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

Question

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!

Answer 1

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

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

Answer 2

Well, you have the fact that $i = \sqrt{-1}$, In this case then we have:

$$i = \sqrt{-1}$$ $$i^2 = -1$$ and since $i = \sqrt{-1}$, we end up having something like this:

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

Answer 3

In general, when you "extend" the definition of something to a larger domain, as when real numbers are extended to the complex numbers, some of the rules in the original domain may not apply in the larger domain.

For example, two real numbers can always be compared, but the law of trichotomy (for any reals $x$ and $y$, exactly one of $x < y$, $x = y$, $x > y$ must be true) does not hold for complex numbers.

Similarly, the law in positive reals that $\sqrt{a} \sqrt{b} =\sqrt{ab} $ does not hold generally for complex numbers. It is an interesting exercise to determine for which complex numbers that equation holds.

Other examples:

Matrix multiplication is not commutative.

Quaternions, Hamilton's extension of complex numbers, are not commutative.

Octonions, the extension of quaternions, are not associative.

Answer 4

In your derivation you do not come along the symbol $\sqrt{-15}$ for no reason (maybe you actually do, because you use some formula for the roots of a quadratic equation, but then you should have the students look at how that comes about), but because you need a numbers whose square is $-15$. That should answer the question what that number multiplied by itself yields ;)

If you are brave enough to investigate this with the students, you can use $\sqrt{-15}\cdot\sqrt{-15}=\sqrt{(-15)^2}=\sqrt{15^2}=15$ to raise the question if maybe it is not possible to consistently work with roots of negative numbers.

And by the way, my favorite way to derive a solution for $x+y=a$, $xy=b$ is $$ (x-y)^2=(x+y)^2-4xy=a^2-4b, $$ hence $x=\frac12((x+y)+(x-y))=\frac12\left(a+\sqrt{a^2-4b}\right)$, $y=\frac12\left(a-\sqrt{a^2-4b}\right)$. Here we are reminded that we need the square of $\sqrt{a^2-4b}$ to be $a^2-4b$. It is also nice, because one can discuss where the apparent asymmetry between $x$ and $y$ suddenly comes from. Of course, if there is a square root of $a^2-4b$, then there is another, namely its negative.

Answer 5

Because $\sqrt{-1}$ can refer to both $i$ and $-i$, whose product is indeed $1$.