1+1=2

For 2 non zero numbers x and y such that:
x = y

mutiply by x gives:
x² = xy

Subtract y² from both sides:
x² - y² = xy - y²

which can be written as:
(x+y)(x-y)=y(x-y)

Dividing by (x-y) gives:
x + y = y

Since x = y, we see that
2y = y

Divide by y gives:
2=1

Therefore:
1+1=1
 
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what if the first 1 was upright and the additional 1 was upside down would the upside down 1 count as a true 1 and therefore be counted into the equation ?
 
If you asked this question at any social event where, say, accountants, engineers, psychotherapists and mathematicians lurk, you may be surprised at the range of answers,

The accountant may ask you if you need to write it off against your tax liability, an engineer may say its two, plus or minus a little bit depending on the tolerances allowed for, the therapist may ask 'just how do you feel about the need for an answer, and the mathematician will say "excuse me there's someone over there i really need to talk to. Now."

And the barman, will tell you that one lemon plus one orange make a St. Clements.

Just my two old pennies worth and thanks for reading this far.

DH
 
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For 2 non zero numbers x and y such that:
x = y

mutiply by x gives:
x² = xy

Subtract y² from both sides:
x² - y² = xy - y²

which can be written as:
(x+y)(x-y)=y(x-y)

Dividing by (x-y) gives:
x + y = y

Since x = y, we see that
2y = y

Divide by y gives:
2=1

Therefore:
1+1=1

This is classic....
 
Wow the one person I tought would bite hasn't. Perhaps everyone else being humourously contrairy have unwittingly used up his arguments before he got to them.
 
is this bit right??


(x+y)(x-y)=y(x-y)

Dividing by (x-y) gives:
x + y = y
 
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