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$$ a = b $$ $$ a^2 = ab $$ $$ a^2 - b^2 = ab - b^2 $$ $$ (a+b)(a - b) = b(a - b) $$ $$ a + b = b $$ $$ 2b = b $$ $$ 2 = 1 $$

Does the self-reference to the original formula make this path of argument invalid? I am confused as to what's going on here.

Would this be a better continuation from $a + b = b$:

$$ a + b = b \implies a = 0 \implies b = 0$$

mathematician
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CinchBlue
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3 Answers3

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The statement $$\mbox{If $xy=xz$, then $y=z$}$$ is false: for example, $0\cdot 1=0\cdot 2$ but $1\not=2$. What is true is that $$\mbox{If $xy=xz$, then $y=z$ or $x=0$.}$$ So when you go from $$(a+b)(a−b)=b(a−b)$$ to $$(a+b)=b,$$ that's not justified; instead, from $$(a+b)(a-b)=b(a-b)$$ all you can conclude is either $(a+b)=b$, or $(a-b)=0$. And indeed, $(a-b)=0$.

Noah Schweber
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  • The thing I have always found odd about this particular fake proof is that even after that faulty step, it fails to arrive at a contradictory state. It only arrives at a state where $a + b = b$, implying $a = b = 0$, and then has to divide by $0$ again to arrive at the contradiction. – Caleb Stanford Nov 23 '16 at 02:03
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You're dividing by zero in the fourth line because $a=b$.

iam_agf
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1

Notice that

$$a=b\implies a-b=0$$

So the third step is not allowed since you can't divide by zero.

Simply Beautiful Art
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