When does 1+1=3?

[dougguk]

Anyone know as I'm stumped?

 

[Seb]

when your missus is bakin? (not granary)

 

[Zip]

Anyone know as I'm stumped?

1 girl+1 guy=1 baby=3 people

 

[Seb]

1 girl+1 guy=1 baby=3 people

Thats what I said!

 

[jezsoup]

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

 

[Seb]

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

 

[jezsoup]

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.

 

[shimy182]

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

 

[spirit]

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

 

[daveyj27]

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.

 

[Ukadder]

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

 

[kaiowas]

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

#DIV/0!

 

[D.P.]

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.

 

[LeperousDust]

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

 

[Von Luck]

..

 

[Garp]

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

WTF? That doesn't read right.

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?

 

[carvegio]

WTF? That doesn't read right.

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

 

[carvegio]

working mod 1 its all equivalent

 

[kitten_caboodle]

in large values of 1.

As demonstrated above

 

[Psyk]

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

Well that would be just like saying 1 + 2 = 12 in decimal. It's to do with the representation of the numbers rather than the numbers themselves.