Can we introduce new operations that make quintics solvable? I have heard from various sources that the typical arithmetic operations (addition, subtraction, multiplication, division, rational exponentiation) are not sufficient to express in general the roots of a quintic polynomial. This is due to something along the lines of their inability to "express the necessary symmetries."
grabrovi0u
Answered question
2022-10-30
Can we introduce new operations that make quintics solvable? I have heard from various sources that the typical arithmetic operations (addition, subtraction, multiplication, division, rational exponentiation) are not sufficient to express in general the roots of a quintic polynomial. This is due to something along the lines of their inability to "express the necessary symmetries." Would it be possible to introduce new arithmetic operations that allow do "express the necessary symmetries" and therefore allow a quintic formula to be written?
Answer & Explanation
Layne Murillo
Beginner2022-10-31Added 14 answers
Step 1 You can invent your own notation: say r(a,b,c,d,e;n) for the n-th smallest root of the quintic with coefficients a,b,c,d,e. You may object that this is kind of a cop-out, and you'd be right, but it's also basically what we do with quadratics. We get excited when we find out that we can "solve" by writing , but all we've really done is say "the solution to x2=2 is a number that, when squared, gives 2". Which is not so insightful. Step 2 What's interesting about is that once we can "solve" equations like we find that we don't need anything new to solve - we can transform all such equations into the simpler form. Moreover, we can develop methods of reasoning with these solutions, so that e.g. and so forth (but note that there isn't really a great deal we can say about , for example). We don't so much say "these are the solutions of this equation" but "this is a relationship between the solutions of this equation and this other one", and that is something that takes genuine mathematical work and insight. Anyway, it sounds like Julian's answer is pretty much what you wanted, but I thought I'd provide some additional colour.