# Solve a radical equation where the root is a negative integer?

1. Jun 11, 2004

### Imparcticle

Is it possible to solve a radical equation where the root is a negative integer?

2. Jun 11, 2004

### AKG

I believe you're referring to the "radicand" (the number under the square root sign). Can you be more specific as to what is trying to be solved, what the equation is, where is the radical, etc? I suppose it would be helpful for you to know that there are complex and imaginary numbers, so if you had:

$$x = \sqrt{-1}$$

then:

$$x = i$$

Where $i$ is the imaginary number defined as the square root of -1.

3. Jun 11, 2004

### Imparcticle

I was referring to the "root". For example, 1/24=2 when the denominator of the exponent is two. "Two" is the root, "4" is the radicand. Can "two" be negative?

Can you explain to me the logic behind imaginary numbers which apparently result when the radicand is negative and the root is an even integer?

4. Jun 11, 2004

### arildno

A complex number consists of two real numbers (a,b) along with the summation operation:
(a,b)+(c,d)=(a+c,b+d)
and the multiplication:

Note that any complex number on the form (a,0) fulfills every property that a real number has!

An imaginary number is on the form (0,b).
If we make the multiplication (0,1)*(0,1) we get:
(0,1)*(0,1)=(0*0-1*1,0*1+0*1)=(-1,0)
This is the meaning of (0,1) as the "root" -1.

5. Jun 11, 2004

### AKG

No. By convention I guess, $\sqrt{4}$ and $4^{1/2}$ are $+2$. If you had a question like $x^2 = 4$, then the answer would be $x=\pm 2$. Similarly, whereas $\sqrt{4} = 2$, $\pm \sqrt{4} = \pm 2$.

An imaginary number is simply any real number times $i$. So, if $\sqrt{5.5225} = 2.35$, then $\sqrt{-5.5225} = \sqrt{5.5225} \times \sqrt{-1} = 2.35i$. You may want to check out the Mathworld.com article on Complex Numbers.

6. Jun 11, 2004

### HallsofIvy

If an exponent is negative, for example 9-1/2, then it is equal to the reciprocal: 9-1/2= 1/(91/2)= 1/3.

7. Jun 12, 2004

### geometer

I dislike the term "imaginary numbers." These numbers exist as much as any other number exists. They are needed to solve equations like x2 + 1 = 0, and they have many everyday applications. Any wave function involves imaginary numbers, so TV, radio, and microwave ovens exist because of the existence of imaginary numbers.

There is a very elegant branch of mathematics known as complex variable theory which deals in detail with the properties of complex numbers (a number with a real and imaginary part). This theory can be used to greatly simplify the solution of some problems that would be just about impossible to solve otherwise. As a simple example, try to prove the common trig identity Cos(A + B) = CosACosB - SinASinB using standard methods. Using Euler's Formula from Complex Variable Theory, it becomes almost trivial.

8. Jun 12, 2004

### arildno

While sharing your dislike of that term (for the same reason), I dislike the term "irrational numbers" even more..

9. Jun 12, 2004

### geometer

haha - There is at least some justification for the term irrational since these numbers can't be represented as the ratio of two integers.

10. Jun 12, 2004

### arildno

Not really, the original word "ratio" means reason, and the term "irrational numbers" was explicitly formed to mean numbers that were "unreasonable"..

11. Jun 12, 2004

### Nicomachus

What are you going on about? Sources? The reason I ask is because this contradicts reality. The word "ratio" is from the Latin to calculate or reckon. Just because something sounds "cool" doesn't mean its right. Obviously, the common usage of rational follows from this definition.
*Nico

12. Jun 12, 2004

### arildno

Well, that's the explanation I've "always" heard.
It sort of fitted the documented opposition in history from the evolution of new numbers (as in the Pythagoreans' ambivalence to the discoveries of irrational numbers, European mathematicians opposition to negative numbers in the 16'th century, and, about the same time, or a bit later, the coining of term "imaginary" in opposition to "real").
It might well be a myth in the case of irrational numbers, so, assuming that, thx.

13. Jun 12, 2004

### jcsd

How does a real number know that it's not an imaginary number imagining that it's a real number? Of course that would be irrational.

14. Jun 12, 2004

### arildno

Now that's REALLY COMPLEX, jcsd

15. Jun 12, 2004

### jcsd

Few numbers are PERFECT though

16. Jun 12, 2004

### arildno

But most of them are FRIENDLY

17. Jun 12, 2004

### jcsd

Yes I've ceratinly found some numbers are very AMENABLE to my needs *groan*

18. Jun 12, 2004

### arildno

I think this thread has degenerated into a PRIME example of stupid math jokes..