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

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

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

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

 

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

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

add 3/2i to both sides

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...

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

add 3/2i to both sides

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 <=====> [math] i^2 = 1\Longleftrightarrow -1 = 1[/math]

 

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 , [math] x^2<0[/math] ln(-2) = 0 e,t,c ,e,t,c in the following way:

 

 

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

 

and since [math] -1\neq 1[/math] we conclude that:

 

 

[math] x^2<0[/math]

2.4 + 2.4 = 4.8

2.4 ≈ 2

4.8 ≈ 5

2 + 2 = 5

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.

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

Or you could simply love Big Brother.

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

 

I'm thinking aloud here.

 

([math]\neq[/math] 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 December 24, 200916 yr by alan2here

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

There are two methods to prove 2 + 2 = 5

 

The Orwellian Method

The Radiohead Method

It's also a joke.

 

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

=Uncool-

Edited December 26, 200915 yr by uncool

Or you could simply love Big Brother.

 

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

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

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.