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Square roots

Yankel

Active member
Jan 27, 2012
398
Hi,

I have a very basic question that suddenly hit me regarding square roots.

Why this is equal
\[\sqrt[3]{(1+x^{3})^{2}}=(1+x^{3})^{^{\frac{2}{3}}}\]

but this isn't

\[\sqrt{(x-2)^{3}}\neq (x-2)^{\frac{3}{2}}\]

(well according to Maple it isn't)

I understand why the first one is correct, but I assumed to believe that also the second one is equal and now I am confused.
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Hi,

I have a very basic question that suddenly hit me regarding square roots.

Why this is equal
\[\sqrt[3]{(1+x^{3})^{2}}=(1+x^{3})^{^{\frac{2}{3}}}\]

but this isn't

\[\sqrt{(x-2)^{3}}\neq (x-2)^{\frac{3}{2}}\]

(well according to Maple it isn't)

I understand why the first one is correct, but I assumed to believe that also the second one is equal and now I am confused.
Wolfram Alpha has no problem with it. (Note though that there is an issue when x - 2 < 0. I don't know why.)

-Dan
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
They are equal when $(x-2)^3 \geq 0$, but that is not defined for all $x \in \mathbb{R}$.

Consider $\sqrt{(-1)^6}$. What is the result of this operation? If you work inside out, you'll get $$\sqrt{(-1)^6} = \sqrt{1} = 1.$$ On the other hand, if you apply the exponents rule, you get $$\sqrt{(-1)^6} = (-1)^{\frac{6}{2}} = (-1)^3 = -1.$$ Is mathematics contradicting itself? Could our whole world be CRUMBLING BEFORE THE MIGHT OF EXPONENTIATION? Not really. The subtlety is that the operations are defined for nonnegative real numbers, letting the theory work smoothly. When we take in account negative real numbers as well, we take the order exponentiation - root to enable such operations.

In some cases it is not even possible to do so: in the real numbers there is no thing as $\sqrt{(-1)^5}$ because it is not defined.

Hope this has helped. Cheers! :D
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,774
(Note though that there is an issue when x - 2 < 0. I don't know why.)

-Dan
What would a calculator make of $(-1)^{0.66667}$?

Oh, and my favorite:
$$-1=(-1)^{\frac 23 \cdot \frac 32}=((-1)^{\frac 23})^{\frac 32}=1^{\frac 32}=1$$

EDIT: Ah, Fantini was quicker than me!
 
Last edited:

Yankel

Active member
Jan 27, 2012
398
Thank you, but if the issue here is the expression under the square root being positive or negative, then how come the first expression is equal ?

1+x^3 is not positive for every x in R, and yet, Maple seem to think it's Ok.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
You used the right term: square root. What you have first is a cubic root, which is defined for all real numbers. :D