I'll present a proof that \(0 = 1\). Of-course there is an error in the proof. Can you find it?

## \(\large \textbf{Proof}\)

Let us start with a vector field \(\overrightarrow{v} = \dfrac{1}{r^2} \hat{r}\).

Let us find the Divergence of this field. \[ \overrightarrow{\nabla} . \overrightarrow{v} = \frac{1}{r^2} \frac{\partial}{\partial r} \left( r^2 \dfrac{1}{r^2} \right) = \boxed{0}\]

Now let us calculate the flux of this field. \[\oint_{S}{ \overrightarrow{v}. \overrightarrow{da}} = 4\pi\]

Green’s Theorem state that \[\int_{V} {\left (\overrightarrow{\nabla} . \overrightarrow{v} \right ) d\tau} = \oint_{S}{ \overrightarrow{v}. \overrightarrow{da}} \]

Applying it we see L.H.S = 0 since \(\overrightarrow{\nabla} . \overrightarrow{v} = 0 \) and R.H.S = \(4\pi\).

This implies \( 0 = 4\pi\).

Dividing by \(4\pi\) we get

\[\boxed{ 0 = 1 }\]

Hence proved.

## Comments

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TopNewest@Rajdeep Dhingra is nishant abhangi better than u?? – Meera Somani · 2 months, 4 weeks ago

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Green's theorem does not hold as the divergence is not continuous inside a ball containing the origin. – Abhishek Sinha · 1 year, 3 months ago

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– Rajdeep Dhingra · 2 months, 1 week ago

Yes , correct. To correct this error we have to use a Diract Delta Function.Log in to reply

Hint 1:There is some thing fishy with the divergence. – Rajdeep Dhingra · 1 year, 3 months agoLog in to reply