Question

I proved 2=0

$\large \sf 2=1+1$
$\large \sf 2=1+\sqrt{1}$
$\large \sf 2=1+\sqrt{(-1)(-1)}$
$\large \sf 2=1+\sqrt{(-1)}\sqrt{(-1)}$
$\large \sf 2=1+(i)(i)$
$\large \sf 2=1+(i)^2$
$\large \sf 2=1+(-1)$
$\large \sf 2=1-1$
$\large \sf 2=0$


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Answer 1

$\large\underline{\sf{Solution-}}$

The fallacy in the above statement is

$\red{\rm :\longmapsto\: \sqrt{( - x)( - y)} \: \ne \: \sqrt{xy} }$

As the correct Statement is

$\red{\rm :\longmapsto\: \sqrt{( - x)( - y)} \: = \: - \: \sqrt{xy} }$

Let first prove this,

So, Consider

$\rm :\longmapsto\: \sqrt{( - x)( - y)}$

can be rewritten as

$\rm \:  =  \: \sqrt{( - x)} \times \sqrt{( - y)}$

$\rm \:  =  \: \sqrt{ -1 \times x} \times \sqrt{ -1 \times y}$

$\rm \:  =  \: \sqrt{ - 1} \times \sqrt{x} \times \sqrt{ - 1} \times \sqrt{y}$

$\rm \:  =  \: i \times \sqrt{x} \times i \times \sqrt{y}$

$\rm \:  =  \: {i}^{2} \times \sqrt{x} \times \sqrt{y}$

$\rm \:  =  \: - \sqrt{xy}$

So,

$\red{\rm :\longmapsto\: \sqrt{( - x)( - y)} \: = \: - \: \sqrt{xy} }$

Now,

$\rm :\longmapsto\:2 = 1 + 1$

$\rm :\longmapsto\:2 = 1 + \sqrt{1}$

$\rm :\longmapsto\:2 = 1 \red{- \sqrt{( - 1)( - 1)} }$

$\rm :\longmapsto\:2 = 1 + \red{ [- \sqrt{( - 1)( - 1)} \: ]}$

$\rm :\longmapsto\:2 = 1 + \red{ [- \sqrt{( - 1)} \times \sqrt{( - 1)} \: ]}$

$\rm :\longmapsto\:2 = 1 + \red{ [- i \times i \: ]}$

$\rm :\longmapsto\:2 = 1 + \red{ [- {i}^{2} \: ]}$

$\rm :\longmapsto\:2 = 1 + \red{ [- ( - 1) \: ]}$

$\rm :\longmapsto\:2 = 1 + \red{ [1 \: ]}$

$\rm :\longmapsto\:2 = 2$

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