Problem #PRU-100255

Problem

Take any two non-equal numbers $a$ and $b$, then we can write $$a^2-2ab+b^2=b^2-2ab+a^2$$ Using the formula $(x-y{)}^2=x^2-2xy+y^2$, we complete the squares and rewrite the equality above as $$(a-b{)}^2=(b-a{)}^2.$$ As we take a square root from the both sides of the equality, we get $$a-b=b-a.$$ Finally, adding to both sides $a+b$ we get $$\begin{array}{rl}a-b+(a+b)& =b-a+(a+b)\\ 2a& =2b\\ a& =b.\end{array}$$ Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)