0

Ok so let me explain my self. We all know that the $\sqrt{X}$ is defined only for $X\ge 0$ (not considering complex numbers). But what if I write $X^{2/4}$ which is the 4th root of $X^2$, and now it's defined on any $X$? (The $\mbox{2n}$th root is defined only for non negative numbers and $X^2$ is non negative obviously so the $\mbox{2n}$th root is defined for this example $n=2$).

So long story short:

$X^{1/2}$ defined for non negative $X$.

$X^{2/4}$ defined for All $X$.

$1/2 = 2/4.$

What am I missing here?

Xam
  • 6,119
Itay.V
  • 406
  • Why $(x^2)^\frac{1}{4}$ and not $(x^\frac{1}{4})^2$? The second one is not defined for negative real numbers (works if you allow complex numbers). The reason is the order of operations. You need to calculate the exponent first – Andrei Nov 15 '16 at 21:55
  • By the title, it sounds like a duplicate of $(-32)^{\frac{2}{10}}\neq(-32)^{\frac{1}{5}}$?. – barak manos Nov 15 '16 at 21:56
  • Generally: You always have to take into account the value ranges. If they differ for two different ways of calculations then these calculations are not any more the same for the value range where they differ. E.g. one calculation can be an extension of another calculation. – user90369 Nov 16 '16 at 09:53

1 Answers1

2

$1/2$ is the same as $2/4$, so $X^{1/2}$ is the same as $X^{2/4}$. However, this is not necessarily the same as $(X^2)^{1/4}$. The "laws of exponents" must be modified when dealing bases that are not positive reals.

Robert Israel
  • 448,999