Problem #PRU-100277

Problems Set theory and logic Mathematical logic


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 \(2 \neq (-2)!\) Therefore, from \((a-b)^2 = (b-a)^2\) we cannot conclude \(a-b = b-a\).