Where is the error?
$-27=-27$
$81-108=9-36$
$9^2 - 2\cdot 9\cdot6 = 3^2 - 2\cdot3\cdot6$
$9^2 - 2\cdot9\cdot6 + 6^2 = 3^2 - 2\cdot3\cdot6 + 6^2$
$(9-6)^2 = (3-6)^2$
$9-6 = 3-6$
$9=3$
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Mauro ALLEGRANZA
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2$5 \implies 6$ is wrong. – Zerox Dec 18 '22 at 19:12
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1Step 6: $\sqrt{(3-6)^2}\neq3-6$ – A. Goodier Dec 18 '22 at 19:13
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1Note that, $$a^2=b^2\implies |a|=|b|$$ not $a=b$ – lone student Dec 18 '22 at 19:14
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Typo on line 3: you wrote $2\cdot9\cdot6$ on both sides of the equation but on the right side it should instead be $2\cdot3\cdot6$. This error is repeated on line 4. On line 5 the error disappears, presumably because you copied that line correctly. – David K Dec 18 '22 at 19:17
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Why is this being downvoted? – Kamal Saleh Dec 18 '22 at 19:35
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In fake proofs like that you only have to find out where we get the invalid equation. – Peter Dec 19 '22 at 06:56
2 Answers
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$a^2=b^2$ doesn't imply $a=b$. So the flaw happens while going from 5th to 6th step.
Oshawott
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Remember that taking the root gives you two solutions: $$y^2=x\implies y=\pm\sqrt{x}$$So In between $5.$ and $6.$ is where the flaw is made.
Whenever you have fallacies like this try to look at where the math starts to not make sense.
Kamal Saleh
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