Why non-real means only the square root of negative? Once in 1150 AD, an Indian mathematician Bhaskara wrote in his work Bijaganita (algebra) that,

There is no square root of a negative quantity, for it is not a square

However later on in 1545 an Italian mathematician Gerolamo Cardano while solving the problem $x+y=10$ and $xy=40$ obtained $x=5+\sqrt{-15}$ and $y=5-\sqrt{-15}$, although he discarded this result saying that it was "useless". But later on mathematicians like Albert Girard, Euler and W. R Hamilton introduced these complex roots with purely mathematical definition.
So this was the tale of imaginary numbers which tells us that concept of imaginary numbers was initially adopted to compensate the theory of polynomial roots (number of roots is equal to the degree of polynomial). However later on some mathematicians proposed it's practical application through geometrical interpretation and other ways.
Now I want to know that in mathematics how one can be so sure that the square root of negative numbers can be the set to be designated as non-real numbers. Or in nut shell can't there be any other definition of non-real numbers.
For example, $x^{\gamma k}=-x^{k}$ (where $\gamma$ is something similar to $\iota$ as in complex number)
or $\log{(-x)}=\gamma \log{x}$ (provided $x>0$)
Notice that I had taken the values in the functions where the input is not lying in the domain. Concept of imaginary numbers was similar to this (i.e. $\sqrt{-x}=\iota \sqrt{x}$). So in this way I'll be able to create hundreds or probably thousands of such non-real sets.
I am sorry if I am loosing some logic in this but this is more like a curiosity than a question.
 A: If you posit $\log(-x)=\gamma \log(x)$ for all $x$, and if you want to allow the usual operations like division, you are going to be forced to conclude that $\gamma=\log(-x)/\log(x)$ for all $x$, and in particular that
$$\frac{\log(-2)}{\log(2)}=\frac{\log(-3)}{\log(3)}$$
But the various logs in this equation already have definitions, and according to those definitions, the equation in question is not true (for any of the various choices of $\log(-2)$ and $\log(-3)$.  
Therefore, your $\gamma$ can exist only if you either ban division or change the definition of the log.  Likewise for your other proposal $x^{\gamma k}=-x^k$.  
This is why you can't just go adjoining new constants willy-nilly and declaring them to have whatever properties you want.  In the case of $i$, the miracle is that you can define it in a way that does not require you to revise the existing rules of arithmetic.  Such miracles are rare.  
A: It seems you understand the difference between a pure imaginary number and a complex number.  A pure imaginary number is the square root of a negative.  Also, a complex number has both real and pure imaginary part.  To wit-
$$c=a+bi$$
defines a general complex number.  I take it on faith that the square root of a negative number is not a real number.  The square root of a negative cannot be positive, negative or zero.  Here are some number system extensions.  The integers to the rationals to the real algebraic numbers to the real numbers, and finally the complex numbers.  Hope this helps.
A: From my understanding of the question(and my limited knowledge);  First constructing the complex numbers;
Complex numbers can be represented as two ordered real pairs; $(a,b)$. The operation of addition is made component by component, while the multiplication is defined $(a,b)(c,d)=(ac-bd, ad+bc)$. The conjugation is an operation defined by $(a,b)^{*}=(a,-b)$. However there is no restriction to construct new algebras in a similar way(It's called the Cayley-Dickson construction). You can define further algebras if you add enough rigor to them this way, then check if the equality you seek is only applicable in $\Bbb C$.
But you're restricted to create only algebras where you double the dimensions of the last algebra this way... Don't stress over it. There are more algebra systems like the Hyperreals, Surreals etc.
A: "There is no square root of a negative quantity, for it is not a square"
This statement very precisely defines why there can be no real solution to √(-1)
Can you find any number which, when squared, gives -1 as a result?
A: Yes. you can also declare $j^2=1$ where $j \neq 1$ and you'll get the double numbers.
Or $\epsilon ^2=0$, $\epsilon \neq 0$ and you'll get the dual numbers.
