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I think this actually misses the biggest point. Complex numbers are special because they are an algebraic closure of the real numbers (what makes the real numbe

I think this actually misses the biggest point. Complex numbers are special because they are an algebraic closure of the real numbers (what makes the real numbers interesting should be more obvious), and because Zorn's lemma tells us that such a closure is unique (up to isomorphism). In other words, complex numbers are precisely what you need to add to real numbers in order to always be able to solve polynomial equations. TL;DR: complex numbers are the unique algebraic extension to the real numbers

akvadrako 9y ago
What do you mean by closure?
- mfukar 9y ago
  Algebraic closure of a field. See the fundamental theorem of algebra.
- enord 9y ago
  Simply put: It's when you extend the mathematical system by applying an operation to all eligible elements in the system (for example by redefining which elements are amendable to a given operation, like sqrt(-1)) Logic example (transitive closure): given "A->B", "B->C" in a sytem then your system can be extended to also contain "A->C" if you decide to interpret "->" as transitive.
- joppy 9y ago
  A field is algebraically closed if whenever you have a polynomial with coefficients in that field, all the roots of the polynomial also belong to the field. For example: The field of rational numbers is not closed since it does not contain the roots of x^3 - 3 (since we can prove the cube root of 3 is irrational). The real numbers are not closed since they do not contain the roots of x^2 + 1. The complex numbers however, are closed. An algebraic closure of a field is an embedding of that field into a closed field. For example, the rational numbers into the complex numbers, or the real numbers into the complex numbers. The second case is special: while rational numbers have many possible closures, the only closure of the real numbers is the complex numbers.
- betterunix2 9y ago
  It means that all polynomials with coefficients in the real numbers have roots in the complex numbers. Maybe a different example would help. Imagine we are working with Integers modulo 3 (denoted Z/3Z). Z/3Z is a field, meaning we have all the expected arithmetic (+, -, , /) and we can look at polynomials. Here is a polynomial that is irreducible i.e. does not have roots in Z/3Z: x^2+1 We can construct an extension field in which this has a root; let's call the root i and write the extension Z/3Z(i). Here is where finite fields like Z/3Z differ from the real numbers: extensions like Z/3Z(i) still have irreducible polynomials. On the other hand, in the real numbers, x^2+1 is irreducible, but if we construct an extension where it is reducible, R(i) i.e. the complex numbers, we get a field where there are no irreducible polynomials (that is the basic idea of the fundamental theorem of algebra).
- nabla9 9y ago
  What others said, but more abstract and simple way to say it: Algebra is set of objects and operations on them. Operation is function that gets number of objects as argument and maps it to some other thing A set has closure under an operation(s) if using those operations on the set members always gives you a member in the same set. Example: Natural numbers (0,1,2, ...) are closed under addition and multiplication, because you can't add two numbers and get something that is not natural number. Natural numbers are not closed under subtraction or division because you can get a result that is not a natural number.
  - katastic 9y ago
    >A set has closure under an operation(s) if using those operations on the set members always gives you a member in the same set. THANK YOU. That line was key and just clicked for me. I'm actually liking everyone's completely-different ways of explaining the same things. When one, or five, are confusing, the sixth may be key to filling in the holes.
mathgenius 9y ago
He does mention this.
monochromatic 9y ago
Doesn’t use the word closure but it does mention this. > Complex numbers were introduced because they cropped up naturally in the solution of cubic equations. What might we need to introduce to solve higher order polynomial equations? What if we let the coefficients of our polynomials be complex numbers themselves? > The astonishing answer is: nothing. Once you've allowed yourself complex numbers it turns out that you don't need anything else for higher order polynomials, or even polynomials with complex coefficients. The fundamental theorem of algebra tells us that any polynomial of degree n with complex coefficients (and that includes real ones, since a real number is a complex number with an imaginary part of zero) has n complex roots. > That's a beautiful mathematical result.
deleted 9y ago
[deleted]
zardo 9y ago
>TL;DR: complex numbers are the unique algebraic extension to the real numbers The simplest, (or maybe most compact) extension. You can also achieve closure by going from R to R^2 and adding the Geometric Product. This is also closed, but it includes vectors in R^2 as well as the complex numbers.

akvadrako 9y ago

What do you mean by closure?

mfukar 9y ago
Algebraic closure of a field. See the fundamental theorem of algebra.
enord 9y ago
Simply put: It's when you extend the mathematical system by applying an operation to all eligible elements in the system (for example by redefining which elements are amendable to a given operation, like sqrt(-1)) Logic example (transitive closure): given "A->B", "B->C" in a sytem then your system can be extended to also contain "A->C" if you decide to interpret "->" as transitive.
joppy 9y ago
A field is algebraically closed if whenever you have a polynomial with coefficients in that field, all the roots of the polynomial also belong to the field. For example: The field of rational numbers is not closed since it does not contain the roots of x^3 - 3 (since we can prove the cube root of 3 is irrational). The real numbers are not closed since they do not contain the roots of x^2 + 1. The complex numbers however, are closed. An algebraic closure of a field is an embedding of that field into a closed field. For example, the rational numbers into the complex numbers, or the real numbers into the complex numbers. The second case is special: while rational numbers have many possible closures, the only closure of the real numbers is the complex numbers.
betterunix2 9y ago
It means that all polynomials with coefficients in the real numbers have roots in the complex numbers. Maybe a different example would help. Imagine we are working with Integers modulo 3 (denoted Z/3Z). Z/3Z is a field, meaning we have all the expected arithmetic (+, -, , /) and we can look at polynomials. Here is a polynomial that is irreducible i.e. does not have roots in Z/3Z: x^2+1 We can construct an extension field in which this has a root; let's call the root i and write the extension Z/3Z(i). Here is where finite fields like Z/3Z differ from the real numbers: extensions like Z/3Z(i) still have irreducible polynomials. On the other hand, in the real numbers, x^2+1 is irreducible, but if we construct an extension where it is reducible, R(i) i.e. the complex numbers, we get a field where there are no irreducible polynomials (that is the basic idea of the fundamental theorem of algebra).
nabla9 9y ago
What others said, but more abstract and simple way to say it: Algebra is set of objects and operations on them. Operation is function that gets number of objects as argument and maps it to some other thing A set has closure under an operation(s) if using those operations on the set members always gives you a member in the same set. Example: Natural numbers (0,1,2, ...) are closed under addition and multiplication, because you can't add two numbers and get something that is not natural number. Natural numbers are not closed under subtraction or division because you can get a result that is not a natural number.
- katastic 9y ago
  >A set has closure under an operation(s) if using those operations on the set members always gives you a member in the same set. THANK YOU. That line was key and just clicked for me. I'm actually liking everyone's completely-different ways of explaining the same things. When one, or five, are confusing, the sixth may be key to filling in the holes.

mathgenius 9y ago

He does mention this.

monochromatic 9y ago

Doesn’t use the word closure but it does mention this. > Complex numbers were introduced because they cropped up naturally in the solution of cubic equations. What might we need to introduce to solve higher order polynomial equations? What if we let the coefficients of our polynomials be complex numbers themselves? > The astonishing answer is: nothing. Once you've allowed yourself complex numbers it turns out that you don't need anything else for higher order polynomials, or even polynomials with complex coefficients. The fundamental theorem of algebra tells us that any polynomial of degree n with complex coefficients (and that includes real ones, since a real number is a complex number with an imaginary part of zero) has n complex roots. > That's a beautiful mathematical result.

deleted 9y ago

[deleted]

zardo 9y ago

>TL;DR: complex numbers are the unique algebraic extension to the real numbers The simplest, (or maybe most compact) extension. You can also achieve closure by going from R to R^2 and adding the Geometric Product. This is also closed, but it includes vectors in R^2 as well as the complex numbers.