MHB Square Root Rules for Fractions: x∈[3,∞)

[Amer]

\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 ?

 

[alyafey22]

Re: about square root

Generally for the case

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

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

• a $\geq$ 0,b>0
• $a\leq 0$,b<0

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

 

[Amer]

Re: about square root

What made me ask this question the website wolfarmalpha plot the function
f(x) = \frac{\sqrt{x-3}}{\sqrt{x}}
View attachment 1433

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 ?

 

[alyafey22]

Re: about square root

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]

Re: about square root

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

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

-Dan

 

[MarkFL]

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

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

we find $x$ in:

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

And over this domain, we may state:

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

 

[Amer]

Re: about square root

ZaidAlyafey's response shows that there is, in fact, a distinction between [math]\sqrt{\frac{x - 3}{3}}[/math] and [math]\frac{\sqrt{x - 3}}{\sqrt{x}}[/math]. 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:

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

we find $x$ in:

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

And over this domain, we may state:

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

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]

Re: about square root

...
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 ?

That would indeed be a better approach, as this way there are no imaginary factors to divide out. (Yes).