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3 years, 2 months ago

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 | 3 years, 2 months ago
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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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rickg | 3 years, 2 months ago
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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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drivel | 3 years, 2 months ago
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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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bugsi | 3 years, 2 months ago
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Only if X is also real.

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ryan loran | 3 years, 2 months ago
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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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reknaw | 3 years, 2 months ago
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Yes.

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