-2

According to this, complex number is algebraically closed, i.e. every polynomial has complex root. What if we allow other type of equations?

I ask this question because equations seemingly can extent number sustem. (From $x+2=1$, we go from natural number to integer, $2x=1$ we have rationals, $x^2+1=0$ we have complex number). Seems that tetration will be the following path, I want to know if there are equation involving tetration (e.g. $x^{x^x+1}-x^x+x^3=1$)that we can create within the complex domain, has no complex root.

Community
  • 1
JSCB
  • 13,698
  • 15
  • 67
  • 125
  • 1
    Polynomials have by definition integer exponents, so the "polynomial" you stated is not actually one. (BTW it does have solutions, see here: http://www.wolframalpha.com/input/?i=%28x%2Bi%29%5E%287%E2%88%923i%29%2Bx%5Epi%3D0) – Anonymous Pi Aug 18 '15 at 16:22
  • 1
    Your thing isn't even a polynomial; polynomials have nonnegative integer exponents. In any case, $e^x$ has no root. – Akiva Weinberger Aug 18 '15 at 16:22
  • If you want a bigger field then $\mathbb C$, by the way, you can look up the quarternions. They're of the form $a+bi+cj+dk$, with $i^2=j^2=k^2=ijk=-1$. They're nonassociative; $ij=-k$ but $ji=k$. – Akiva Weinberger Aug 18 '15 at 16:33
  • 1
    There seem to be several implicit (and incorrect) assumptions behind the sequence of statements in the question, particularly some basic misunderstandings about the definition of a "polynomial" and the implicit assertion that if an equation has no roots, then "the roots must exist in some larger field". Could you please try to clarify what you're asking? – Andrew D. Hwang Aug 18 '15 at 16:33
  • Polynomials are particularly well-behaved; they generally work like integers. "Irreducible polynomials," which can't be factored, are the analogs of primes (i.e. $x^2+1$ over $\mathbb R$, $x+1$ over $\mathbb C$ — or any field, actually), Euclid's algorithm works, you have unique factorization, there are always GCDs and LCMs, etc. This ends up being the main reason that you can always extend a field with polynomial solutions. – Akiva Weinberger Aug 18 '15 at 16:45
  • Tetration really isn't well-behaved in $\mathbb C$. $i^i$ has more than one value, though there's one called the "principle value." Analytic functions like sine and $e^x$ are friendlier, I guess. – Akiva Weinberger Aug 18 '15 at 16:48
  • Hey, here's an equation for you: $x^2=0\ &\ x\ne0$ ! This gets you the dual numbers. The new thing is called $\epsilon$. (Wikipedia treats them as an extension of $\mathbb R$, but there's no reason not to do the same to $\mathbb C$.) Unfortunately, this is no longer a field; you can't have $\frac1\epsilon$. You do have some nice properties with differentiation; for example, $p(x+\epsilon)=p(x)+p'(x)\epsilon$. – Akiva Weinberger Aug 18 '15 at 16:59

1 Answers1

5

Will we find a equation with no complex root, and hence escape from complex number?

Those are two separate questions. $e^u=0$ has so complex solution, but there is no apparent way to attach a new solution and extend the complex numbers by it, while retaining useful properties like $e^u e^{-u} = 1$.

If you did add such a $u$, so as to "escape" to something even larger than the complex numbers, what would $u^3$ or $\sin(u)$ be? Are they related to each other in any way?

It's always possible to invent bigger algebraic systems, but the complex numbers are the maximum possible example of the main properties that they possess (commutative, field, complete so that calculus can be done, Archimedean). Bigger may not be better because it comes at the cost of losing some of the useful algebraic and analytic properties of the complex numbers.

ASCII Advocate
  • 2,462