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



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

EinsteinJr  May 2, 2015

Best Answer 

 #3
avatar+91477 
+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. :)

Melody  May 3, 2015
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3+0 Answers

 #1
avatar+91477 
+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

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

Melody  May 2, 2015
 #2
avatar+886 
0

Exactly. Here's your cookie, Melody:

Cookie

EinsteinJr  May 2, 2015
 #3
avatar+91477 
+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

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