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About square root

Amer

Active member
Mar 1, 2012
275
[tex]\sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}} [/tex]

that is not true for all x, it is true for [tex]x\in [3,\infty) [/tex]
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
Jan 17, 2013
1,667
Re: about square root

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
Mar 1, 2012
275
Re: about square root

What made me ask this question the website wolfarmalpha plot the function
[tex]f(x) = \frac{\sqrt{x-3}}{\sqrt{x}} [/tex]
Capture.JPG

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

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
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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[tex]f(-1) = \frac{\sqrt{-4}}{\sqrt{-2}} [/tex] which is not a real number
Unfortunately this is a real number [tex]f(-1) = \frac{\sqrt{-4}}{\sqrt{-2}} =\frac{2i}{\sqrt{2} i}=\sqrt{2}[/tex]
 

topsquark

Well-known member
MHB Math Helper
Aug 30, 2012
1,123
Re: about square root

[tex]\sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}} [/tex]
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
 
Last edited:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
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
Mar 1, 2012
275
Re: about square root

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
[tex] \sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}} [/tex]

if x< 0
[tex] \sqrt{\frac{x-3}{x}} = \frac{\sqrt{3-x}}{\sqrt{-x}} [/tex]

right ?
 

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
Re: about square root

...
I think we should split it into two parts if x>=3
[tex] \sqrt{\frac{x-3}{x}} = \frac{\sqrt{x-3}}{\sqrt{x}} [/tex]

if x< 0
[tex] \sqrt{\frac{x-3}{x}} = \frac{\sqrt{3-x}}{\sqrt{-x}} [/tex]

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