Substitution to Eliminate Linear TermsDate: 2/9/96 at 10:51:49 From: Anonymous Subject: Modern algebra topics I have a few questions about Modern Algebra: 1. When finding the roots of a cubic equation we substitute x = y - b/3 to eliminate the square term. Are there any such substitutions for eliminating the linear term in both quadratic and quartic equations? 2. Given a set S and a relation R on S, how do I show the equivalence classes are disjoint and that their union is S? Thanks, Joe D'Amato Date: 6/27/96 at 12:35:7 From: Doctor Ceeks Subject: Re: Modern algebra topics In any polynomial in x of any degree n, substitution of y-k for x yields a polynomial in y whose coefficients are polynomials in k. In particular, the coefficient of the linear term y will be a polynomial in k generally of degree n-1. Any value of k which makes this coefficient 0 will work. (And there will always be a solution...though possibly complex...thanks to Gauss' Ph.D. thesis!) For quadratics, this process is known as completing the square. (Note that your question asks something different from what you did for the cubic, where you eliminated the quadratic term, not the linear term. Also, note that you cannot simultaneously eliminate both the linear and quadratic terms in general, although in special cases you can.) For your second question: every element of S belongs to an equivalence class, so the union is definitely all of S. On the other hand, if x belongs to the equivalence classes A and B, then since all a in A are equivalent to x and all b in B are equivalent to x, by transitivity, all a in A are equivalent to all b in B, and therefore A = B. -Doctor Ceeks, The Math Forum |
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