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 where but in ratio where 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: 5  Re: Line segment
Strictly speaking a ratio is a fraction so "ratio " 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 , we can construct a segment of length 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 . Extend that and mark off a length one and you have a line segment marked of lengths of , a ratio of " 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 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 and so is rational. The numbers that are "algebraic of degree 1" are precisely the rational numbers. is not rational but satisfies the equation so is algebraic of order 2. For the same reasons, is algebraic of order 3 (which is not a power of 2). There exist real numbers, such as or e that are not "algebraic" of any order. They are the "transcendental numbers". Since all rational numbers are "algebraic of order we can construct a ratio of to 1 which is the same as "p to q" or what you called just . Since is "algebraic of order " we can construct a ratio of to 1 as I indicate above. But is algebraic of order 3 which is NOT a power of 2. We cannot construct a ratio of " to 1". And, as above, and e are ot algebraic of any order so we cannot construct ratios of " 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 where is algebraic number of order mostly 2. But what happen when I want to constract a line segment with ends and and a point on it so when is irrational? Does have to be as above?


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