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Posted

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 :)

Posted

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...

  • 2 months later...
Posted
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]

  • 3 weeks later...
Posted

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.

  • 3 weeks later...
  • 4 months later...
Posted (edited)

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 by alan2here
Posted

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.

  • 3 months later...
Posted

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

  • 4 weeks later...
Posted

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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