# When does 1+1=3?

#### dougguk

Suspended
Anyone know as I'm stumped?

when your missus is bakin? (not granary)

dougguk said:
Anyone know as I'm stumped?

1 girl+1 guy=1 baby=3 people

1.4 + 1.4 = 2.8

or to 1 significant figure, 1.4 rounds down to 1 and 2.8 rounds up to 3

1 + 1 = 3

god did you have to be all boring and mathematical about it.

Seb said:
god did you have to be all boring and mathematical about it.

Sorry, just come off from a 13 hour night shift loading and unloading a steal laser cutter, all sense of imagination and fun and life got sapped out of me!

Ill try think of soemthing more imaginative...... Nope sorry.

a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1

proof that 1 can equal 2

shimy182 said:
a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1

proof that 1 can equal 2

or that a = 0 ?

and htf u manage to get away with (a-b) ) dividing by zero naughty naughty boy

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Seb said:
god did you have to be all boring and mathematical about it.

His way of putting the answer was far more interesting than yours though mate.

When talking about binary bytes with the first 1 = 1 and the 2nd 1 = 2? thus byte 2 + byte 1 = 3?

For very numbers of 1, e.g 1.5 , a common problem with floating point maths.

Alternatively if you override the '+' function to be be a funtion of arrity 2 equal to the sucessor function of the addition function of the 2 attributes.

e..g. x + y = s(x 'plus' y) = x 'plus' y + 1
such that 1 + 1 = s(1 'plus' 1) = 3, where s is the succesor function.

If you had a different set of numbers with different operators you can do anything you want.
So 1 + 1 = 3 can happen.

..

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shimy182 said:
a = b
a^2 = a*b
a^2-b^2 = a*b-b^2
(a+b)(a-b) = b(a-b)
(a+b) = b
a+a = a
2a = a
2 = 1

proof that 1 can equal 2

Surely this part is wrong:

(a+b)=b
a+a=a

There has to be at least an intermediary step surely?
(a+b)=b
all you've done there is state that b is the same as (a+b)
a+(a+b)=a+b
hmm not sure how you can make any steps from (a+b)=b that puts you in the right direction?

Garp said:

Surely this part is wrong:

(a+b)=b
a+a=a

a=b so he just subbed the a where b is, so this step is correct.

But a=b, so a-b is zero and he divided by zero, thats where it goes wrong

working mod 1 its all equivalent

in large values of 1.

As demonstrated above