Proof of the Simple Theorem: x^2 >= 0 for Real x

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The discussion centers on the theorem stating that for any real number x, x^2 is greater than or equal to zero. An incorrect proof is presented, suggesting that if x^2 were less than zero, then using x=3 leads to a contradiction, thus concluding x^2 must be non-negative. However, the proof is flawed because it incorrectly applies a specific case (x=3) to generalize the statement for all real numbers. The correct approach requires demonstrating that the assumption leads to a contradiction for all x, not just a single instance. This highlights the importance of rigorous logical reasoning in mathematical proofs, particularly in the use of reductio ad absurdum.
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Theorem: Suppose that x is real, then x^2 > or = 0.
Incorrect Proof: Suppose not. Then x is real and x^2<0. Consider x=3. 9<0, a contradiction. Therefore x^2 >=0.
 
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I'm sorry, but what exactly is the brain teaser at hand here?
 
Werg22 said:
I'm sorry, but what exactly is the brain teaser at hand here?
Why the proof is incorrect, even though seemingly it looks like it is.
 
I don't see how could this fool anyone... Given the assumption P -> -Q , --Q(3) does not give rise to P -> Q. One needs to show that the assumption P -> -Q gives rise to --Q (a statement for all x being considered, not just 3) and therefore conclude P -> Q by reductio ad absurdum.
 
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