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Proof:

Using simple Algebra it can be proved that -20 $\neq$ -20. Is it Algebraically correct, can I treat this as a puzzle only?

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Willie Wong
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Avinesh
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    It is not algebraically correct. – TZakrevskiy Dec 06 '13 at 06:57
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    Well, it is not algebraically correct =]. Look at T. Bongers' answer. – LASV Dec 06 '13 at 06:58
  • I need to edit my question, especially when i said it is algebraically correct. – Avinesh Dec 06 '13 at 06:59
  • People this is just a puzzle which I have put not to get downvote but rather suggestions. Id appreciate someone deleting the question rather than downvote. Please think logically! – Avinesh Dec 06 '13 at 07:11
  • $\displaystyle{\large\sqrt{,x^{2},}, = \left\vert,x,\right\vert}$ – Felix Marin Dec 06 '13 at 08:39
  • In other words, the problem stems from the fact that the function $f:\mathbb R\to\mathbb R$ mapping $x$ to $x^2$ is not injective. I could invent my own "proof" using the same idea. Since I am crazy I will assume $1=1$, of all things! Then, $\cos(0)=\cos(2\pi)$, and so $\cos^{-1}(\cos(0))=\cos^{-1}(\cos(2\pi))$, or $0=2\pi$. What could I have possibly done wrong?! – Karl Kroningfeld Dec 06 '13 at 09:30

1 Answers1

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The assertion that

$$\sqrt{\left(4 - \frac 9 2\right)^2} = \sqrt{\left(5 - \frac 9 2\right)^2} \implies 4 - \frac 9 2 = 5 - \frac 9 2$$ is false; in particular, this step assumes that

$$|x| = |y| \implies x = y$$ which is certainly false. Hence this is not algebraically correct.