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# Is any real number positive ?

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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

#3
+91510
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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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#1
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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
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EinsteinJr  May 2, 2015
#3
+91510
+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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