What's wrong with my calculation?

[Anzas]

whats wrong here?

 

[jcsd]

$$(-4)^{\frac{2}{4}} \ne \sqrt[4]{(-4)^2}$$

Because both the rhs and the lhs are specifically the principle values.

 

[YourLooks]

I think there is no definition of $$\sqrt{-4}$$ if complex analysis is not concerned right in the first place, so you have to stop right at that first place too, or your continuing process is then all meaningless, which is what the ugly guy in that picture is doing...:shy:

 

[BobG]

The fourth root of 16 should be plus or minus two.

If it weren't specifically defined that a had to be greater than 0 and not equal to one for $$f(x)=a^x$$ to be a valid function, you could have all kinds of possibilities:

$$-8^{\frac{1}{3}}$$ would be "75% chance of -2 and 25% chance of +2". A '1/3' exponent would give you -2, while a '2/6' exponent would give you plus or minus 2, etc.

 

[JasonRox]

The fourth root of 16 should be plus or minus two.

If it weren't specifically defined that a had to be greater than 0 and not equal to one for $$f(x)=a^x$$ to be a valid function, you could have all kinds of possibilities:

$$-8^{\frac{1}{3}}$$ would be "75% chance of -2 and 25% chance of +2". A '1/3' exponent would give you -2, while a '2/6' exponent would give you plus or minus 2, etc.

It depends on the convention, I guess.

 

[jcsd]

The fourth root of 16 should be plus or minus two.

If it weren't specifically defined that a had to be greater than 0 and not equal to one for $$f(x)=a^x$$ to be a valid function, you could have all kinds of possibilities:

$$-8^{\frac{1}{3}}$$ would be "75% chance of -2 and 25% chance of +2". A '1/3' exponent would give you -2, while a '2/6' exponent would give you plus or minus 2, etc.

We had this one before f(x) = x^(n/p) is specifically the principle branch of the function as is the radical symbol.

 

[Alkatran]

Let's do it the other way then:

16^(1/4) = (16^(1/2))^(1/2) = (+-4)^(1/2) = 2 or -2 or 2i or -2i

 

[newPatrick]

<1> Let x=-4 < 0, then let y1=sqrt(x) > 0 and y2=sqrt(x) < 0 (sqrt definition)and suppose these are correct. This means y^2=x. Paradox! y^2 < 0
<2> Rewrite the proof induction in the picture in this vars (x,y) form y=sqrt(x) = x^(2x(1/4))= (x^2)^(1/4)= ((-x)^2)^(1/4) = -x/2 (minus because x<0), this means y>0
If we suppose sqrt(negative) is allowed. From 2, y2 < 0 value is lost as Alkatran and many posts above indicated. But the truth is that sqrt(negative) isnot allowed, such assumption is impossible.
From1,2 say that picture gives a seemingly logical but incorrect induction

 

[matt grime]

newPatrick, why are the complex numbers ordered? (in <1> you use < on complex quantities.)

 

[Alkatran]

Here's the basic gist of what this guy did:

x = x
x^2 = x^2
sqr(x^2) = x or -x
x = -x

It's not a correct logic, or is NOT =

 

[HallsofIvy]

16, like any number, has 4 fourth roots: 2, -2, 2i and -2i.

In the real numbers we can specifically identify THE square or fourth root as the POSITIVE root. The complex numbers do NOT form an ordered field so we cannot distinguish complex roots in that way.