Tuesday, October 2, 2012

The two meanings of $\sqrt{4}$

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What is wrong with the following expression?

$$4=\sqrt{16}=\sqrt{-4 \times -4}=\sqrt{-4}\times \sqrt{-4}=-4.$$

It seems easy. $\sqrt{-4}$ doesn't make sense. Right? Well, yes and no.

If we just consider the usual real numbers, then square of any number is positive and $\sqrt{-4}$ simply does not exists.

However, there is something called complex numbers.  For complex numbers, square can be negative.  In this case, there are two ways of interpreting $\sqrt{-4}$.

1.  $\sqrt{-4}=2 i$, where $i$ is the imaginary unit.  In this case, the equality $\sqrt{-4 \times -4}=\sqrt{-4}\times \sqrt{-4}$ does not hold.

2. $\sqrt{-4}=\{ \pm 2 i \}$, i.e. it is a set.  We would also reinterpreting $\sqrt{16}=\{\pm 4\}$. In this case, we have
$$\{\pm 4\}=\sqrt{16}=\sqrt{-4 \times -4}=\sqrt{-4}\times \sqrt{-4}=\{\pm 2\}\times\{\pm 2\}=\{\pm 4\},$$
here we use element-wise multiplication of two sets.

In general, there are two different meanings of $\sqrt{x}$.

1. $\sqrt{x}$ is the unique positive square root of $x$, if $x>0$; $\sqrt{0}=0$; $\sqrt{x}=\sqrt{-x}i$ if $x<0$;  any one of the square root of $x$ if $x$ is not a real number.  Under this meaning, we have $\sqrt{ab}=\sqrt{a}\times\sqrt{b}$ only if at least one of $a$ and $b$ are nonnegative.

Under this meaning, the formula for the quadratic equation $ax^2+bx+c=0$ is $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.  As we have extended the definition of $\sqrt{\cdot }$, this formula now makes senses for any complex numbers $a\ne 0$, $b$ and $c$.

2. $\sqrt{x}$ is the set consists of the two roots of $x$.  Indeed, we can generalize it to $\sqrt[n]{x}$ being the set of all $n$-th roots of $x$.  For this meaning, we have $\sqrt[n]{ab}=\sqrt[n]{a}\times\sqrt[n]{b}$ for all complex numbers $a$ and $b$.

The formula for the quadratic equation is now simply $\frac{-b+\sqrt{b^2-4ac}}{2a}$, which is indeed the set of the two solutions.