Root Then Power Or Power Then Root?

Algebra Level 2

True or false :

$\text{For all real }z\text{, } \sqrt [ n ]{ { z }^{ m } } ={ (\sqrt [ n ]{ z } ) }^{ m }.$

False True

3 solutions

Zeeshan Ali
Jan 20, 2016

The statement given as: $\sqrt [ n ]{ { z }^{ m } } = { (\sqrt [ n ]{ z } ) }^{ m }\quad \forall z\in R$ is always true when z is a non-negative real number. But when $z$ is negative, $\sqrt [ n ]{ z }$ is not real, hence the equality may or may not hold.

And what if z = -27 and n = 3 ? Won't it be real?

Kunal Jadhav - 5 years, 4 months ago

well.. of course it is. But you can't say that for all non-zero integers

Zeeshan Ali - 5 years, 4 months ago

But "is only true when z is a non-negative real number. Because when z is negative, is no more real." says otherwise.

That is probably what he is trying to point out.

Jerry Hill - 5 years, 4 months ago

So 🇲 changes..

Unmilon Saikia - 5 years ago

The fact that the root effectively filters out negative values is not the cause as can be seen in:
(z^(1/n))^m = (z^m)^(1/n)
Where if z = -1; n = m = 2; gives {-i, +i}^2 = (1)^(1/2) {-1, -1} = {-1, 1}

Fred Eisele - 3 years, 3 months ago

But it says non negative in the statement

Marta Jovanovic - 3 years, 1 month ago

Jesús Rodríguez
Jan 23, 2016

When z=-1, n=2, m=2, the relation is false. The left part is 1 and the right part is -1. So, it makes the statement false.

Gauss Uddin
Jan 22, 2016

Put some real values in equation and match them

Root Then Power Or Power Then Root?

3 solutions

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