We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
741
3
avatar+869 

$$\\
\mathbb{R}\textup{ is the set of real numbers. For any number }x\textup{ in }\mathbb{R:}\\
x^2\geq 0\\
(x^2)^{1/2}\geq 0^{1/2}\\
x^{2\times1/2}\geq 0\\
x^1\geq 0\\
x\geq 0$$



Are all these calculations correct ? Or is there some form of miscalculation ?

 May 2, 2015

Best Answer 

 #3
avatar+104479 
+5

Thanks EinstienJr,

did you already know the problem with the question?

If you did you really should say so up front because otherwise, one day you will  ask a 'real' question and I will think,

I won't answer that because I am buy and I expect that he/she already knows the answer. :)

 May 3, 2015
 #1
avatar+104479 
+5

There is a bit of a problem here Einstein,

I have been struggling with exactly the same problem on complex number questions lately.

 

$$\\(x^2)^{1/2}>0\qquad true\\\\$$

 

consider

$$\\x^2=36\\
\sqrt{x^2}=\pm 6\\
$I put the square root in so both answers are valid$\\$$

 

so

$$\\(x^2)^{1/2}\ge0\\
\pm x \ge0\\
$either $x\ge0\;\;or\;\;-x\ge 0\\
$either $x\ge0\;\;or\;\;x\le 0\\
$Hence x is any real number $$$

.
 May 2, 2015
 #2
avatar+869 
0

Exactly. Here's your cookie, Melody:

Cookie

 May 2, 2015
 #3
avatar+104479 
+5
Best Answer

Thanks EinstienJr,

did you already know the problem with the question?

If you did you really should say so up front because otherwise, one day you will  ask a 'real' question and I will think,

I won't answer that because I am buy and I expect that he/she already knows the answer. :)

Melody May 3, 2015

12 Online Users