tag:blogger.com,1999:blog-9004452969563620551.post1637762160659998456..comments2019-05-15T12:04:42.529+05:30Comments on Paramanand's Math Notes: Gauss and Regular Polygons: Euclidean Constructions PrimerUnknownnoreply@blogger.comBlogger5125tag:blogger.com,1999:blog-9004452969563620551.post-49455480085614225502013-03-20T14:11:12.739+05:302013-03-20T14:11:12.739+05:30Thanks, paramanand! When you switched from $x_{1}$...Thanks, paramanand! When you switched from $x_{1}$ to $x$ my thought process switched from specific to general, but really the discussion stems from a specific example so I should have realized that you are exposing a contradiction in the case at hand.rasrasternoreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-85197882721766584342013-03-20T14:10:22.326+05:302013-03-20T14:10:22.326+05:30Hello rasraster, Since there will be at least one ...Hello rasraster,<br />Since there will be at least one value of $y$ which is rational, it will lead to at least one value of $x = (1 - y^{2})/y$ as rational. But from the equation $x^{2} + 2x - 1 = 0$ we see that both the values of $x$ are irrational. That’s why we can’t have even a single value of $y$ as rational.paramanandhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-15466037899920480722013-03-20T14:09:27.769+05:302013-03-20T14:09:27.769+05:30@paramanand Thanks for the response. I follow it ...@paramanand<br /><br />Thanks for the response. I follow it until the last sentence. Having a factor that is linear in y and a rational constant means that there is at least one rational root, but I don’t see how it implies that all values of y must be rational. Could you please explain? Thanks.rasrasternoreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-40999683058797017442013-03-20T14:07:01.332+05:302013-03-20T14:07:01.332+05:30@rasraster Let me explain by using a simple examp...@rasraster<br /><br />Let me explain by using a simple example. If $x = \sqrt{2} - 1$ then $x^{2} + 2x - 1 = 0$.<br /><br />Let us then assume that $x_{1}, x_{2}$ are roots of $x^{2} + 2x - 1 = 0$ and $x_{1} &gt; x_{2}$. Next suppose we need to solve the following equation:<br /><br />$y^{2} + x_{1}y - 1 = 0$<br /><br />Then the approach is to replace $x_{1}$ by $x$ in the above equation and then we have the following system of equations:<br /><br />$x^{2} + 2x - 1 = 0\,\,\,\,\cdots (1)$<br />$y^{2} + xy - 1 = 0\,\,\,\,\cdots (2)$<br /><br />From the second equation above we can write $x = (1 - y^{2})/y$ and putting this value of $x$ in $(1)$ we get<br />$\displaystyle \left(\frac{1 - y^{2}}{y}\right)^{2} + 2\cdot\frac{1 - y^{2}}{y} - 1 = 0$<br /><br />This is what I have termed as “elimination” in my post and the objective here is a find an equation for $y$ with rational coefficients only.<br /><br />After simplification we get an equation of degree 4 and it may happen that it gets factorized as a production of two quadratic expressions over rationals (this is the case I am referring to as “the elimination of coefficients from equations of type and type will result in a quadratic equation with rational coefficients”) or it may remain irreducible. However it will never happen that it gets split as a product of a linear factor and a cubic factor over rationals. The reason is simple that the linear factor will make $y$ rational and hence $x = (1 - y^{2})/y$ will also turn out to be rational which is not the case.paramanandhttps://www.blogger.com/profile/03855838138519730072noreply@blogger.comtag:blogger.com,1999:blog-9004452969563620551.post-18302786263964698022013-03-20T14:03:53.858+05:302013-03-20T14:03:53.858+05:30Could you please explain what you mean by: “Thus ...Could you please explain what you mean by:<br /><br />“Thus we if have only two equations to solve and we wish to remove the irrationalities involved in the coefficients of 2nd equation then we need to eliminate them from both the equations”<br /><br />and<br /><br />“If an equation of type splits as a factor of two linear polynomials having coefficients from the field F1, then the roots of this equation will also be in F1, and in this the elimination of coefficients from equations of type and type will result in a quadratic equation with rational coefficients.”<br /><br />The key word in question for me is “eliminated.” Is this simply a mathematical operation, or does one redo the construction so that the irrationality disappears, or is it some other process? Could you please show an example of what you are referring to?<br /><br />Thanks!rasrasternoreply@blogger.com