# False Proof – 1 = 2

This is the first in a series of “false proofs.” Despite their falsity, they will be part of the Proof Gallery. The reason for putting them there is that often times a false proof gives insight into the nature of the problem domain. We will be careful to choose problems which do so.

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).