Why for a real number $x$ and an even integer $n, x^{1/n}$ is not real if $x<0$?

Question

Why for a real number $x$ and an even integer $n$, $\;x^{1/n}$ is not real if $x<0$? What I'm really asking is how do you define a complex number without referring to imaginary numbers.

How do you define a complex number without circular logic that involves a complex number
($a+bi$ where $i$ is the imaginary unit, and $a, b$ are real).

Explain without assuming I'm defining functions.

Answer 1

Let $x<0$,$n=2k$

Assume $x^{\frac{1}{2k}}$ to be a real number

What this means that there exist a Real number $p=x^{\frac{1}{2k}}$ such that $p^{2k}=x$

But $p^{2k}=(p^k)^2$, and The square of any real number is always non-negative, Hence $(p^k)^2$ is also non-negative, which in fact means $p^{2k}$ is non-negative

But $p^{2k}=x$, Which follows x must be non-negative, Which contradicts the assumption that There exist a real number $x^{\frac{1}{2k}}$,Such that $x<0$

Hence our assumption is wrong, There does not exist a a real number $x^{\frac{1}{2k}}$,Such that $x<0$

Now $p$ is not real

That means there is a whole set of such cases where all such $p's$ are not real

Now Imaginary numbers were named as such numbers, Whose $2^{nd}$ or $4^{th}$ or even powers would lead to a negative real number, Historically, They were first conceived by Italian Mathematician Gerolamo Cardano, While solving a general cubic equation

Roots such as $a+b\sqrt-1$ Were expressed, Now, These itself had no real meaning in those times, But when these roots were put into the polynomial, The result would give out "$0$",Meaning they had mathematical usefulness, Such complete expressions were later named complex numbers, But the imaginary parts were indeed already used before, For a short history lesson, This might be useful, Note that $i$,whose notation as $\sqrt-1$ was given by Rafael Bombelli, always was used historically for its mathematical usefulness as a number whose square is $-1$

It was later in time that geometrical interpretations were given to such expressions and later called complex numbers as $a+ib$, Note $i$ as complex number $0+1i$ was later seen as "complex", after these definitions were rigorously defined, Hence, historically complex numbers were consequence of Applying mathematical operations on imaginary and Real numbers

Answer 2

"How do you define a complex number without circular logic that involves a complex number is $a+bi$ where $i$ is a complex number, $a$ real $b$ real."

First, you write ${\bf R}[x]$ for the ring of all polynomials with real coefficients.

Then you let $J$ be the ideal of all multiples of the polynomial $x^2+1$.

Given a polynomial $p(x)$ in ${\bf R}[x]$, you define the coset of $J$ containing $p(x)$ to be the set of all polynomials of the form $p(x)+j(x)$ where $j(x)$ runs through $J$, that is, it's the set of all polynomials that are $p(x)$ plus a multiple of $x^2+1$. We'll use the notation $\overline{p(x)}$ for this coset.

Notice that any polynomial $p(x)$, when you divide by $x^2+1$, gives you a quotient and a remainder, and the remainder is a polynomial of degree one, $ax+b$ for some real numbers $a,b$. You have $$ p(x)=(x^2+1)q(x)+ax+b $$ ($q(x)$ is the quotient when you do the division). Notice that in this situation, $\overline{p(x)}=\overline{ax+b}$; since $p(x)$ and $ax+b$ differ by a multiple of $x^2+1$, they define the same coset of $J$.

So, the set $\bf C$ of all cosets of $J$ is the set of all $\overline{ax+b}$ with $a,b$ real.

Well, $\bf C$ is the complex numbers, defined here with no circularity. The element $\overline x$ satisfies $\overline{x}^2=\overline{-1}$ because $\overline{x^2+1}=\overline0$. So $\overline x$ plays the rôle of $i$.