-5

\begin{align} -20 &= -20\\ 16-36 &= 25-45\\ 4^2-4\times 9&=5^2-5\times 9\\ 4^2-4\times 9+81/4&=5^2-5\times 9+81/4\\ 4^2-4\times 9+(9/2)^2&=5^2-5\times 9+(9/2)^2\\ \end{align}

Considering the formula $a^2+2ab+b^2=(a-b)^2$, one has \begin{align} (4-9/2)^2&=(5-9/2)^2\\ \sqrt{(4-9/2)^2}&=\sqrt{(5-9/2)^2}\\ 4-9/2&=5-9/2\\ 4&=5\\ 4-4&=5-4\\ 0&=1 \end{align}

Animesh Sahu
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    $\sqrt{x^2}=|x|$ so $\sqrt{a^2}=\sqrt{b^2}$ does *not* imply $a=b$ but rather $|a|=|b|$. And, indeed, $|4-9/2|=|5-9/2|=1/2$ but $4-9/2 \ne 5-9/2$ and the rest doesn't follow. – dxiv Dec 31 '16 at 03:44
  • I don't see why this question keeps getting downvotes. I'll give it a +1 just because it is very clearly titled "*spot the mistake*", it is written down in full detail, and correctly tagged as `fake-proofs`. – dxiv Dec 31 '16 at 04:12
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    Possible duplicate of [Best Fake Proofs? (A M.SE April Fools Day collection)](http://math.stackexchange.com/questions/348198/best-fake-proofs-a-m-se-april-fools-day-collection) – Moo Dec 31 '16 at 04:16

2 Answers2

4

Here let me simplify your process:

$$ \begin{align} (-2)^2 = 4 &\implies \sqrt{(-2)^2} = \sqrt{2^2} \\ &\implies -2 = 2 \\ &\implies -2 + 2 = 2 +2 \\ &\implies 0 = 4 \end{align} $$

QED. Do you see the mistake?

gowrath
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  • So, I'm simply give the answer: we cannot root the negative integer,as in question √(4-9/2)2=√(5-9/2)2 explaination: LHS:4-9/2=(8-9)/2 so √((8-9)/2)2 √(-1/2)2 √(1/4) (1/2) RHS:5-9/2=(10-9)/2 so √((10-9)/2)2 √(1/2)2 √(1/4) (1/2) LHS=RHS – Animesh Sahu Dec 31 '16 at 17:01
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$\sqrt {a^2}\ne a $. You should never have been taught that it does.

Instead $\sqrt {a^2}=|a| $ .

So....

$(4-9/2)^2=(5-9/2)^2$

$\sqrt {(4-9/2)^2}=\sqrt {(5-9/2)^2}$

$|4-9/2|=|5-9/2|$

$4 - 9/2 = \pm 5 \mp 9/2$

$4 = 5$ or $4 = -5 +2* 9/2=-5+9$

$4-4=5-4$ or $4-4 =-5-4+9$

$0=1$ or $0 = 0$

fleablood
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  • So, I'm simply give the answer: we cannot root the negative integer,as in question `√(4-9/2)2=√(5-9/2)2` explaination: LHS:4-9/2=(8-9)/2 so √((8-9)/2)2 √(-1/2)2 √(1/4) (1/2) RHS:5-9/2=(10-9)/2 so √((10-9)/2)2 √(1/2)2 √(1/4) (1/2) LHS=RHS – Animesh Sahu Dec 31 '16 at 17:00
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    Not quite. $(5-9/2)^2 = 1/4$ and $(4-9/2)^2 = 1/4$ and that's fine. But if $a^2 = b^2$ that does NOT mean $a = b$. The problem $x^2 = 25$ has *two* answers. $x = 5$ is one answer but $x = -5$ is another. If $c > 0$ then there is one *positive* number $r$ so that $r^2 = c$ and there is one *negative* number that is equal to $-r$ so that $(-r)^2 = c$. So $a^2 = b^2$ does not mean $a = b$ but it means $|a| = |b|$ but either a or b could be positive or negative. So you have $(4-9/2)^2 = (5-9/2)^2$ which is fine But 4 - 9/2 < 0 while 5 - 9/2 > 0. so... to be continued... – fleablood Dec 31 '16 at 17:36
  • ... continued.... and you have $(4-9/2)^2 = (5-9/2)^2$ which is fine (they both equal $1/4$). And you have $\sqrt{(4- 9/2)^2} = \sqrt{5-9/2)^2}$ which is *also* fine (they both equal $1/2$). But you have $\sqrt{(4-9/2)^2} = 4-9/2 = - 1/2$ which is *NOT* fine. This isn't true. $\sqrt{(-5)^2} = \sqrt{25} = 5 \ne -5$. $\sqrt{a^2} \ne a$--- you should never have learned that. $\sqrt{a^2} = |a|$. $|a| = 0$ if $a \ge 0$ but $|a| = -a$ if $a \le 0$. So you can conclude $|4- 9/2| = |5-9/2|$ which is fine. They both equal 1/2. – fleablood Dec 31 '16 at 17:43