Difference between revisions of "Fallacious proof/All numbers are equal"

 
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{{incomplete|explanation}}
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It does not follow in general from <math>x^2 = y^2</math> that <math>x = y</math>.  Instead, we have either that <math>x = y</math> or <math>x = -y</math>.  In the given situation, we have that <math>a - \frac{t}{2} = \frac{a - b}{2} = -\left(\frac{b - a}{2}\right) = - \left(b - \frac{t}{2}\right)</math>, and so this was the fallacious step.
Just a thought. When you square root both sides, since there are positive and negative square roots of a number, something must have been left out...
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''[[Fallacious_proof#All_numbers_are_equal | Back to main article]]''

Latest revision as of 15:00, 21 July 2009

It does not follow in general from $x^2 = y^2$ that $x = y$. Instead, we have either that $x = y$ or $x = -y$. In the given situation, we have that $a - \frac{t}{2} = \frac{a - b}{2} = -\left(\frac{b - a}{2}\right) = - \left(b - \frac{t}{2}\right)$, and so this was the fallacious step.

Back to main article