Problem #PRU-5090

Problem

Let’s prove that $1=2$. Take a number $a$ and suppose $b=a$. After multiplying both sides we have $a^2=ab$. Subtract $b^2$ from both sides to get $a^2-b^2=ab-b^2$. The left hand side is a difference of two squares so $(a-b)(a+b)=b(a-b)$. We can cancel out $a-b$ and obtain that $a+b=b$. But remember from the start that $a=b$, so substituting $a$ for $b$ we see that $2b=b$, dividing by $b$ we see that $2=1$.