# The imaginary Unit

1. Dec 11, 2013

### aaaa202

When i is defined by an equation which has 2 solutions, how does it make sense to do algebra with complex numbers?

2. Dec 11, 2013

### 1MileCrash

How is i defined by an equation which has 2 solutions?

Oh, I understand what you mean. I don't think it matters, because of the definition, everything true for i will be true for -i too. It becomes purely a notational issue, we have two solutions, and they are additive inverses, just pick a symbol for one and put a "-" in front of the other to show that it is the additive inverse. It doesn't matter.

Last edited: Dec 11, 2013
3. Dec 12, 2013

### aaaa202

I feel kind of stupid, but I don't understand this argument that if everything is true for i it will be for -i too. Could you give some examples? I feel it's kind of the same as choosing a right or lefthanded coordinate system but then again I have not understood that either.

4. Dec 12, 2013

### HallsofIvy

It doesn't! We don't define "i" by "$i^2= -1$"- although elementary treatments may introduce it that way.

Rather, we define the complex numbers as pairs of real numbers, (a, b), with addition defined by (a, b)+ (c, d)= (a+ c, b+ d) and multiplication defined by (a, b)(c, d)= (ac- bd, ad+ bc). That way (a, 0)+ (c, 0)= (a+ c, 0) and (a, 0)(c, 0)= (ac, 0) sp we can associate the "complex number" (a, 0) with the real number, a. Also, if we define i= (0, 1) then we have $i^2= (0, 1)(0, 1)= (0(0)- 1(1), 0(1)+ 1(0))= (-1, 0)$ so that "$i^2= -1$". We can then say (a, b)= (a, 0)+ (0, b)= a(1, 0)+ b(0, 1)= a(1)+ b(i)= a+ bi.

The point is that the way a mathematical concept is introduced to the student is not necessarily the way it is formally defined by mathematicians.

5. Dec 12, 2013

### aaaa202

Okay so maybe you can help me understand where it goes wrong in the following (which motivated this thread for a start):

1. 1/i = sqrt(1)/sqrt(-1)
2. 1/i * i/i = sqrt(1/-1)
3. i / -1 = sqrt(-1)
4. -i = i

What is the problem by simply defining i=sqrt(-1) and where does it go wrong in the above when doing so?

6. Dec 12, 2013

### Mentallic

The problem is going from line 1 to 2,

$$\frac{\sqrt{a}}{\sqrt{b}}=\sqrt{\frac{a}{b}}$$

Is a rule that applies only to positive a and b values.

7. Dec 12, 2013

### 1MileCrash

OK, I think it will become obvious to you.

3i + i = 4i
3(-i) + (-i) = 4(-i)

For example, or even
e^(-i(pi)) = -1

Where i/-i is only referenced once.

If we make the 'basis' for the 1 dimensional vector space -i instead of i, the change is only superficial.

Essentially, I believe that -i and i may be regarded as equals (not that they are equal to one another, but that they are on the same footing.)

It doesn't matter which one is which, as long as both exist. I don't believe you can say ANYTHING that differentiates I from -i... but I may be wrong.

8. Dec 12, 2013

### aaaa202

I see. But what exactly is the difference then between defining i=(0,1) and i=√(-1). Is the only problem that the last one could make you mistakingly think that you can apply usual rules for square roots? And noting that this is not the case, are the 2 definitions the equivalent?

9. Dec 12, 2013

### 1MileCrash

All you've done is arbitrary chosen (0,1) to be i instead of (0, -1). If (0,1) = i, then (0,-1) = -i, and then we're back where we started: -what's the difference?-

Last edited: Dec 12, 2013
10. Dec 12, 2013

### lostcauses10x

11. Dec 12, 2013

### rbj

i'm in agreement, Hall, with 1Mile. originally the concept of "imaginary" numbers and then "complex" numbers did come from, i believe, the solution to quadratics (or higher-order polynomials) set to 0 when there were no "real" numbers (the set of rational and irrational numbers that are ordered on the "real" number line) that satisfy those equations.

saying that

$$(a, b) + (c, d) = (a+c, b+d)$$

makes sense just from a normed, linear space POV. but when you say that

$$(a, b) \times (c, d) = (ac-bd, ad+bc)$$

as a definition, we can just as well say that is motivated by or a result of the definition:

$$i^2 = -1$$ .

i might recommend the OP to take a look at the wikipedia article on it, especially the 2nd section. i think that explains it pretty well without just requiring the OP to accept $(a, b) \times (c, d) = (ac-bd, ad+bc)$ as a definition for how complex numbers multiply.

12. Dec 12, 2013

### pwsnafu

This is not true. It comes from the the Tartagalia-Cardano formula, i.e. the cubic formula. As you know cubics always have a real solution, however from time to time the formula generates solutions with square roots of negative numbers. For example $x^3 - 15x = 4$ produces
$x = \sqrt[3]{2 + \sqrt{-121}} + \sqrt[3]{2 - \sqrt{-121}}$
but this equals 4.

Another example is in 1702, when Leibniz demonstrated to Hyugens that
$\sqrt6 = \sqrt{1+\sqrt{-3}}+\sqrt{1-\sqrt{-3}}$
who exclaimed "defies all human understanding".

All of this predates the notion of complex numbers.

13. Dec 12, 2013

### rbj

there's a reason i parenthetically inserted "or higher-order polynomials".

but, from a pedagogical POV, i don't understand why it's necessary to go above quadratics to introduce someone conceptually to imaginary and complex numbers. you don't need to go to cubic or quartic to see the need for imaginary and complex numbers.

14. Dec 12, 2013

### pwsnafu

You wrote "set to 0 when there were no "real" numbers (the set of rational and irrational numbers that are ordered on the "real" number line) that satisfy those equations." That part is wrong. It's the real solutions that caused the discovery. Back then $x^2 + 1 = 0$ simply had no solutions.

But you weren't making a point of pedagogy. You were making a claim about the history of mathematics.

15. Dec 12, 2013

### rbj

no. you cannot represent me nor what i was saying. you cannot even presume to.

it was purely a point of pedagogy from the beginning. as best as i can tell, you were making an historical point. but i was covering my arse anyway.

16. Dec 12, 2013

### pwsnafu

What? You wrote
Explain please how that is not about history. You explicitly used the phrase "originally...did come from". How am I supposed to parse that as a pedagogical point?

Look, if you are claiming that teaching complex numbers is easier through quadratics, then I agree with you. But that was not what you wrote in the first paragraph.

Last edited: Dec 12, 2013
17. Dec 12, 2013

### lostcauses10x

Folks is this arguing helping?

18. Dec 12, 2013

### pwsnafu

Yeah. I'm done. Moving on.

As Halls pointed out complex numbers are formally defined as pairs, with $i = (0,1)$. This space of numbers is of course $\mathbb{C}$.

Let's create a new number system. I'll define $j = (0,-1)$ and call this space $\mathbb{D}$ which is composed of numbers $\alpha + \beta j$.

What 1Mile is saying that if I make a statement (theorem) that is true on $\mathbb{C}$, there is an equivalent statement that is true on $\mathbb{D}$. And we can use conjugation to write it down.

19. Dec 12, 2013

### rbj

they are qualitatively equivalent (both have equal claim to squaring to -1), but not quantitatively, since they are not zero and are negatives of each other.

$\mathbb{C}$ is not $\mathbb{R}^2$. Halls was complete when he stated the rules of addition and multiplication of these pairs. my point is that the "formally defined as pairs" is not the original meaning (in case you're wondering, i mean the original pedagogical meaning), but can be made equivalent as long as you define the rules of addition (trivial, about the only way you can) and multiplication (less trivial, there are other ways to define multiplication of 2-vectors which are not compatible with complex numbers). i really don't see any point to it or advantage over the common pedagogy (geez, i have to be careful to remind snafu, lest the words be misconstrued).

20. Dec 13, 2013

### R136a1

Getting all argumentative is getting us nowhere, rbj. I think all of pwsnafu's replies have been very accurate.

21. Dec 13, 2013

### rbj

you have every right to think that.

(i'm here for information. $\mathbb{C}$ is not the same as $\mathbb{R}^2$. call it argumentative or not, but if anyone "simplifies" the concept of complex numbers to reducing the concept to only "pairs" of real numbers, they need to read that quote from Einstein: "Things should be as simple as possible, but no simpler.")

22. Dec 13, 2013

### lostcauses10x

The set of complex numbers is not reals squared. True.
And no one has said this.
The statement is ordered pairs of real parts.

Yet by leaving out i it can confuse some, and is easier for others. The questions is "is it equivalent with the definitions for operation defined? I have no doubt of this, yet...

Yet the statement
does define the rules of the complex system, not the number i.

i of course is undefined in the reals, yet is a constant such as the real part of "a+bi" or bi is such that a defines how many "i"s are added together.

Of course this can confuse folks very easy, and part of the whys "i" were and are avoided by a lot.

23. Dec 13, 2013

### rbj

but the sets are one-to-one with each other. (what did they call that? "entire"? can't remember.)

well, dunno if i agree with that. i quoted someone making, without the reference to "$\mathbb{R}^2$", an equivalent statement.

complex numbers are more than just ordered pairs of real parts. $\mathbb{R}^2$ are also ordered pairs of real parts. while there is a simple and complete one-to-one mapping of $\mathbb{R}^2$ to and from $\mathbb{C}$, it is tempting for some to think that they are the same, but $\mathbb{C}$ requires an operation or rule that $\mathbb{R}^2$ does not. and it makes it different.

and that is solely what my disagreement with Halls is about. it's about pedagogy, and i see little advantage, conceptually, to the pedagogy that leaves out $i^2=-1$.

24. Dec 13, 2013

### lostcauses10x

First to have the ordered pairs, takes two sets of reals; not just one: with the rules as stated in this thread. The intersection of coarse is at (0,0) with one set vertical to the other.
End result is that "i" is equivalent to (0,1)
Which is why I referred to the link I posted.

The problem with boards and such talks is that to state all involved is not practical. Just to get into a talk about a term or terms can lead a topic way off.

And I learned the old way myself. I have always had a bit of trouble when they say (a+bi) that b is a real.
automorphism? There is one non-trivial automorphism between $\mathbb{C}$ and itself. The mapping is complex conjugation (invert the imaginary part).