# Now I see!

1. Aug 4, 2006

### Alkatran

Now this bugs me. First there were no negative numbers. Then there were no square roots to negative numbers. Then every real number had two square roots, but no word on imaginary numbers. Luckily for me, I had a calculator that told me the square root of i wasn't some other type of number or I would have thought of it that way (hey, I was in grade 10!). That meant all complex numbers had two square roots.

Not too long ago I thought "My that's a nice pattern, two square roots, 1 cube root, 2 quad...". Then something occure to me: why didn't every complex number have 3 complex roots, 4 4th roots, etc? Wouldn't that make more sense? The only way to be sure was to try it!

So I grabbed a piece of paper and found 3 complex numbers that, when cubed, equalled 1. "God damnit!" I thought. This was "there's no square root to negative numbers" all over again! I'd just taken it for granted that cube roots were unique because we were always told that in school! *shakes fist at high school*

Obviously, people have known this for a very long time. Why weren't we told this as soon as we were introduced to imaginary numbers? Is there any reason teachers hide this type of information and, in effect, lie by not telling the whole truth?

2. Aug 4, 2006

### Data

The reason that they "hide" information like that is simply that most people can't handle things if you give them everything all at once. Trying to explain to someone in third grade "well, negative numbers don't have REAL square roots, but we can define a new number $i$ with $i^2 = 1$ and then they'll all have two square roots," just isn't practical with most of them, because they usually don't even have much of a grasp on how basic operations work yet.

Further, the details of definitions and proofs are just irrelevant to most people (including many gradeschool teachers!). If you are interested in mathematics and its formalisms, then you have to go further than what is taught in general.

It is sort of annoying that even after telling you what complex numbers are, it's usually still not pointed out important things that they add to what you've already learned (like n nth roots for all nonzero complex numbers). And there are of course many other examples of silly omissions like that.

Last edited: Aug 4, 2006
3. Aug 4, 2006

### SGT

In reality, every nth degree equation has n solutions. Some or all solutions may be real and some complex. In particular, the nth degree equation
$$x^n = R$$, where R is any real number will have one real solution and n-1 complex solutions if n is odd and 2 real and n-2 complex if n is even.

4. Aug 4, 2006

### Data

careful, you need R to be positive to have real solutions if n is even!

5. Aug 5, 2006

### quasar987

I find this wording of the fondamental thm of algebra confusing! I seem to be the only one too, but without specifying that n counts the order of multiplicity, Alkatran could spend quite some time trying to find the 2nd solution to z²-2iz-1=0 other than i!

6. Aug 5, 2006

### mathwonk

i suspect most teachers tell you what they themselves know and understand, which is always limited.

i am abut to teach basic grad algebra and i was just looking at the course description for basic grad algebra at harvard. i myself would like to take the cousre from Noam Elkies there. But i will teach whatever i can.

7. Aug 5, 2006

### mathwonk

and quasar is right, unless you check that the discriminant is non zero, you do not know that all n roots of a nth degree equation are different. did anyone ever tell you what the discriminant of an equation is?

8. Aug 5, 2006

### Alkatran

I don't mind not learning about imaginary numbers in grade 3. I do mind being told "negative numbers don't have square roots." This is incorrect. You should say something along the lines of "They do, but it's very complicated for third graders, you'll learn about them later."

Also, I'm aware there are n (not necessarily unique) solutions to an nth degree polynomial. I prefer the fact that, given n points, you can always find an (n-1)th polynomial that crosses all of them (assuming you allow things like 0*x^2 + 1 to be 2nd degree).

On a related note, I remember our teacher trying to teach us about negative numbers. She used the analogy of a bunch of houses with address 0, 1, 2, 3 and 4, but then they build houses to the left of house 0. What were the house numbers? Of course, this went badly because we spent the whole time saying things like "just change the existing numbers" and "just use 5, 6, 7..." How hard is it to go in class and say "what is the answer to 0-1? What is before 0? I give you: negative numbers!"

9. Aug 5, 2006

### mathwonk

you are right, you were given poor instruction, now get over it, and accept once for all that you have moved beyond that level, and achieve as much as you can. good luck!

10. Aug 6, 2006

### robert Ihnot

I can't remember the roots of 1 as being surprising. In College Algebra we got into the Fundamental Theorem of Algebra proven by Gauss, and beyond the scope of the course, however as the Fundamental Theorem it was, assumedly, very important.

Thus is X^3=1, we have by factoring (X-1)(X^2+X+1)=0.

However that i had two square roots, that seemed surprising; but then it comes down to the Fundamental Theorem. It does not exclude complex coefficients, even if our problem set did.

11. Aug 6, 2006

### VietDao29

I don't really consider this important. There are just so many things that you can learn. It's just the matter of time before you'll be taught something new. We start from the basic, and then build our knowledge up to learn more advanced things. From where I stand, there's really no point to explain to some 3-rd grade student that negative numbers, or complex numbers do exist. As the matter of fact, they won't pay any attention to that.
And what's the benefit of telling them that they'll learn them later? They still don't know anything.
And further more, teachers just teach what they are supposed to teach, so it's common that some of them are not aware of complex numbers. You can't, of course, demand your teacher to fetch you the knowledge that you'll learn in university, when you are only in high school.
Last but not least, centuries before complex numbers are found, people thought that there is no solution to the equation x2 = -1, and they accepted that fact. And that fact does not do any harm to them, nor you. And you cannot know what people can discover in the future, right? When they discover something new, there may be some conflicts with what used to be thought to be correct.
So, it's normal. Don't make a fuss on it. :)

Last edited: Aug 6, 2006
12. Aug 9, 2006

### Robokapp

3 complex numbers that when cubed equal 1.

I only took AB but this sounds like an Algebra point...Can you post (or someone) those numbers? There is no such thing as an equivalent for i for cubes, is it?

x^3=1
x^3-1=0
(x-1)(x^2+x+1)=0

the first part... (x-1) turns into a...1
so how do you make it complex? unless you're writing 1 as...i^4

lol

Last edited: Aug 9, 2006
13. Aug 9, 2006

### HallsofIvy

You don't have to "make" 1 complex, it already is. A complex number is any number of the form a+ bi where a and b can be any real numbers. In particular, a= 1, b= 0 is a complex number. The set of all real numbers is a subset of the set of all complex numbers.
To find the other solutions to x3= 1, after getting (x-1)(x2+ x+ 1)= 0, use the quadratic formula to solve x2+ x+ 1= 0.
$$x= \frac{-1\pm\sqrt{1^2- 4(1)(1)}}{2}= \frac{-1\pm\sqrt{-3}}{2}$$
The other two roots are
$$x= \frac{-1}{2}+\frac{i\sqrt{3}}{2}$$
and
$$x= \frac{-1}{2}-\frac{i\sqrt{3}}{2}$$

14. Aug 9, 2006

### mathwonk

the non real cube root of 1 with smallest angle is often written as a lower case omega. but not as consistenly as i is used so it is always defined on the spot when so used.

15. Aug 9, 2006

### shmoe

little omega is not universal like you say, I've seen $$\rho$$ used many times (always defined at the time of course)

16. Aug 10, 2006

### myspip

The reason why imaginary numbers are not well-introduced in high school (I have the right experience; I'm in high school but have studies maths at college level in 3 years, however in Sweden tho') is indeed that most students are incapable of handling maths of that level. Even though the basics (adding, subracting and multiplying) of imaginary numbers is simple, and I'm sure that for 20 years ago most students could have handled it. But today the mathematical knowledge of high school students is much lower than that of the students for some decades ago (generally said)...

17. Aug 10, 2006

### bomba923

But can't that be changed via altering schools' mathematics curricula?
(e.g., perhaps via inclusion of parts from more traditional mathematics curricula?)

Last edited: Aug 10, 2006