Firstly, recall some information from beginning algebra:
For a nonnegative real number $\,k\,$, $$ \cssId{s7}{\overbrace{\ \ \ \sqrt{k}\ \ \ }^{\text{the square root of $\,k\,$}}} \cssId{s8}{:=} \ \ \cssId{s9}{\text{the unique nonnegative real number which, when squared, equals $\,k\,$}} $$
(Note: when the distinction becomes important in higher-level mathematics, then $\,\sqrt{k}\,$ is called the principal square root of $\,k\,$.)
Thus, the number that gets to be called the square root of $\,k\,$ satisfies two properties:
So, what is (say) $\,\sqrt{-4}\ $?
There does not exist a nonnegative real number which, when squared, equals $\,-4\ $.
Why not?
Every real number, when squared, is nonnegative:
for all real numbers $\,x\,$, $\,x^2 \ge 0\,$.
Complex numbers to the rescue!
Complex numbers allow us to compute the square root of negative numbers, like $\,\sqrt{-4}\ $.
Remember the key fact:
$\,i:=\sqrt{-1}\,$,
so that $\,i^2=-1\,$
Observe that $\,i\sqrt{\vphantom{h}p}\,$, when squared, does indeed give $\,-p\ $: $$ \cssId{s36}{(i\sqrt{\vphantom{h}p})^2} \ \cssId{s37}{=\ i^2(\sqrt{\vphantom{h}p})^2} \ \cssId{s38}{=\ (-1)(p)} \ \cssId{s39}{=\ -p} $$
Some of my students like to think of it this way:
You can slide a minus sign out of a square root, and in the process, it turns into the
imaginary number $\,i\,$!
Here are some examples:
Recall these two different questions, with two different answers:
Similarly, there are two different questions involving complex numbers, with two different answers:
The following statement is true:
for all nonnegative real numbers $\,a\,$ and $\,b\,$,
$\,\sqrt{ab} = \sqrt{a}\sqrt{b}\ $.
For nonnegative numbers, the square root of a product is the product of the square roots.
Does this property work for negative numbers, too?
The answer is NO, as shown next.
Certainly, anything called ‘the square root of $\,-4\,$’ must have the property that,
when squared, it equals $\,-4\ $.
Unfortunately, the following incorrect reasoning gives the square as $\,4\,$, not $\,-4\,$:
$$
\cssId{s87}{\text{? ? ? ? }}\ \ \ \
\cssId{s88}{(\sqrt{-4})^2} \ \
\cssId{s89}{=\ \ \sqrt{-4}\sqrt{-4}}
\cssId{s90}{\overbrace{\ \ =\ \ }^{\text{this is the mistake}}}\ \
\cssId{s91}{\sqrt{(-4)(-4)}}\ \
\cssId{s92}{=\ \ \sqrt{16}}\ \
\cssId{s93}{=\ \ 4} \ \ \ \
\cssId{s94}{\text{? ? ? ? }}
$$
Here is the correct approach:
$$
\cssId{s96}{(\sqrt{-4})^2} \ \
\cssId{s97}{=\ \ \sqrt{-4}\sqrt{-4}}\ \
\cssId{s98}{=\ \ i\sqrt{4}\ i\sqrt{4}} \ \
\cssId{s99}{=\ \ i^2(\sqrt{4})^2} \ \
\cssId{s100}{=\ \ (-1)(4)} \ \
\cssId{s101}{=\ \ -4}
$$
Similarly, $\,\sqrt{-5}\sqrt{-3}\,$ is NOT equal to $\,\sqrt{(-5)(-3)}\,$.
Instead, here is the correct simplication:
$$
\cssId{s104}{\sqrt{-5}\sqrt{-3}} \ \
\cssId{s105}{=\ \ i\sqrt{5}\ i\sqrt{3}} \ \
\cssId{s106}{=\ \ i^2\sqrt{5}\sqrt{3}} \ \
\cssId{s107}{=\ \ (-1)\sqrt{(5)(3)}} \ \
\cssId{s108}{=\ \ -\sqrt{15}}
$$
Be careful about this!
On this exercise, you will not key in your answer. However, you can check to see if your answer is correct. |
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