that, what, exactly, does "ratio of k", where k is irrational, mean? You could say "ratio of k to r" where k and r are numbers, possibly irrational. Assuming that "ratio of k" means "ratio of k to 1", you can do this by constructing a linesegment of length k+ 1, the cutting it at length k. That's easy to do as long as you can construct a line segment of length k. So, for example, if $k= \sqrt{2}$, we can construct a segment of length $\sqrt{2}$ by starting with a segment of length 1, constructing a right angle at one end, then using compasses to mark of length 1 at one end. The segment connecting the two endpoints has length $\sqrt{2}$. Extend that and mark off a length one and you have a line segment marked of lengths of $\sqrt{2}$, a ratio of "$\sqrt{2}$ to 1". However, there exist numbers that are NOT "constructible" (you cannot construct a linesegment of that length with straightedge and compasses alone). In fact, it is possible to prove that any "constructible number" is "algebraic of order a power of 2". (A number is "algebraic of order n" if and only if it satifies a polynomial equation, with integer coefficients, of degree n but no such equation of lower degree. If x is a rational number, it can be written as $x= \frac{m}{n}$ with integer coefficients so it satisfies nx- m= 0, a polynomial equation of degree 1. Of course, the other way works also. If x is "algebraic of degree 1", it satisfies an equation of the form px+ q= 0 and so $x= -\frac{q}{p}$ and so is rational. The numbers that are "algebraic of degree 1" are precisely the rational numbers. $\sqrt{2}$ is not rational but satisfies the equation $x^2- 2= 0$ so is algebraic of order 2. For the same reasons, $\sqrt[3]{2}$ is algebraic of order 3 (which is not a power of 2). There exist real numbers, such as $\pi$ or e that are not "algebraic" of any order. They are the "transcendental numbers". Since all rational numbers are "algebraic of order $1= 2^0$ we can construct a ratio of $r= \frac{p}{q}$ to 1 which is the same as "p to q" or what you called just $\frac{p}{q}$. Since $\sqrt{2}$ is "algebraic of order $2= 2^1$" we can construct a ratio of $\sqrt{2}$ to 1 as I indicate above. But $\sqrt[3]{2}$ is algebraic of order 3 which is NOT a power of 2. We cannot construct a ratio of "$\sqrt[3]{2}$ to 1". And, as above, $\pi$ and e are ot algebraic of any order so we cannot construct ratios of "$\pi$ to 1" or "e to 1".)
 August 27th, 2012, 12:48 AM #4 Member   Joined: Nov 2010 Posts: 48 Thanks: 0 Re: Line segment Thanks. So I can construct line segments with lengrth $k$ where $k$ is algebraic number of order mostly 2. But what happen when I want to constract a line segment with ends $A$ and $B$ and a point $C$ on it so $\frac{AC}{CB}=k$ when $k$ is irrational? Does $k$ have to be as above?

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# by geometric construction, is it possible to divide a line segment into ratio (2 1):(2-1)

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