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The 47th edition of the Math Teachers at Play Blog Carnival is now up at Math Hombre. Check it out!

Activities, games, lessons, hands-on fun — the Math Teachers at Play blog carnival would love to feature your article about mathematics from preschool to precollege level (through the first year of calculus). You can submit your article online, or email John directly to make sure he gets your submission.

Deadline: This Wednesday night!
The carnival will be published Friday at Math Hombre.

I wrote this note in 2008 to introduce complex numbers. It is posted for the Math Teachers at Play blog carnival.

What is the ${\sqrt[4]{-16}}$?

In the set of Real Numbers, ${\mathbb{R}}$, the solution is undefined. We must consider the set of Complex Numbers, ${\mathbb{C}}$. Complex numbers are numbers of the form of ${a+bi}$, where ${a}$ and ${b}$ are real numbers: ${a,b\in\mathbb{R}}$ and ${i}$ is the imaginary unit. Note ${i}$ is defined by ${i^{2}=-1}$ or equivalently ${i=\sqrt{-1}}$.

Thus, we have:

$\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & (-16)^{1/4}\\ & = & (-1\cdot16)^{1/4}\\ & = & (-1\cdot2^{4)^{1/4}}\\ & = & (-1)^{1/4}\cdot2^{4/4}\\ & = & (-1)^{1/4}\cdot2\\ & = & ((-1)^{1/2})^{1/2}\cdot2\\ & = & (\sqrt{-1})^{1/2}\cdot2\\ & = & i^{1/2}\cdot2\\ & = & 2\sqrt{i}.\end{array}$

We now have ${\sqrt[4]{-16}=2\sqrt{i}}$. So we must ask "What is the ${\sqrt{i}}$?". To evaluate ${\sqrt{i}}$ we will use Euler’s formula, ${e^{ix}=\cos\left(x\right)+i\sin\left(x\right).}$ So by substituting ${x=\frac{\pi}{2}}$, giving

$\displaystyle \begin{array}{rcl} e^{i(\frac{\pi}{2})} & = & \cos\left(\frac{\pi}{2}\right)+i\sin\left(\frac{\pi}{2}\right)\\ & = & 0+i(1)\\ & = & i.\end{array}$

Note we have isolated ${i}$ and now arrived at the following equality: ${i=e^{i(\frac{\pi}{2})}}$. Taking the square root of both sides gives

$\displaystyle \begin{array}{rcl} \sqrt{i} & = & \sqrt{e^{i(\frac{\pi}{2})}}\\ & = & e^{i(\frac{\pi}{2})(\frac{1}{2})}\\ & = & e^{i(\frac{\pi}{4})}.\end{array}$

Thus, we have now have ${\sqrt{i}=e^{i(\frac{\pi}{4})}}$ and following through application of Euler’s formula to ${x=\frac{\pi}{4}}$, gives

$\displaystyle \begin{array}{rcl} \sqrt{i} & = & e^{i(\frac{\pi}{4})}\\ & = & \cos\left(\frac{\pi}{4}\right)+i\sin\left(\frac{\pi}{4}\right)\\ & = & \frac{1}{\sqrt{2}}+i(\frac{1}{\sqrt{2}})\\ & = & \frac{\sqrt{2}}{2}+i(\frac{\sqrt{2}}{2})\\ & = & \frac{\sqrt{2}+\sqrt{2}i}{2}.\end{array}$

So going back to the original problem and substituting ${\sqrt{i}=\frac{\sqrt{2}+\sqrt{2}i}{2}}$, gives

$\displaystyle \begin{array}{rcl} \sqrt[4]{-16} & = & 2\sqrt{i}\\ & = & 2(\frac{\sqrt{2}+\sqrt{2}i}{2})\\ & = & \sqrt{2}+\sqrt{2}i.\end{array}$

Hence, the solution, ${\sqrt[4]{-16}=\sqrt{2}+\sqrt{2}i}$ , is a complex number!

I wrote this note in 2008 to give a down-to-earth example to help in the understanding of negative numbers [2]. It is posted for the Math Teachers at Play blog carnival.

— 1. Introduction —

First, lets get all the formalities out of the way by observing a few algebraic rules regarding positive and negative real numbers. We will consider the familiar arithmetic operations of addition and multiplication [3]. Feel free to skip over this section and continue to section.

— 1.1. Rules Addition (Sum) —

Rules for adding any two real numbers ${a}$, ${b}$

1. The result of addition is called the sum.
2. The sum between ${a}$ and ${b}$ is indicated as:

$\displaystyle a+b$

3. If ${a}$ and ${b}$ are both positive, then the sum is positive.

$\displaystyle 4+3=7$

4. If ${a}$ and ${b}$ are both negative, then the sum is negative.

$\displaystyle -4+(-3)=-7$

5. If ${a}$ is positive and ${b}$ is negative and ${|a|>|b|}$, then the sum is positive.

$\displaystyle 4+(-3)=1$

6. If ${a}$ is positive and ${b}$ is negative and ${|a|<|b|}$, then the sum is negative.

$\displaystyle 3+(-4)=-1$

7. Addition is commutative: ${a+b=b+a}$.

$\displaystyle 4+3=3+4=7$

8. Addition is associative: ${a+(b+c)=(a+b)+c}$.

$\displaystyle 1+(2+4)=(1+2)+4=7$

9. ${a+(-b)=a-b}$.
10. ${a+(-a)=0}$; ${a+0=a}$.

— 1.2. Rules Multiplication (Product) —

Rules for multiplying any two real numbers ${a}$, ${b}$

1. The result of multiplication is called the product.
2. The product between ${a}$ and ${b}$ is indicated as:

$\displaystyle a\times b=a\cdot b=a(b)=(a)b=(a)(b)=ab$

3. If a and b have the same sign, then the product is positive.

$\displaystyle (4)(3)=12$

$\displaystyle (-4)(-3)=12$

4. If a and b have opposite signs, then the product is negative.

$\displaystyle (-4)(3)=-12$

$\displaystyle (4)(-3)=-12$

5. Distributive property: ${a(b+c)=ab+ac}$; ${a(b-c)=ab-ac}$.

$\displaystyle 4(2+1)=8+4=12$

$\displaystyle 4(2-1)=8-4=4$

6. Multiplication is commutative: ${ab=ba}$.

$\displaystyle (4)(3)=(3)(4)=12$

7. Multiplication is associative: ${a(bc)=(ab)c}$.

$\displaystyle 1(2\cdot6)=(1\cdot2)6=12$

8. The product of any real number and -1 is the additive inverse of the real number.

$\displaystyle (-1)a=-a$

$\displaystyle (-1)(-1)=-(-1)=1$

$\displaystyle (-a)(-b)=(-1\cdot a)(-1\cdot b)=(-1)(-1)(a)(b)=1(ab)=ab$

9. ${a\cdot1=1\cdot a=a}$; ${a\cdot0=0\cdot a=0}$.

— 2. Example —

Why is a negative times a negative a positive? Many people have trouble answering this question. While, most will not have an issue memorizing the rule. Understanding the reasoning for the rule is another matter. Many people try to come up with a visualization to help picture what is going on [1]. We shall use money to illustrate the concept. When going over the example try to think of a negative as a loss and a positive as a gain.

— 2.1. The Offer You Can’t Refuse —

Well, you asked for it!

Suppose your are bankrupt. You now own ${\0}$ dollars. You approach Mr. Mob Boss for a loan of ${\4,000}$ dollars. He does so with the condition that you pay him back 4 times that amount. You now have a debt of

$\displaystyle 4\cdot-\4,000=-\16,000$

See rule 1.2.4 for reference.

In other words, you owe Mr. Mob Boss ${4}$ debts of ${\4,000}$. You have lost ${\16,000}$ dollars. You take the ${\4000}$ and invest it all in starting an Italian restaurant. The problem is all your customers are members of La Cosa Nostra, who expect nothing less than a free meal with the finest wine.

Eventually realizing you will never break even, you approach Mr. Mob Boss to tell him the bad news. Unexpectedly, he is so happy about the service you provide that he decides to forgive your ${4}$ debts of ${\4,000}$!

$\displaystyle -4\cdot-\4,000=\16,000$

See rule 1.2.3 for reference.

You have gained ${\16,000}$ dollars, and, thus, you are back where you started…absolutely broke! However, try not to be so "negative". At least you have a job for the rest of your life.