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Why 1=2... (1 Viewer)

Kmahal1990

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Let A be numerically equal to B, and A = 1.

A=B
AB=B^2
AB-A^2=B^2-A^2
A(B-A)=(B-A)(B+A)
A=A+B
Sub A=1,B=1
1=2

Or
A-A=B
B=0
B= 1
1=0

What went wrong there? :S
 
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P

pLuvia

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Well from this line
(A+B)^2 - 2AB=B(A+B)
to this line
A+B-2AB=B
How did you cancel out the (A+B) from both sides?

There's your problem
 

Bank$

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pLuvia said:
Well from this line
(A+B)^2 - 2AB=B(A+B)
to this line
A+B-2AB=B
How did you cancel out the (A+B) from both sides?

There's your problem
yeah u cannot do that cause u should have 2AB/(A+B)
 

drynxz

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A(B-A)=(B-A)(B+A)
A=A+B

those two lines...if b=a then b-a will equal 0...u cannot divide through zero.
 

victorheaven

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u said A=B
how can u divide both sides by A-B?
two sides are identically equal..
 

morganforrest

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drynxz said:
A(B-A)=(B-A)(B+A)
A=A+B

those two lines...if b=a then b-a will equal 0...u cannot divide through zero.
This is the correct answer as if A=B=1 then A - B = 0 and you can't divide by zero or Euler or someone will come and kick ur dog for being a dumbass....

QED
 

SeDaTeD

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Or he could be working over the ring containing just {0}.
 

Scooter89

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yeah you cant divide by zero. consider another simplistic method with the same fault:

X^2 - X^2 = X^2 - X^2

therefore X(X-X) = (X+X)(X-X)

dividing by (X-x), ie dividing by zero, we have..


X = X + X

dividing by X, we have

1 = 1 + 1

ie

1 = 2... see?
 

Slidey

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pLuvia should make a sticky thread about why 1 doesn't equal 0 lol :p

Here's a question: Is it possible that if AB=0, neither A nor B = 0? If not in normal algebra, is there an algebra in which it is possible?
 
Last edited:

LoneShadow

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Slidey said:
pLuvia should make a sticky thread about why 1 doesn't equal 0 lol :p

Here's a question: Is it possible that if AB=0, neither A nor B = 0? If not in normal algebra, is there an algebra in which it is possible?
Any ring which is not an integral domain will have that property. Called zero devisors. for example consider the ring of integers modulo 4: 2*2 = 4 = 0 ; but 2 is not zero.

...it basically depends on the how you define your operations. For all I care you could define the multiplication to be trivial: i.e. a*b = 0 for all a,b.

P.S. Don't confuse high school kids with ring theory Robert :p
 

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