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The Sufficiency of Complex Numbers

Starting with the Natural Numbers {N|1, 2, 3...}, i.e. numbers you can see

in nature, we can solve equations like

x+3=5

If we try different numbers in that equation then we can create an equation

that the Natural Numbers can’t solve, such as

x+4=3

No number you can see in nature will work. To solve that equation we

have to expand our toolbox of numbers and create a new kind of number—

the negative numbers.

Together with the Natural Numbers we call these the

S S

Integers {Z|0 a −a, where a ∈ N}, i.e. the positive and negative Natural

Numbers, and we can use them to solve all equations of this form no matter

what other two numbers we put in the equation. We can also use Integers to

solve equations like this one where we multiply the variable by a number:

2x + 4 = 0

But with the following equation we again run into problems:

2x + 3 = 4

There is no Integer that will solve this problem, so again we must expand our

toolbox of numbers, this time with fractions. Including fractions with the

Integers gives us the Rational Numbers {Q| ab , where a, b ∈ Z}, and we can

use them to solve ALL first-degree equations. But what about second-degree

equations?

x2 + 4 = 3

This equation cannot be solved using Rational Numbers. To be able to

solve all second-degree equations we need another new√kind of number—the

Complex Numbers {C|a + bi, where a, b ∈ R and i = −1}.

By this point the suggested pattern begs the question, ”What higher

order equations might there be that require yet further types of numbers to

be invented?” We’ve had to expand our toolbox of numbers with each new

kind of equation. Three times already we’ve needed new kinds of numbers

to solve higher-order equations. What kinds of numbers might we need to

solve third-degree equations, or fourth, or sixty-fifth degree equations?

Amazingly, none.

All equations of n’th degree—no matter how big n is—are solved by the

set of Complex Numbers.

1

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