Axiom to prove $ \sqrt2 $ is irrational by contradiction. $ \sqrt2 $ is irrational. Proof By contradiction.
Assume $ \sqrt2 $ is not irrational i.e. Assume $ \sqrt2 $ is rational.
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Since $ \sqrt2 $ is rational is false , hence $ \sqrt2 $ must be irrational.
Isn't this assumption $ \sqrt2 $ is rational incomplete? Shouldn't one also prove that $ \sqrt2 $ is also not an imaginary number or one of my axioms state something like there are only either rational or irrational numbers and then proceed with this proof?? 
By contradiction we just prove  $ \sqrt2 $ is rational is false, but $ \sqrt2 $ can still be something that's either imaginary, complex or yet even not discovered.
 A: Strictly speaking, we do not prove that $\sqrt 2$ even exists, i.e., we only prove

There is no rational number $x$ with $x^2=2$.

Similarly, we can prove 

There is no rational number $x$ with $0\cdot x=1$.

But after such a proof, we would not say that "$\frac10$ is irrational".  Instead, we say that $\frac 10$ is not defined. What is the difference?
Historically, $\sqrt 2$ appeard as length of the diagonal of a unit square, which by Pythagoras had the property that $x^2=2$. So the existence in some sense was not under doubt. But once we are given existence of a number with this property (or with another property $P$), there is no difference between "There is no rational number with property $P$" and "The (or any) number with property $P$ is irrational". Note that all non-real complex numbers, for example are irrational, hence questions about real or imaginary or complex can be ignored. If we know, say, from the fundamental theorem of algebra that $X^2-2$ has some complex root, then there is no need to first show that it is real before showing that it is not rational.
A: If we define $\sqrt2$ to be the positive root of $x^2-2=0$, by Newton's method we may show that the sequence $$x_{n+1}=x_n-{f(x_n)\over f'(x_n)}={x_n^2+2\over 2x_n}\quad,\quad x_1=1$$tends to $\sqrt 2$. Since all the terms of this sequence are real, so must be $\sqrt 2$, since a non-zero imaginary part forces $\sqrt 2$ to lie above the real line.
