Fallacious proof/All numbers are equal

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.

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