For $x\geq 0$, is $\sqrt{x}$ the magnitude of $x^{1/2}$?

Question

Motivation

\begin{align} 4^{1/2} &= \begin{cases} \left(2^2\right)^{1/2}\\ \left(\left(-2\right)^2\right)^{1/2} \end{cases} \\ &= \begin{cases} 2\\ -2 \end{cases} \end{align}

And $\sqrt{4}=2$ (by definition)

Question

Can I conclude that $\sqrt{x}$ is the magnitude of $x^{1/2}$ for $x\geq 0$?

$\sqrt x$ and $x^{1/2}$ stand for the same thing and both of them are nonnegative numbers. — Kavi Rama Murthy, Aug 19 '19 at 07:32

score 1 · Accepted Answer · answered Aug 19 '19 at 07:37

1

The rule that you used $x^{ab}=(x^a)^b$ is true only when $x \ge 0$ you can't use it when $x=-2$, and as Kavi Rama Murthy said $\sqrt x=x^{1/2} \ge 0$

answered Aug 19 '19 at 07:37

Fareed Abi Farraj

3,446

It is by convention rule? – Display Name Aug 19 '19 at 07:38
1

Maybe we can call it so, but the reason why it is not true is that if it is true then $-1=(-1)^1=(-1)^{(2).(1/2)}=((-1)^2)^{1/2})=1^{1/2}=1$ – Fareed Abi Farraj Aug 19 '19 at 07:48

For $x\geq 0$, is $\sqrt{x}$ the magnitude of $x^{1/2}$?

Motivation

Question

1 Answers1