Problem #PRU-100277

Problem

Do you remember the example from the previous maths circle?

“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 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 $a-b+(a+b)=b-a+(a+b)$. It simplifies to $2a=2b$, or $a=b$. Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)”

Do you remember what the mistake was? In fact we have mixed up two things. It is indeed true “if $x=y$, then $x^2=y^2$”. But is not always true “if $x^2=y^2$, then $x=y$.” For example, consider $2^2=(-2{)}^2$, but $2\ne (-2)!$ Therefore, from $(a-b{)}^2=(b-a{)}^2$ we cannot conclude $a-b=b-a$.