What's correct?

[Steven G]

Have you had your coffee this morning?
$\dfrac{1}{2}=\dfrac{2}{4}$ or $a\ge 0\text{ then }\sqrt{a}=\sqrt[4]{a^2}$

Yes, you are correct if a≥0. But who said a≥0?? What is a<0? Just because 1/2 and 2/4 are both positive does not imply that ≥0

 

[lex]

I think I am responsible for starting this debate! (Apologies to the OP).
I hesitantly make this contribution:

 

[Steven G]

Lex, I am still confused.
Consider (-4)^(2/4) = 16^(1/4) = 2

Now (-4)^(1/2) = 2i

We get different answers.

Am I making an error somewhere?

 

[lex]

This is the point. When you write $(-4)^{\tfrac{2}{4}}$ you are writing $(-4)^{\tfrac{1}{2}}$.
When you say that $(-4)^{\tfrac{2}{4}}=(16)^{\tfrac{1}{4}}$ you are applying a rule which does not apply: that $(-4)^{\tfrac{2}{4}}=((-4)^{2})^{\tfrac{1}{4}}$

$(-4)^{\tfrac{2}{4}}$ means the principal square root of -4 (i.e. 2i).

$(16)^{\tfrac{1}{4}}$ means the principal $4^{\text{th}}$ root of 16 (i.e. 2).

The principal square root of -4 (2i) is one of the $4^{\text{th}}$ roots of 16, but it is not the principal one.

 

[Steven G]

This is the point. When you write $(-4)^{\tfrac{2}{4}}$ you are writing $(-4)^{\tfrac{1}{2}}$.
When you say that $(-4)^{\tfrac{2}{4}}=(16)^{\tfrac{1}{4}}$ you are applying a rule which does not apply: that $(-4)^{\tfrac{2}{4}}=((-4)^{2})^{\tfrac{1}{4}}$

$(-4)^{\tfrac{2}{4}}$ means the principal square root of -4 (i.e. 2i).

$(16)^{\tfrac{1}{4}}$ means the principal $4^{\text{th}}$ root of 16 (i.e. 2).

The principal square root of -4 (2i) is one of the $4^{\text{th}}$ roots of 16, but it is not the principal one.

Got it. That is important for me to see! Thanks for clearing that up for me!

 

[lex]

@Steven G
Great - glad to hear it. It is a nuisance which keeps turning up! It's worth taking the time to think it through.