dumb question

[psi20]

I don't get this. Do you take the root first? Or take the power first?

 

(-4)^(2/10). Do you take the tenthroot first? If so, you would run into imaginary numbers right? But if you square it first, then you don't have imaginary numbers?

Do you always simplify the fraction? So do you turn it into

(-4)^(1/5)?

 

Each yields a different answer. One's imaginary, one's positive, one's negative.

[JaKiri]

They all yield the same answer.

 

An imaginary number squared will get you to an odd number, so will taking an even root of an even number (Saying that it ends up even is like saying SQRT(-2)^2 ends up even.)

[psi20]

Thanks for clearing up about the imaginary number part, but I still don't get the part about the negative and positive.

 

(-4)^(2/10) = tenthroot(-4)^(2)=tenthroot16

 

(-4)^(2/10) = (-4)^(1/5) = fifthroot(-4)

[NSX]

use i² = -1

[psi20]

Yeah I get the imaginary number part and I see how I did jumped to conclusions too quickly. But I don't get the part about the positive and negative thing I showed up there.

[psi20]

can someone clear this up for me? is the answer positive or negative? is it 1.3195 or is it -1.3195?

[JaKiri]

As an example, lets look at the equation:

 

(x-1)² = x².

 

It's pretty obvious if you multiply it out that x = 0.5.

 

However, if you take the square root of both sides...

 

x - 1 = x

 

1 = 0.

 

This doesn't work, does it?

 

It's a similar problem to the one above, in that you have to get the right sign on the square root, because it can be either positive or negative.

 

If we change it to

 

x - 1 = - x (-x being the square root of x²)

 

2 x = 1

 

x = 0.5, as required.

[psi20]

I don't think your analogy works well, in that mine is just an evaluation and yours is an equation. In equations, you would to +/- for even roots, but in evaluations you don't. Like the square root of 4 is just 2. There is no -2. But if x^2 = 4, x = +/- 2.

 

Let me try to clear up my question.

 

In p^(m/n),

would you do (nth-root of p) to the m-th

or would you do nth-root of (p to the mth)?

 

In p^(mn/mo),

would you do (m*nth-root of p) to the m*o-th

or would you do m*nth-root of (p to the m*o-th)

or would you do (nth-root of p) to the o-th

or would you do nth-root of (p to the o-th)

 

It can yield different answers. And it shouldn't.

[JaKiri]

psi20 said in post # :

I don't think your analogy works well, in that mine is just an evaluation and yours is an equation. In equations, you would to +/- for even roots, but in evaluations you don't. Like the square root of 4 is just 2. There is no -2. But if x^2 = 4, x = +/- 2.

 

I don't see why there is an elementary difference between equations and 'evaluations'. The square root of 4 is ALWAYS +-2, it's just not expressed verbally for simplicity (you'd probably not get away with saying it's 2 at a higher level).

 

Furthermore, my analogy does hold, because it's precisely the same thing at work.

[psi20]

Oh, that's not what my math teacher told me. But it seems that sometimes math teachers teach things wrong. So the square root of 4 IS +/- 2? That changes everything.

[JaKiri]

psi20 said in post # :

Oh, that's not what my math teacher told me. But it seems that sometimes math teachers teach things wrong. So the square root of 4 IS +/- 2? That changes everything.

 

Maths is all about being correct, and to say the square root of 4 is 2 (as opposed to A square root) just isn't right.

[Dave]

Think about the meaning of 'square root'. If you try and find the square root or some number n, it's another number p such that p^2 = n.

 

For n = 4, there exist two such numbers: -2 and +2. It's all a matter of definition.