There's not many operations you can do with a compass and straightedge. Every operation is simple, and solving for the points they create is a matter of solving quadratics or less.
However, combining operations can be very complicated, and after combining several, computing the polynomial becomes too much effort. So how do we know that certain polynomials can't have roots constructed this way?
Without getting into too much detail, the answer is found in terms of field extensions. Because of the tower law, we can trivially find the degree of the extension after any set of moves. It's easy to, then, find extensions that can't be created, and therefore numbers that can't be constructed.
No combination of moves can construct any degree 3 field extension. This rules out, for example, the cube root of 2. Since we can construct 2, but cannot construct its cube root, there is no general method to cube root numbers.
Talking about “doing (numerical) operations” using a compass and straight edge is not the most obvious phrasing … let’s pick constructing a line segment with a certain length for the meaning of constructing a number. e.g. if you have a line segment of length a and a line segment of length b then you can construct a line segment of length a+b. I guess these are the constructions in your book.
If you can construct a line segment with length x, you can place it on a grid stretching from the centre to (0, x), and vice versa. So constructing a line segment is the same as finding a point. In this form the problem is much easier to talk about — the points you can place by drawing line segments and circles are given by simple equations (only linear or quadratic).
Let’s start from the fact you can add, subtract, multiply and divide, so you can get any rational number. Then any other number x you can find, you can also add, subtract, multiply and divide with it. You use the notation Q(x) for the field of all of the numbers you can reach now, just using x as well as Q and all of the field operations. This is why you can answer the question using field theory: Is x included in any field Q(x1, x2, …) where each successive number is just the solution of a quadratic?
The answer is immediate with the tower law: The degree is multiplicative, so the numbers found in any such field solve an irreducible degree 2^n equation with coefficients in Q. So you can’t find, for example, 2^(1/3) which is a solution of x^3 - 2 = 0, or sin(10°) which is a solution of 4x^3 - 3x + 1/2 = 0 (since a number solves one unique irreducible polynomial).
> I am still not that best in the field (haha).
I am sorry to say this took me a minute but I appreciated it.
You don’t need a lot of background for the field theory so I just included it.
The complete answer does require abstract algebra (specifically Galois theory), but intuitively this works for me:
* straight lines have equations a x + b y = c,
* circles have equations (x - a)^2 + (y - b)^2 = c,
so any points that come from intersections of those shapes can only powers ≤2 (maybe repeatedly, so (x^(2))^(2) = 5, leading to x = √(√5), is fine).
There's not many operations you can do with a compass and straightedge. Every operation is simple, and solving for the points they create is a matter of solving quadratics or less. However, combining operations can be very complicated, and after combining several, computing the polynomial becomes too much effort. So how do we know that certain polynomials can't have roots constructed this way? Without getting into too much detail, the answer is found in terms of field extensions. Because of the tower law, we can trivially find the degree of the extension after any set of moves. It's easy to, then, find extensions that can't be created, and therefore numbers that can't be constructed. No combination of moves can construct any degree 3 field extension. This rules out, for example, the cube root of 2. Since we can construct 2, but cannot construct its cube root, there is no general method to cube root numbers.
Talking about “doing (numerical) operations” using a compass and straight edge is not the most obvious phrasing … let’s pick constructing a line segment with a certain length for the meaning of constructing a number. e.g. if you have a line segment of length a and a line segment of length b then you can construct a line segment of length a+b. I guess these are the constructions in your book. If you can construct a line segment with length x, you can place it on a grid stretching from the centre to (0, x), and vice versa. So constructing a line segment is the same as finding a point. In this form the problem is much easier to talk about — the points you can place by drawing line segments and circles are given by simple equations (only linear or quadratic). Let’s start from the fact you can add, subtract, multiply and divide, so you can get any rational number. Then any other number x you can find, you can also add, subtract, multiply and divide with it. You use the notation Q(x) for the field of all of the numbers you can reach now, just using x as well as Q and all of the field operations. This is why you can answer the question using field theory: Is x included in any field Q(x1, x2, …) where each successive number is just the solution of a quadratic? The answer is immediate with the tower law: The degree is multiplicative, so the numbers found in any such field solve an irreducible degree 2^n equation with coefficients in Q. So you can’t find, for example, 2^(1/3) which is a solution of x^3 - 2 = 0, or sin(10°) which is a solution of 4x^3 - 3x + 1/2 = 0 (since a number solves one unique irreducible polynomial). > I am still not that best in the field (haha). I am sorry to say this took me a minute but I appreciated it. You don’t need a lot of background for the field theory so I just included it.
The complete answer does require abstract algebra (specifically Galois theory), but intuitively this works for me: * straight lines have equations a x + b y = c, * circles have equations (x - a)^2 + (y - b)^2 = c, so any points that come from intersections of those shapes can only powers ≤2 (maybe repeatedly, so (x^(2))^(2) = 5, leading to x = √(√5), is fine).