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exspiro exspiro
 

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M$1.00  Funded By Mahalo ? |  March 23, 2009 08:45 PM

if x does not equal 0, is x^2>0

Since x does not equal 0 by trichotomy, either x>0 or x<0 (that is, 0-x = -x), For second case use -x*-x = x*x
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pmacdon1 pmacdon1
 
 March 23, 2009 10:00 PM
Ok I'm gonna have to go against the grain and say NO.

@rickg had it right that if it is a real number, then x^2 > 0.

However your question doesn't specify that x is a real number.

Any non-real number squared will give you a negative number (I think)

For example:
i = square root (-1)
if x = i then x^2 = -1


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Voted as best: williamwaco, cp24
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ryan loran ryan loran
 
March 23, 2009 09:12 PM
Yes, is x does not equal 0, x^2 will always be greater than zero. You basically outlined the answer in your question- any negative value of x becomes positive when squared.

Decimal values can approach zero, but will never reach zero. For instance, .2 squared is .04, point .04 squared is .0016, and so on. This number will get very small, but will always be above zero.

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rickg rickg
 
March 23, 2009 09:14 PM
You seem to have the outline of the proof in the question. Yes, assuming x is a real number not equal to zero, then x^2 is positive definite.

Equivalently, say that for all real numbers x, x^2 >= 0, and equality occurs if and only if x=0..

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Voted as best: pazaq
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reknaw reknaw
 
March 23, 2009 09:38 PM
Yes.

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bugsi bugsi
 
March 24, 2009 12:26 AM
Only if X is also real.

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drivel drivel
 
March 24, 2009 02:29 AM
It depends. If we talking about real numbers then yes, if we are talking about complex numbers then the answer is no.

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