What's correct?

Have you had your coffee this morning?
[imath]\dfrac{1}{2}=\dfrac{2}{4}[/imath] or [imath]a\ge 0\text{ then }\sqrt{a}=\sqrt[4]{a^2}[/imath]
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
 
I think I am responsible for starting this debate! (Apologies to the OP).
I hesitantly make this contribution:
exp1.jpgexp2.jpg
 
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?
 
This is the point. When you write [imath](-4)^{\tfrac{2}{4}}[/imath] you are writing [imath](-4)^{\tfrac{1}{2}}[/imath].
When you say that [imath](-4)^{\tfrac{2}{4}}=(16)^{\tfrac{1}{4}}[/imath] you are applying a rule which does not apply: that [imath](-4)^{\tfrac{2}{4}}=((-4)^{2})^{\tfrac{1}{4}}[/imath]

[imath](-4)^{\tfrac{2}{4}}[/imath] means the principal square root of -4 (i.e. 2i).

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

The principal square root of -4 (2i) is one of the [imath]4^{\text{th}}[/imath] roots of 16, but it is not the principal one.
 
Last edited:
This is the point. When you write [imath](-4)^{\tfrac{2}{4}}[/imath] you are writing [imath](-4)^{\tfrac{1}{2}}[/imath].
When you say that [imath](-4)^{\tfrac{2}{4}}=(16)^{\tfrac{1}{4}}[/imath] you are applying a rule which does not apply: that [imath](-4)^{\tfrac{2}{4}}=((-4)^{2})^{\tfrac{1}{4}}[/imath]

[imath](-4)^{\tfrac{2}{4}}[/imath] means the principal square root of -4 (i.e. 2i).

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

The principal square root of -4 (2i) is one of the [imath]4^{\text{th}}[/imath] 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!
 
@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.
 
Top