# Why were Imaginary numbers invented

1. Dec 14, 2004

### daster

Why was $i$ invented?

2. Dec 14, 2004

### chroot

Staff Emeritus
3. Dec 14, 2004

### Tom Mattson

Staff Emeritus
It was invented to be a solution to the equation x2+1=0.

Before you write this off as a contrivance, consider the following. Imagine a civilization that has only the counting numbers {1,2,3,...}. In their mathematics, there exists solutions to equations such as x-2=1 (let's assume these people had algebra). But an equation of the form x+1=1 has no solution in this system. So, they invent the number 0 and so extend their number system to the whole numbers {0,1,2,3,...}

So now all algebraic equations have solutions, right? Wrong. If they are confronted with the equation x+3=0, they find that their number system is inadequate. So they invent negative numbers and so extend their system to the integers {...,-2,-1,0,1,2,...}. But again they hit a speed bump when confronted with equations such as 2x-1=0, whose solutions are not integers. So, they extend their number system to the rationals {0,1/2,-4/7,4.5,-133212,...}, which are all numbers that can be expressed as the ratio of 2 integers.

So that covers everything, right? Wrong again! Consider the equation x2-2=0. The solutions (there are 2) are not rational (this can be proven in number theory), and so they are not in our civilization's existing number system. So they include irrationals in their number system, which are all the numbers that can be represented as decimals, but are not rational. This number system is called the real number system {0,1.2, -21/2,e,&pi;,3/4}, which is the set of all numbers that can be represented by a decimal.

So you might be tempted to think that their number system is adequate to provide the solution to any algebraic equation now, right? Well, maybe not, because I gave the answer at the top of this post. As I said, the equation x2+1=0 still has no solution in this system, so they extend their system to include the imaginary numbers, which are multiples of i=(-1)1/2. So the number system is extended to the complex numbers {2.3, -1+3i, &pi;i, (43-112i)/3,(53)1/2i,...}.

Naming the reals "reals" and the imagniary numbers "imaginary" is probably the biggest misnomer in all of mathematics. It leads people to view the so-called "reals" as more credible than the so-called "imaginaries", and to view the latter with suspicion. That is most unfortunate, because the reals are no more "real" (ontologically) than the imaginaries. I mean, it's not as though you can stub your toe on "the number 3".

Hopefully my little anecdote (which parallels roughly the development of number systems in human civilizations) will show that the imaginary numbers are just a natural extenstion of our number system, very much in the spirit in which it had been extended previously with various subsets of the reals.

4. Dec 14, 2004

### daster

So, $i$ was invented to counter the "root of a negative number" problem, which I understand. My real question is:

Why isn't there another type of 'imaginary' number that deals with division by zero?

$$q=\frac{1}{0}$$

5. Dec 14, 2004

### cyby

q = 1/0 would lead to a lot of inconsistencies in our current axiomatic system, no? Furthermore, I'm sure you've seen proofs which demonstrate the problems with defining 1/0...

6. Dec 14, 2004

### Tom Mattson

Staff Emeritus
Because the notion is ill-defined. If you allow division by zero, you can "prove" that any 2 numbers are equal to each other, as follows:

Let x=y.
x2=xy
x2+x2=x2+xy
2x2=x2+xy
2x2-2xy=x2+xy-2xy
2x2-2xy=x2-xy

Now pull out a common factor on each side: (x2-xy)

2(x2-xy)=1(x2-xy)

And now divide by x2-xy to get:

2=1

!!!

What happened was in the division step, when I divded by x2-xy, I really divided by zero (because x=y). It is not possible to have an algebra in which division by zero is allowed. But such ill-defined notions do not crop up when allowing imaginary numbers.

7. Dec 14, 2004

### daster

2(x2-xy)=1(x2-xy)
2q(x2-xy)=1

8. Dec 14, 2004

### Tom Mattson

Staff Emeritus
That doesn't fix anything, because if q=1/0 (the multiplicative inverse of 0), then it is still the case that q0=1, and we still have 2=1.

Not at all! It's the good students who ask probing questions.

9. Dec 14, 2004

### Hurkyl

Staff Emeritus
Because, unlike the other examples, a number system cannot include q without throwing out some rather basic properties that are generally taken as essential.

By definition, 1/0 = q means 0q = 1. So...

1 = 0q = (0+0)q = 0q + 0q = 1 + 1 = 2

So, if we wanted to accept this new number, q, we'd have to throw out one of these familiar properties of numbers:

0+x = x
a(b+c) = ab+ac
2 is not 1

So, one cannot have 1/0=q in any meaningful arithmetic sense.

10. Dec 15, 2004

### Galileo

As the others said, it leads to inconsistencies in our developed number system.
The same thing could happen to i though.
i is defined by the equation $i^2=-1$.
Well, sure we can define whatever we want, why the hell not? But is it meaningfull and at all possible?
When we want to combine these new numbers with the real numbers we arrive at expressions of the form a+bi. With a and b real numbers.

The nice things are: Standard multiplication yields another number of the same form (another complex number).
We can add, subtract, multiply and divide (except by 0).
By taking b=0 in a+bi, we are simply dealing with a real number.
Multiplication and division have some really nice geometrical properties which make them veru useful.

It can also be shown that an extension to the complex numbers by means of finding solutions to algebraic equations is no longer possible. Every algebraic equation has a solution in the complex numbers. (We call it algebraically closed).

11. Dec 15, 2004

### Olaf.of.Ísland

Actually, that helps a lot. I have always had this "thing" about imaginary numbers. What a terrible naming convention.

12. Dec 15, 2004

### HallsofIvy

By the way, imaginary and complex numbers really became important with Cardano's formula for solving cubic equation. For some cubic equations having only real solutions, Cardano's formula would require working with complex numbers (the imaginary parts cancelled out at the end).

13. Dec 15, 2004

### cire

I still remember my math teacher using the word imaginary to refer to complex numbers. That word had an impact on me. Are they imaginary or real? Do they exit or not? Leibniz describe as “that amphibian between existence and nonexistence”. I was taught that complex numbers are square root of negative numbers that appear in solving the quadratic equation when . Ever since I looked at complex number as square root of negative number

When you star seeing prosperities like square root of i you start thinking that complex numbers are no solely square root of negative numbers, or imaginary.
It was not the quadratic formula what forced to take complex numbers seriously it was the cubic. It makes sense that the line couldn’t intersect the parabola thus complex numbers are imaginary, they are just imagination because geometrically they don’t exit. But it is clear that the line always intersects the cube; . Therefore the equation x^3=3px+2q has always at least one real solution.

x= (q+(q^2-p^2)^1/2)^1/3 + (q-(q^2-p^2)^1/2)^1/3 This is the analogous of the quadratic formula for the cube.

Bombelli’s example x^3=15x+4 yield to x=(2+11i)^1/3 + (2-11i)^1/3 but inspecting in the equation x = 4 is the solution . He had a thought that he called it “wild thought”:

x=(2+11i)^1/3 + (2-11i)^1/3=2+ni +2-ni=4 :tongue2:

This problem perfectly real, required complex numbers arithmetic to their solution! Historically, it wasn’t the quadratic equation what led to take complex numbers seriously.
On the other hand, can you imagine the Schrödinger equation without i, the probabilities will have and exponential decay in time!

14. Dec 15, 2004

### Hurkyl

Staff Emeritus
I just want to emphasize for those who don't know...

The cubic formula doesn't merely employ the use of complex numbers -- there exist real numbers that cannot be written, except in terms of complex numbers!

(Of course, I mean in terms of integers,+,-,*,/, and roots)

15. Dec 15, 2004

### cire

$$\x= \sqrt[3]{q+\sqrt{q^2-p^3}} +\sqrt[3]{q-\sqrt{q^2-p^3}}$$
sorry this is the correct fromula fro the cube

16. Dec 16, 2004

### Moonbear

Staff Emeritus
That's probably the best explanation I've ever heard of imaginary numbers! It's unfortunate someone gave them the name "imaginary" because it does conjure up thoughts of fiction. I sure wish my algebra teachers had explained it that way years ago.

17. Dec 16, 2004

### Chronos

Two cents. I picture imaginary numbers as relational. The concept of zero is where imaginary numbers enter the equation. I could, for example, arbitrarily set the speed of light as speed zero in my mathematical construct: and that would require imaginary numbers to deduce maxwell's equations or GR. IMHO, negative quantities of anything do not exist in this universe.

18. Dec 16, 2004

### matt grime

"IMHO, negative quantities of anything do not exist in this universe."

you must have a very understanding bank manager

19. Dec 16, 2004

### robert Ihnot

I was going to question the question, "Why was i invented?"

After all are we sure it was not just waiting to be discovered? However it seems that Tom Mattson answered that very well: It was invented to be a solution to the equation x2+1=0.

But then Tom goes on to say: That is most unfortunate, because the reals are no more "real" (ontologically) than the imaginaries.

You know that is confusing, was i just waiting to be discovered?

Kronecker is famous for saying, God created the integers, all else is the work of man. Well, I guess according to Kronecker, i was invented.

Last edited: Dec 17, 2004
20. Dec 17, 2004

### Alkatran

I don't know if you can really 'invent' math, except when you set up the original axioms. Someone had to invent 1+1=2, but they would only have discovered 2+1 = 3, if you get my drift.

It's the difference between Einstein seeing that the universe is 4 dimensional, and some observers' spaces are tilted relative to others and me typing my dinner plate.