anonymous
  • anonymous
2+2=5. how can i prove this?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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terenzreignz
  • terenzreignz
LOL Who wants to know? :D
DLS
  • DLS
\[\Huge \infty+\infty=\infty\] \[\Huge \frac{2}{\cancel 0}+\frac{2}{\cancel 0}=\frac{5}{\cancel 0}\]
anonymous
  • anonymous
its a proof alone.. but can we just expand this to equate this?

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ParthKohli
  • ParthKohli
\[\dfrac{2}{0} \ne \infty\]
terenzreignz
  • terenzreignz
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
terenzreignz
  • terenzreignz
You guys know this stuff's not meant to be taken seriously, right? -.-
ParthKohli
  • ParthKohli
\[x^2 = \underbrace{x + x + x + x\cdots x + x}_{x \rm ~times}\]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 \(2 + 2 = 5\)
terenzreignz
  • terenzreignz
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
terenzreignz
  • terenzreignz
Trolls

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