Question

2+2=5. how can i prove this?

Answer 1

LOL Who wants to know? :D

Answer 2

\[\Huge \infty+\infty=\infty\]
\[\Huge \frac{2}{\cancel 0}+\frac{2}{\cancel 0}=\frac{5}{\cancel 0}\]

Answer 3

its a proof alone..
but can we just expand this to equate this?

Answer 4

\[\dfrac{2}{0} \ne \infty\]

Answer 5

Let \(\large a,b \ne 0\)
such that
\[\large a = b\]
Then
\[\large a^2 = ab\]
then
\[\large a^2 \color{red}{-b^2}= ab\color{red}{-b^2}\]
Then
\[\large (a+b)(a-b)=b(a-b)\]

Cancelling (a-b) we get
\[\large a+b=b\]

Since a = b
\[\large b+b = b\]\[\large2b = b\]

Divide both sides by b
\[\large 2 = 1\]




NOW
\[\Large \color{blue}{2+2=4}\]

We add 2 to both sides
\[\Large \color{blue}{2+2\color{red}{+2}=4\color{red}{+2}}\]
And then we also subtract 2 from both sides
\[\Large \color{blue}{2+2+2\color{red}{-2}=4+2\color{red}{-2}}\]

Since 2 = 1, we can replace the -2 at the right with a -1
\[\Large 2+2+2-2 = 4+2\color{red}{-1}\]

This part cancels out
\[\Large 2+2\cancel{+2-2}=4+2-1\]

Leaving
\[\Huge 2+2 = 5\]

QED

LOL

Answer 6

You guys know this stuff's not meant to be taken seriously, right? -.-

Answer 7

\[x^2 = \underbrace{x + x + x + x\cdots x + x}_{x \rm ~times}\]Differentiating both sides,\[2x = \underbrace{1 + 1 + 1 + 1\cdots 1}_{x ~ \rm times}\]\[\implies 2x = x\]\[\implies 2 = 1\]Or similarly,\[1 = 2\]Now adding \(3\) to both sides,\[4 = 5\]But \(2 + 2 = 4\), so \(2 + 2 = 5\)

Answer 8

ehh... methinks the challenge here is picking the most creative way to prove this stuff :P

And then the next challenge is finding the (inevitably present) flaw in the proof...
Have fun, guys :D

Answer 9

Trolls