Problem: Show 1 = 2.

“Solution”: Let $ a=b \neq 0$. Then $ a^2 = ab$, and $ a^2 – b^2 = ab – b^2$. Factoring gives us $ (a+b)(a-b) = b(a-b)$. Canceling both sides, we have $ a+b = b$, but remember that $ a = b$, so $ 2b = b$. Since $ b$ is nonzero, we may divide both sides to obtain $ 2=1$, as desired.

Explanation: This statement, had we actually proved it, would imply that all numbers are equal, since subtracting 1 from both sides gives $ 0=1$ and hence $ a=0$ for all real numbers $ a$. Obviously this is ridiculous.

Digging into the algebraic mess, we see that the division by $ a-b$ is invalid, because $ a=b$ and hence $ a-b = 0$.

Division by zero, although meaningless, is nevertheless interesting to think about. Much advanced mathematics deals with it on a very deep and fundamental level, either by extending the number system to include such values as $ \frac{1}{0}$ (which still gives rise to other problems, such as $ \frac{0}{0}$ and $ 0 \cdot \infty$), or by sidestepping the problem by inventing “pseudo” operations (linear algebra) and limiting calculations (calculus).