The following example uses division by zero to "prove" that 2 = 1, but can be modified to prove that any number equals any other number.1. Let a and b be equal non-zero quantitiesA = B2. Multiply through by aA² = AB3. SubtractA² - B²4. Factor both sides(A - B)(A + B) = B(A - B)5. Divide out (A - B)A + B = B6. Observing thatB + B = B7. Combine like terms on the left2B = B8. Divide by the non-zero b2 = 1The fallacy is in line 5: the progression from line 4 to line 5 involves division by a − b, which is zero since a equals b. Since division by zero is undefined, the argument is invalid. Deriving that the only possible solution for lines 5, 6, and 7, namely that a = b = 0, this flaw is evident again in line 7, where one must divide by b (0) in order to produce the fallacy (not to mention that the only possible solution denies the original premise that a and b are nonzero). A similar invalid proof would be to say that since 2 × 0 = 1 × 0 (which is true), one can divide by zero to obtain 2 = 1. An obvious modification "proves" that any two real numbers are equal.Many variants of this fallacy exist. For instance, it is possible to attempt to "repair" the proof by supposing that a and b have a definite nonzero value to begin with, for instance, at the outset one can suppose that a and b are both equal to one:A = B = 1However, as already noted the step in line 5, when the equation is divided by a − b, is still division by zero. As division by zero is undefined, the argument is invalid.

math is some pretty cool stuff to be honest. have you ever seen minute physics on youtube? his videos are great!

EDIT: I tried to say that the fail is at line 3, not line 5.So basically A²=AB --> A²-B² ????I got lost, how in hell Apples² = Apples(pears) becomes Apples²-Pears² ?????the correct way to solve A²=AB would be in case A²/AB=0 which actually is A/B=0I believe your above statment it's kinda weird

From the GIF you posted here:"this image can't be displayed for the momment, please go to www.example.com to see it."

A²=AB will never be AB=ABNot in real/normal math.Maybe: A²/A=B which could be:A=B But that brings you back to the start

Which will then lead you back to the first equation, A = B.But copherfield has a point. I also don't understand step 2 ~ 3.A² = ABSubtract A² - B² From A² = AB.It'll give you-B² = A² - B² + AB, then0 = A² + AB, then0 = A(A +B),then,