Question

Hey guys, up for proving something?

Answer 1

Prove: \[\LARGE 2+2=5\]

Answer 2

you must have divided by 0

Answer 3

http://www.youtube.com/watch?v=XXhbThOITLQ

Answer 4

Watch and learn yo

Answer 5

Divide by 0?

Answer 6

i think ur 1 is tryin 2 b d invisible man
oops, sry..invisible number

Answer 7

Hmm... To prove that I'll start off with something that is true:
say...
\[-20=-20\]
rewriting them a little:
\[16-36=25-45\]
I'll rewrite them again:
\[(2+2)^{2}-(2+2)(9)=5^{2}-(5)(9)\]
if I multiply and divide by two, and reorder:
\[(2+2)^{2}-(2)(2+2)(\frac{ 9 }{ 2 })=5^{2}-(2)(5)(\frac{ 9 }{ 2 })\]
just playing a little with squares and fractionary statments:
\[(2+2)^{2}-(2)(2+2)(\frac{ 9 }{ 2 })+(\frac{ 9 }{ 2 })^{2}=5^{2}-(2)(5)(\frac{ 9 }{ 2 })+(\frac{ 9 }{ 2 })^{2}\]
Now with that, I can complete some squares:
\[(2+2-\frac{ 9 }{ 2 })^{2}=(5-\frac{ 9 }{ 2 })^{2}\]
by cancelative property then I can remove the 9/2 and take square roots on both sides to end with:
\[2+2=5\]
There is a mistake here though, try find it.

Answer 8

u divided only d no 9 by 2 instead of d entire 2(2+2)(9)

Answer 9

dont we have two possibilities

\[2+2-\frac{9}{2}=+(5-\frac{9}{2})\] and

\[2+2-\frac{9}{2}=-(5-\frac{9}{2})=-5+\frac{9}{2}\]

and the combination of this equations wen added gives
\[2+2=0\implies 2=0\] wich we divided and multiplied with ealier

Answer 10

no its not zero its
\[2(2+2)=9\implies 2+2-\frac{9}{2}=0\]
meaning the 2nd last equation yields
\[( 2+2-\frac{9}{2})^2=(5-\frac{9}{2})^2=0\]

Answer 11

this is crazy

Answer 12

I know, right?

Answer 13

Thank you @Owlcoffee :)

Answer 14

@Jordannyah Will this work for ya?

Answer 15

the paradox is the same
\[x=y\\x^2=xy\\x^2-y^2=xy-y^2\\(x+y)(x-y)=y(x-y)\\x+y=y\\x+x=y\\2x=y\implies 2=1\implies 2+3=1+3\implies 5=4\]
but here its obvious the error was dividing by zero \[x-y=0\]

Answer 16

the problem with @Owlcoffee proof is using the wrong implication that if\[a^2=b^2\implies a=b\]instead of
\[a=\pm b\]

Answer 17

you discovered me!
:( .... *hides under rock*