Why I cannot do this: $|x| = \sqrt{x^2} = (x^2)^{1/2} = x^{2 \ \times \ 1/2} = x$

Question

Why is it wrong?

$$|x| = \sqrt{x^2} = (x^2)^{1/2} = x^{2 \ \times \ 1/2} = x$$

Because the chain equation is not true at the next-to-last step when $ \ x \ $ is negative. — , Jun 20 '21 at 21:48
It's frowned upon but also because the square root function only accepts/returns non-negative numbers. In your case, $x$ can be anything. — LogicalGooGoo, Jun 20 '21 at 21:49
The identity $(a^b)^c=a^{(bc)}$ is only guaranteed when the base $a$ is a positive real number. Overlooking the assumptions under which various laws hold can lead to all sorts of apparent paradoxes. — Barry Cipra, Jun 20 '21 at 22:00
And not even "$a$ is a positive real number" will guarantee $(a^b)^c=a^{bc}$ -- you need additional conditions such as "$b$ and $c$ are real numbers". Otherwise you can get into trouble with, for example, $(e^{2\pi i})^{1/4}$. — Troposphere, Jun 20 '21 at 22:55

score 2 · Answer 1 · answered Jun 20 '21 at 22:44

As mentioned in the comments, $(a^b)^c=a^{bc}$ is not always true when $a$ is negative. For example, take $x=-1$ in your proof:

$$[(-1)^2]^\frac{1}{2}=1^{\frac{1}{2}}=\sqrt{1}=1$$

$$(-1)^{2\times \frac{1}{2}}=(-1)^1=-1$$

$$\therefore [(-1)^2]^\frac{1}{2} \neq (-1)^{2\times \frac{1}{2}}$$

1 Answers1