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This question already has an answer here:

I have some idea of where the mistake in this 'proof' may be, but can't quite formalize.

We start with the trivially correct statement, $1 - 3 = 4 - 6$. Then, completing the square on the LHS, we add $\frac{9}{4}$ to both sides, giving us $1 - 3 + \frac{9}{4} = 4 - 6 + \frac{9}{4}$, which is still accurate. Then, we factor both sides into $\left(1 - \frac{3}{2}\right)^2 = \left(2 - \frac{3}{2}\right)^2$. I was confused particularly by this step, especially as to why this works on the right-hand side, but upon verifying it, it seems to hold.

The proof then, I think incorrectly, takes principal roots of both sides and writes that $1 - \frac{3}{2} = 2 - \frac{3}{2}$, which implies that $1 = 2$.

I believe the mistake is in not writing $1 - \frac{3}{2} = \pm \left(2 - \frac{3}{2}\right)$, but this statement is still inaccurate unless we take take the negation on the RHS, though I'm not sure why, other than realizing this with arithmetic, we would think to only take its negation.

So, my questions are:

(a) How does the factoring on the right-hand side work? I know that the standard method of factoring when completing the square is to take the square of the first number, sign of the second, and square of the third. But I don't quite understand 'why' this works or, for that matter, why it would work on the right-hand side, other than the fact that $4$ and $\frac{9}{4}$ are squares, though I don't understand the significance of the $-6$ here: if this is all that is required, that $-6$ could in principle be anything at all, which is surely not the case.

(b) Is it correct that the mistake in the proof is in taking principal square roots? What is the logic behind leaving out the $\pm$ here?

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marked as duplicate by Jam, Daniel Fischer Jul 13 '18 at 17:24

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  • $\begingroup$ How come $-1/2$ is a "principal root" of $1/4$? $\endgroup$ – Lord Shark the Unknown Jul 13 '18 at 16:27
  • $\begingroup$ In fact they didn't take the principal square root on both sides! The principal square root is positive. $\endgroup$ – David C. Ullrich Jul 13 '18 at 16:27
  • $\begingroup$ $x^2=y^2 \implies x=y\,\lor x=-y$ $\endgroup$ – Jakobian Jul 13 '18 at 16:31
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There's a lot of misdirection in that proof - the entire start of the proof is irrelevant.

If you start with $(2-1)^2=(1-2)^2$, can you conclude that $1-2=2-1$?

The principal value of $\sqrt{x^2}$ is not $x$, because $x^2=(-x)^2$ and thus we'd have $$-x=\sqrt{(-x)^2}=\sqrt{x^2}=x$$ for all $x.$

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For every real number $x\quad$ $\sqrt{x^2} = \lvert x\rvert$
That is the mistake - omitting absolute value. You believe correctly.

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(a): There is a well-known equality $a^2-2ab+b^2=(a-b)^2$. In this case, $a=2$ and $b=\dfrac{3}{2}$, so we have $a^2-2ab+b^2=4-6+\dfrac{9}{4}$, because $-2ab=-2\times 2\times \dfrac{3}{2}=-6.$

(b): For $a,b\in\mathbb{R}$, $a^2=b^2$ does not imply that $a=b$. It implies that $a^2-b^2=0$ or $(a-b)(a+b)=0$, so $a=b$ or $a=-b$, which can be informally written as $a=\pm b$.

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