# 2 + 2 = 5

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i keep seeing this problem and i dont see how it posible

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It's not. It's usually a "proof" involving division by zero, or some such nonsense...

And I'm not sure this is the right section for this, although I'll leave that to the mods

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Or for extremely large values of 2, and bad communication.

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Don't be biased; it could also be a very small value of 5, and bad interpretation.

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there is a fallacy in that equation.

Merged post follows:

Consecutive posts merged

for some reason the picture didnt show up so here is a link

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The versin of the proof that I've always seen was 1=2, but I'm sure it could be rigged up for 2 + 2 = 5

1=1

-1=-1

-1/1 = 1/-1

root both sides

i/1 = 1/i

3/2i + 1/i = i/1 + 3/2i

multiply both sides by i

3/2 + 1 = -1 + 3/2

2.5 = .5

Ok. so it didn't equil 1=2, but its the principle of the thing...

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The versin of the proof that I've always seen was 1=2, but I'm sure it could be rigged up for 2 + 2 = 5

1=1

-1=-1

-1/1 = 1/-1

root both sides

i/1 = 1/i

3/2i + 1/i = i/1 + 3/2i

multiply both sides by i

3/2 + 1 = -1 + 3/2

2.5 = .5

Ok. so it didn't equil 1=2, but its the principle of the thing...

obviously when you root both sides the result is not :i/1 =1/i ,because :

i/1 =1/i <=====> $i^2 = 1\Longleftrightarrow -1 = 1$

Now the statement, -1 =1 in any line in any proof can result in any conclusion right or wrong

For example can result to false statements ,like:

5=7 ,1>4 , $x^2<0$ ln(-2) = 0 e,t,c ,e,t,c in the following way:

-1 =1 $\Longrightarrow [(-1 =1 )$or $(x^2<0)]\Longleftrightarrow[(-1\neq 1)\Longrightarrow (x^2<0)]$

and since $-1\neq 1$ we conclude that:

$x^2<0$

2.4 + 2.4 = 4.8

2.4 ≈ 2

4.8 ≈ 5

2 + 2 = 5

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onequestion: that's called rounding error for a reason.

what you should have said for the last line is

2 + 2 ≈ 5

which basically means it doesn't equal 5 but its close enough if your tolerance for error is large enough.

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yes lol i know i realized that obviously its just i didn't want other people to notice since the question originally said "=" not "≈"

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Or you could simply love Big Brother.

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A lookup post could help maybe not this one but a lot of thease sorts of things.

I'm thinking aloud here.

($\neq$ 0)/0 = infinite = inf

(>0)/0 = inf positively large

(<0)/0 = inf negitively large

0/0 = undefined

inf (- or /) inf = undefined

inf positively large (+ or *) inf positively large = inf positively large

inf negitively large - inf negitively large = inf negitively large

inf (+ or -) finite = inf

undefined (operator or function) anything generally = undefined

undefined * 0 = 0

undefined ^ 1 = 1 or -1

anything \ self = 1

Maybe this table is going to get too big.

inf = undefined in a way. It's either a range or a set of two ranges.

sorry, got a bit off topic.

Edited by alan2here

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my teacher showed a way of doing this, using rounding

2+2=5

2.4+2.4= 4.8

round it up and down

2+2=5

2.4 is rounded down to 2 and 4.8 is rounded up to 5.

ffs

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There are two methods to prove 2 + 2 = 5

The Orwellian Method

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It's also a joke.

2 + 2 = 5, for very large values of 2...

=Uncool-

Edited by uncool

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Or you could simply love Big Brother.

Yeah, I was wondering if I was the only one having a 1984 flashback.

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if numbers would be real,

2.x + 2.x = 5.xx

2 + 2 ≈ 5

or if numbers have powers,

2^x + 2^y = 5 ..ex: 2^0 + 2^2 = 5

or that it is a false degoma,

2 + 2 = 5 (FALSE)

..etc

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If in the case 2=2.5

2+2=5 <=> 2.5+2.5=5

Or quite simply the meaning of those numbers is different than what we are used to. So the symbol 2 does not mean two, nor does the symbol 5 mean five.

It could also be that the person is using some sort of other numbering system like binary or hex, simply a different base than 10 that we are used to and did not communicate it.

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