How Can This Be True?

[dimon]

What do you think about this? Do you see anything wrong with this?

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 = 0 ?

WOW!

 

[JakeD]

What do you think about this? Do you see anything wrong with this?

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 = 0 ?

WOW!

If A=B then A-B = 0 and you've divided by zero in the third to last step, which is not a valid operation.

Put another way, if X = 0, then 1X = 2X is true, but 1 = 2 does not follow from dividing by X. That's why dividing by zero is not a valid operation: there is no way to define it without getting nonsense.

 

[soroban]

Hello, dimon!

Here's a variation with a different punchline . . .


(1) We have: . . . . . . . . . . $\displaystyle A\:=\:B$

(2) Multiply by $\displaystyle A:$ . . . . . . .$\displaystyle A^2\:=\:AB$

(3) Subtract $\displaystyle B^2:\;\;\;\;A^2\,-\,B^2\:=\:AB\,-\,B^2$

(4) Factor: $\displaystyle \;\,(A\,+\,B)(A\,-\,B) \:= \:B(A\,-\,B)$

(5) Divide by $\displaystyle (A-B):\;\;A\,+\,B\: = \:B$

(6) In (1) we had: $\displaystyle A\,=\,B$
$\displaystyle \;\;\;$ so (5) becomes: $\displaystyle \;A\,+\,A\:=\:A$

(7) Simplify: . . . . . . . . . $\displaystyle 2A\:=\:A$

(8) Divide by $\displaystyle A$: . . . . . . . .2 = 1 . . . ta-DAA!

 

[dimon]

that's even better

 

[Denis]

that's even better

dimon, does that mean if I give you a buck, you'll give me two :?: :roll:

 

[dimon]

Sure Denis,
you have got yourself a deal. If, you can ever prove that deviding by zero is a valid operation.

 

[Denis]

Sure Denis,
you have got yourself a deal. If, you can ever prove that deviding by zero is a valid operation.

3 / 0 = 3

"/ 0" means you're not dividing nuttin'

so you erase "/ 0", and you're left with 3;
if you have 3 beers and divide them up with nobody, then you still have 3 beers...

You owe me 2 bucks :idea:

 

[dimon]

So then dividing 3 "beers" by 0 we would get x such that x * 0 = 3, not possible.
try again