There’s the quadratic formula and the one for 3 and 4 factor ones (apologies if I’m wording this wrong), but I just heard that apparently there isn’t anything like a quintic(?) formula and so on. Why is this?? Googling gives me a bunch of confusing terminology that’s difficult to parse.
In: 75
There are formulae, but what there aren’t are formulae that only involve addition, subtraction, multiplication, division, and nth roots (eg, square roots or cube roots).
As for why, that’s a question that vexed almost 300 years of mathematicians, until it was finally proven by a guy called Abel, and later another guy called Galois, who provided a general framework for this kind of question before being killed in a duel at the age of 19.
Lesson 1: if you like mathematics, don’t get involved in 19th century French politics.
Here’s roughly how Galois’ proof works.
**The first big insight** is that if you have a system of numbers, (eg, the real numbers, you can take a polynomial that doesn’t have a solution (eg, x^2 + 1 = 0), then pretend it does have solutions, and get a bigger number system. Eg, if we say “actually, x^2 + 1 = 0 has a solution, and we call it i”, then the real numbers turn into the complex numbers.
Galois came up with a general system for “extending” number systems with roots of polynomials, and used the system to shoot down a whole lot of questions that had been plaguing mathematicians since the Ancient Greeks, eg “how do you trisect an angle with ruler and compass”? (Galois: you can’t, and here’s why). or “how do you use ruler and compass to make a cube whose volume is twice that of a given one?” (Galois: you can’t, and here’s why).
Eg, we can extend the rational numbers with a solution to x^3 – 2 = 0, getting numbers of the form a + b 2^(1/3) + c 2^(2/3).
**The second big insight** is that sometimes, the roots of a polynomial are all pretty much indistinguishable. Not the *same*, but indistinguishable. Eg, x^2 + 1 has two roots, i and -i. But these two roots have all the exact same properties. The engineers use j instead of i to represent “the” root of x^2 + 1. But what if they were actually using j for -i all along? There’s no way anyone could know. If you take any chunk of maths, and replace all the i with -i, and all the -i with i, then it comes out exactly the same.
We could call this a “symmetry” of the roots of x^2 + 1. And there are, it turns out, two symmetries: {i -> -i and -i -> i}, and {i -> i and -i -> -i}.
If we collect all the symmetries of the roots of a polynomial together into a collection (mathematicians call this a “group” of symmetries), then there are all kinds of possibilities for what that “group” of symmetries might look like.
**The third big insight** is some interesting things about what kinds of “groups” of symmetries are possible – in general, and of roots of polynomials.
One important type of group is when you just rotate a number of things: For example {A -> B, B -> C, C -> D and D -> A}. The letters A, B, C and D just get rotated.
There are other, more complicated groups of symmetries: for example, if we look at all possible permutations of three things, there are 6 of them: first, there are three cycles:
* {A->A, B->B, C->C}, (kind of a trivial cycle, I know).
* {A->B, B->C, C->A},
* {A->C, B->A, C->B},
And there are also three swaps:
* {A->A, B->C, C->B},
* {A->C, B->B, C->A},
* {A->B, B->A, C->C}.
If A, B and C these were roots of polynomials, it would represent a situation (like the complex numbers i and -i), where the roots could be swapped or cycled any old way, and nobody would be able to tell the difference.
Not all polynomials are like this. For example,
x^3 + 2 = 0 over rational numbers has three roots, but if you decide to shuffle them, then as soon as you decide what to replace the first one with, then your decisions about the others are forced. It turns out that all the shuffles of the roots are cycles.
This is typical of polynomials of the form x^n – a = 0.
And it turns out this cuts both ways: if your group of roots has only cycles (or powers of cycles), then your polynomial has to be x^n – a = 0 in disguise.
**The fourth big Insight:**
Not every group of root shuffles has just cycles, but many of them can be built out of things that are just cycles.
For example: We could build up this group:
* {A->A, B->B, C->C},
* {A->B, B->C, C->A},
* {A->C, B->A, C->B},
* {A->A, B->C, C->B},
* {A->C, B->B, C->A},
* {A->B, B->A, C->C}
by saying “let’s just have all the cycles of A, B and C”:
* {A->A, B->B, C->C},
* {A->B, B->C, C->A},
* {A->C, B->A, C->B},
(so this would correspond to taking a cube root) and then saying “let’s also have the cycles that swap A and B” (so that would correspond to a square root).
So if you had a polynomial whose 3 roots could be shuffled in every possible way, it would be possible to solve it by doing a cube root, and then a square root, with possibly some normal addition, multiplication etc between, before, and after.
**The fifth big Insight:**
So, what polynomials can be solved with roots? Galois showed this is the same question as “which groups of shuffles can be built up out of cycles?”
And people who studied group theory already knew enough about that second question to give the famous answer about quintics.
It turns out that any group of shuffles of 1, 2, 3 or 4 roots can be built out of cycles, but as soon as you add that 5th root, there are some groups of shuffles that can’t; and so, the 5th degree polynomials corresponding to that group can’t be solved with just roots.
As you can see, there’s a lot .. a lot of somewhat heavy maths behind it, which I’ve tried (and probably failed) to ELI5.
Galois was a genius, and spent the night before his duel frantically writing letters to friends, and writing down pages and pages of advanced maths that was still in his head and nowhere else. This was at the age of 19. If he had stayed out of French politics, or carefully avoided the girl his political enemies sent to entrap him, he might be a household name – who knows what else he would have discovered, given a chance at a long and productive life.
Latest Answers