Solvable polynomials ed

Introduction ed

In school you learn: second order polynomials \(p(x) = ax^2 + bx + c = 0\) can be solved via\[ x_{1,2} = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Similarly, formulas for 3rd and 4th order polynomials have been found using more and higher roots. But the pattern breaks for 5th order...

(explicit example for \(a,b,c\)) Playing a game: you don't know any numbers, but know how to calculate. Start with knowing \(p\), so we know \(a,b,c\). Include \(b^2\) etc. and the root...

Fields ed

Same game, more extreme. You know \(0\) and \(1\). We can write \(1+1, 1+1+1, 1+1+1+1,\dots\), so we need to include \(2,3,4,\dots\) (all positive integers). Also the negative, because \(0-1,0-1-1,\dots\). So, at least, we need all integers \(\mathbb Z\).

We can also look at multiplication, but whatever integers we multiply, the result is again an integer, so multiplication does not force us to include more numbers. What about division? We get \(\frac{1}{2}, \frac{1}{3}, \frac{-13}{5},\dots\), this will "fill" all spaces between integers and we get the rational number \(\mathbb Q\) of all fractions \(\frac{m}{n}\).

\(\mathbb Q\) is finally "self consistent". Any operation (\(+,-,\cdot,/\)) will only give you numbers, that you already know. \(\mathbb Q\) is called a field. and it is the smallest field starting with \(0\) and \(1\).

Roots ed

What is a square root \(\sqrt{2}\)? You're looking for a number \(x\) that gives \(2\), when squared. I.e. it solves the equation\[x^2 = 2 \, .\]