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#### Amer

##### Active member
$$\sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}}$$

that is not true for all x, it is true for $$x\in [3,\infty)$$
I want to teach my students that the exponents distribute over fractions unless we have a case like that square root or any even root.
what do you think ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

Generally for the case

$$\displaystyle \sqrt{\frac{a}{b}}$$ we require that

$$\displaystyle \frac{a}{b}\geq 0$$ which means either

• a $\geq$ 0,b>0
• $a\leq 0$,b<0

For the first case we can state $$\displaystyle \sqrt{\frac{a}{b}}=\frac{\sqrt{a}}{\sqrt{b}}$$

#### Amer

##### Active member

What made me ask this question the website wolfarmalpha plot the function
$$f(x) = \frac{\sqrt{x-3}}{\sqrt{x}}$$

although
$$f(-1) = \frac{\sqrt{-4}}{\sqrt{-2}}$$ which is not a real number
the domain of the function is $$[3,\infty)$$
f should start from x=3 to infinity, Am I right ?

#### ZaidAlyafey

##### Well-known member
MHB Math Helper

Wolfram gives the plot of the function in the complex plane , we have to add the condition that $x \geq 3$ to only focus on real part.

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$$f(-1) = \frac{\sqrt{-4}}{\sqrt{-2}}$$ which is not a real number
Unfortunately this is a real number $$f(-1) = \frac{\sqrt{-4}}{\sqrt{-2}} =\frac{2i}{\sqrt{2} i}=\sqrt{2}$$

#### topsquark

##### Well-known member
MHB Math Helper

$$\sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}}$$
ZaidAlyafey's response shows that there is, in fact, a distinction between $$\displaystyle \sqrt{\frac{x - 3}{x}}$$ and $$\displaystyle \frac{\sqrt{x - 3}}{\sqrt{x}}$$. What level are your students at? It might be better to sweep this particular kind of example under the rug.

-Dan

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#### MarkFL

Staff member
I see this as a domain issue. If we consider the inequality:

$$\displaystyle \frac{x-3}{x}\ge0$$

we find $x$ in:

$$\displaystyle (-\infty,0)\,\cup\,[3,\infty)$$

And over this domain, we may state:

$$\displaystyle \sqrt{\frac{x-3}{x}}=\frac{\sqrt{x-3}}{\sqrt{x}}$$

#### Amer

##### Active member

ZaidAlyafey's response shows that there is, in fact, a distinction between $$\displaystyle \sqrt{\frac{x - 3}{3}}$$ and $$\displaystyle \frac{\sqrt{x - 3}}{\sqrt{x}}$$. What level are your students at? It might be better to sweep this particular kind of example under the rug.

-Dan
They are high school students 11th class

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I see this as a domain issue. If we consider the inequality:

$$\displaystyle \frac{x-3}{x}\ge0$$

we find $x$ in:

$$\displaystyle (-\infty,0)\,\cup\,[3,\infty)$$

And over this domain, we may state:

$$\displaystyle \sqrt{\frac{x-3}{x}}=\frac{\sqrt{x-3}}{\sqrt{x}}$$
I think we should split it into two parts if x>=3
$$\sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}}$$

if x< 0
$$\sqrt{\frac{x-3}{x}} = \frac{\sqrt{3-x}}{\sqrt{-x}}$$

right ?

#### MarkFL

$$\sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}}$$
$$\sqrt{\frac{x-3}{x}} = \frac{\sqrt{3-x}}{\sqrt{-x}}$$