$a = b$
$a^2 = ab$
$a^2 + a^2 = a^2 + ab$
$2a^2 = a^2 + ab$
$2a^2 - 2ab = a^2 + ab - 2ab$
$2a^2 - 2ab = a^2 - ab$
$2(a^2 - ab) = 1(a^2 - ab)$
$2 = 1$
What is the error in this proof??
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Because $a^2-ab = 0$, you can't divide both sides by it to get your last equation. Think about $1\cdot 0 = 2\cdot 0$ doesn't imply $1=2$.
Lazy Lee
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This is a very famous question which is listed in Michael Spivak's "Calculus". This proof is meant to teach the student that you cannot divide by zero. But, also to follow proofs logically. If $a^2 = ab$ then, $a=b$ , but that was already stated on the first line.
Where does zero come into play?
This line right here: $2a^2−2ab=a^2+ab−2ab$
Looking at the LHS, if we take $2a^2$ and subtract $2ab$ the result is zero. As we have already decided that $a=b$. On the RHS $a^2 +ab$ = $2a^2 $, if we subtract $2ab$ the result is zero.
$2a^2−2ab=a^2−ab$ here, the result is still zero on both sides. We have something, and are subtracting it.
$2(a^2−ab)=1(a^2−ab)$ here we have factored out the coefficients, but inside the brackets the terms are still equal to zero, and as we know: $2 * 0 = 0$, $1 * 0 = 0$.
Lastly, we divide by zero, giving us $2 = 1$. If the student had been following the proof, they would have realized everything from the fifth line on wards was equal to zero.