Question

Can you show this question to be true or false? The question was ruled false by Wolfram. Question is in first post.

Answer 1

Wolfram says false.

Can you show that the statement is true?

Answer 2

could you type input code?

Answer 3

2*(-2)^(1/5) = -2 * (2)^(1/5)

Answer 4

WA is a blind machine and it's dumb. This is certainly true. Perhaps it's gotten stuck looking at complex valued results.

Answer 5

i think that because 2*[fifth root of (-2) ] have more than one solution [other complex solutions]

Answer 6

and mainly root of negative value are not in particular scalar values for weird reason [ huh funny that woli knows that!]

Answer 7

I know to exercise caution when dealing with even indices of radicands and still as a bit "jumpy" with odd roots of negative numbers.

For those interested, the origin of the question I posted comes from a problem posted last
night.

You can view that thread here:

http://openstudy.com/users/directrix#/updates/56ecb5c0e4b0e02a5e0b4e70

Answer 8

i get it now, wolfram [unlike school teachers] understand that root of something negative is not scalar value =) [it actually determine a magnitude and orientation mostly] .
but that doesn't mean its completely wrong, but letting the negative out is the trivial solution.

@Kainui your opinion ?

Answer 9

Yeah I agree I just think any time I see \(\sqrt[n]{a}\) I immediately assume we're working with real numbers just because when I think of complex numbers we're probably going to be seeing \(a^{1/n}\) instead.

It is just a more natural and less awkward notation, it would be like seeing this and talking about PEMDAS with integrals -- I've never ever ever ever seen this happen except when I'm joking around haha

\[\int_0^1 (1 \div e^x ) dx\]

Answer 10

lol

Answer 11

I think the main problem is this is what's uncertain:

\[-1 = \sqrt[5]{-1}\]

You can check that (-1) is a fifth root of -1 by seeing that \((-1)^5=-1\) although there are other solutions.

Answer 12

if it's not always true, maybe it is false

https://www.wolframalpha.com/input/?i=(-1)%5E(1%2F5)+%3D+-1

Answer 13

It's definitely true on the domain and range of real numbers. For the same reason, we wouldn't say this graph is undefined at a point:

\[f(x) = \frac{1}{x^2+1}\]

But really there are singularities at x=i and x=-i that you have to avoid in the complex plane if you don't want to divide by zero that aren't in the domain. So as long as we restrict ourselves to real numbers (which we do) there's no issue :D

Answer 14

exactly!... beside consider the orientation
when (-1)^1/5=-1 it means a direction

Answer 15

I suppose it might as well be said in the complex plane for \(n \in \mathbb{Z}\),

\[-1 = e^{i \pi(2n+1)}\]
So:
\[\sqrt[5]{-1} = e^{i \pi (2 n+1)/5}\]

And this sucker right here is periodic in n, so you can look at just when n=0,1,2,3,4 to see all the solutions, of which n=2 gives us -1.

Answer 16

who said real only?

Answer 17

It's in the context of the question. The kid is learning radicals and he probably doesn't even know such a distinction even exists so he wouldn't state it.

Answer 18

lol!

right

Answer 19

It's false for complex numbers?

Answer 20

When is it false?

Answer 21

\[
2\sqrt[5]{-2} \overbrace{=}^{1}2\sqrt[5]{-1}\sqrt[5]{2} \overbrace{=}^{2} -2\sqrt[5]{2}
\]Which step is disputed?

Answer 22

High School students do learn about DeMoivre's Theorem and n th roots.

While a fifth root of -1 = -1, there are four more.

Should an Algebra Two student be expected to know that?

The student should, however, know that a positive number has two square roots.

Answer 23

The operation \(\sqrt[5]{\ldots}\) is a function that returns a specific root.

Answer 24

@wio i agree that its a function that return a specific value iff domain is donated by R or subset of it.

Answer 25

\[
\sqrt[5]{-1} = e^{i\pi/5}
\]

Answer 26

This is why Wolfram correctly gives false, I think

Answer 27

and a whole bunch of repeating

Answer 28

However, this is also why I do not like complex numbers. I don't think they should be considered numbers to begin with

Answer 29

lol @IrishBoy123

Answer 30

lol ikkels!

massive brains all talking bollocks

Answer 31

tots!

Answer 32

The "wolf" returns the real 5th root of -64 (not in simplest form)
and renders a list of the five Complex roots, also.

http://www.wolframalpha.com/input/?i=(-64)%5E(1%2F5)%3D

Answer 33

Warning: The following remark may come off as that uttered by a rube.

If I can show that given one of the two radical expression equivalent to the other using rules of Algebra II, then I am calling the statement "true."

Answer 34

@wio I still shiver from the day I learned that Complex Numbers are not ordered.

Answer 35

Complex numbers are really a type of vectors. I think number should be restricted to ordered sets.

For complex numbers \(\sqrt[n]{\ldots}\) should give the minimum \(\theta\) solution.

Answer 36

After the above interesting ruminations, there still is no answer to the question:

Can you show the statement to be true or false?

Answer 37

cuz -1 cant gone out the root \[-1\sqrt{1}\neq \sqrt{-1}\] @Directrix i think

Answer 38

@dinamix This is a radical with an odd index and a negative radicand.

Answer 39

lol why

Answer 40

To all posters and future posters in this thread:

Is there anyone who will say True to the posted problem which Wolfram rules False?

Is there anyone who agrees with Wolfram and will post that in this thread?

Answer 41

but @Directrix my statement is true ? for u or no

Answer 42

The left side \(2 \sqrt[5]{-2}\) is a complex number of the form a+bi.
The right side \(-2 \sqrt[5]{2}\) is a real number of the form c+0i.

Hence, `False is correct`. Agree with wolf.

Answer 43

@hartnn How would you explain that to an Algebra II student who can correctly show that the two original expressions are equivalent?

Answer 44

I would tell them, that your teacher expects you to assume -1 for \(\sqrt[n]{-1}\) for now, but actually that is not true, and you'll learn more about it when you learn complex numbers.

Answer 45

@dinamix

I agree but do not know the relevance of your statement to the posted problem.

Answer 46

@Directrix i think it's postulate mathlete suppose it like i^2 = -1 i think my mind broken now @Directrix @hartnn

Answer 47

-1 is given by Wolfram as one of the five fifth roots of -1 but not the principal fifth root.

@hartnn If we state that we will remain within the set of Reals, would you agree that an odd root of -1 is equal to - 1 ?

Answer 48

For what ever it is worth:
Mathematica version 9.0.1.0 Home Edition
Refer to the attachment, a screen capture.

Answer 49

Finding an `odd root of -1`, is same as solving \(x^{odd} +1=0\)
which has multiple solutions.

Answer 50

By the way, double == is treated the same as 2 = 4/2 in normal math notation.
A single = sign is a permanent assignment such as f = 40x .
Any reference to "f" in the Mathematica code will be associated with 40x.
1+f
is interpreted as 1+40x in any calculations.

Answer 51

I think some calculators use logs to compute things like (-8)^(1/3)

If you let x = (-8)^(1/3) then we'd have -8 = x^3

Let's use logs to isolate x


-8 = x^3

Ln(-8) = Ln(x^3)

Ln(-8) = 3*Ln(x)

3*Ln(x) = Ln(-8)

Ln(x) = (1/3)*Ln(-8)

x = e^[(1/3)*Ln(-8)]


since Ln(-8) is undefined, this means x is undefined. But that's not the case since the cube root of -8 is -2.


I'm not 100% sure about this theory but I read a similar idea off this page

http://mathforum.org/library/drmath/view/54376.html

so it's possible. Though not all calculators are the same

Answer 52

Welp, this is a long post that I don't understand :3

Answer 53

It appears this web calculator

http://web2.0calc.com/

uses logs (my best guess anyway) to calculate fractional exponents. Since it spits out an error. See `attachment1.jpg`

---------------------------------------------

but a calculator like graphcalc

http://www.graphcalc.com/

(desktop downloadable calculator) doesn't appear to use logs since it seems to handle something like `(-8)^(1/3)` just fine. So I'm guessing it doesn't use logs. See `attachment2.jpg`

Answer 54

The command "(-2)^(1/5) in reals" gives the expected result when considering solutions in real numbers. I think it's safe for a student without knowledge of complex numbers to say it's true. Just like they would correctly say some equations have no solutions even if complex solutions exist.

Answer 55

i am going with the wolf

Answer 56

Wolfram is comparing the principal roots. Then the equality clearly doesn't hold

Answer 57

http://www.wolframalpha.com/input/?i=principal+root

Answer 58

Me: Not understanding squad, it is like a foreign language.

Answer 59

Way to make me feel stupid guys <3 :D.

Answer 60

There are 5 fifth roots of -2 for the wolfram to choose for the comparison. Your question itself is ambiguous for any computer program. Wolfram is resolving the ambiguity by picking the principal roots for comparison. (A principal root is a root with the smallest positive angle.)

Answer 61

Squat*


Fff- even my grammar sucks. -_-.

I am just going to say; wolf is right.

Why: Don't ask, i JUST think it because he just changed the integers around. Tomato tumato