[Math] Why $\sqrt {-1}\cdot \sqrt{-1}=-1$ rather than $\sqrt {-1}\cdot \sqrt{-1}=1$. Pre-definition reason!

complex numberseducationmath-history

It is for years that I teach complex numbers following a historical route. I start with the famous problem of Cardano:

Find two numbers whose sum is equal to 10 and whose product is equal
to $40$.

After "solving" we try to check the "things" we have found actually satisfies the conditions of the problem. To do so, we need to multiply $\sqrt {-15}$ by $\sqrt {-15}$. "Relying on" and at the same time "extending" the rule of real numbers, we have two choices: take the product to be $-15$ or $15$. We choose the first choice since it gives the solutions of the original problem. I wonder whether there is a better pre-definition explanation of this choice.

Update Please consider that you all know the right answer! Go back in time for a moment and remember that the Great Euler took $\sqrt {-1}$ . $\sqrt {-4} = 2$. See for example here. And remember that I follow a historical route. And I have no $i $, no nothing yet. So later on, I use something like $\sqrt {-4} = \sqrt {4} \sqrt {-1} = 2\sqrt {-1}$. Now from a student's point of view, I guess everything seems like an ad hoc game: we use whatever we need whenever we want!

Best Answer

Consider the roots of $x^2=-1$. If its roots are presented as $\sqrt{-1}$ and $-\sqrt{-1}$ then we can say that $(\sqrt{-1})^2=-1$.


Look at here to see why the notation of $\sqrt{-1}$ is confusing.

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