If $i^2=-1$ by definition then, how can this contradiction be justified?

Question

$i=\sqrt{-1}$	$i^2=(\sqrt{-1})^2$
	$i^2=(\sqrt{-1})*(\sqrt{-1})$
	$i^2=\sqrt{(-1)*(-1)}$
	$i^2=\sqrt{1}$
	$i^2=1$

If $i^2=-1$ by definition then, how can such a definition be contradicting the distributive properties of algebra?

Answer 1

The properties you used hold only for $a,b \geq 0$ in $\mathbb{R}$. You must be careful when dealing with negative numbers in square roots, as it can lead to wrong conclusions.

Keep in mind that $\sqrt{x^2} \neq \sqrt{x}^2$ because the first one means $\left |x\right |$, however this is not true with the second one. Thus, what you're proving here is really $\left |-1\right |$ = $\left |1\right |$, which is true.

Answer 2

Arithmetic rules differ between real and complex numbers. The imaginary unit isn't the real square root of $-1.$ The imaginary unit is one solution of the equation $x^2+1=0.$ This makes a difference since the equation is real, but the solution is not, which is why you cannot solve it by real number methods.

A sure method to avoid mistakes like that is to calculate in $$ \mathbb{C}\cong \mathbb{R}[x]/(x^2+1) $$ say, with real polynomials modulo $x^2+1.$ The next better possibility is to work with polar coordinates. In any case, $\sqrt{-1}$ should not be used IMO.