Mistake in this proof of $1=2$.

Question

Can anybody can tell Me what is wrong with this process? I don't believe this is true but I couldn't find any mistakes in the flow of equations.

LET $A=B$

THEN : Equation 1.

$A=B$

Multiply both sides by “$A$” :

$A(A)=A(B)$
$A^2 =AB$

Subtract both sides by “$B^2$“

$A^2-B^2=AB-B^2$

Factor both sides

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

Eliminate “$(A-B)$”

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

Equation 2.

$A+B = B$

From Equation 1 .

$A=B$

Substitute Equation 2 by 1

$B+B=B$

THEN

$2B=B$

Divide both sides by $B$:

$2=1$

Answer 1

since $$A = B$$

$$A-B = 0$$

Upon dividing by $A-B$, you are dividing by 0 .

Answer 2

While you subtract both sides by B^2 ,ye resulting answer is zero. You can not cut zeros on both sides. Hence mistake is there. Try it by substituting A=B=1 .You will get to know your mistake.

Answer 3

Here's a nice way to see what you did wrong: choose a value for $A$ and $B$. Let's say $A=B=42$ (after all, the proof seems to work for any $A,B$...)

OK, so start with $42=42$. Sounds reasonable.

Next, multiply both sides by $A$, so by $42$, we get

$$42\cdot 42= 42\cdot 42$$

looking good so far.

Now subtract $B^2$ from both sides:

$$42^2-42^2=42\cdot 42-42^2$$

yup, still just fine.

Factoring both sides, we get

$$(42-42)(42+42)=42(42-42)$$

which is $$0\cdot(42+42)=0\cdot 42$$

still looks ok...

Now, eliminate $(42-42)$...so we need to eliminate $0$ and we get

$$42+42=42$$

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WAIT A SECOND!. That's not true! $42+42$ is not equal to $42$!

So, the mistake must have happened in the last step! It turns out we can't go from $$a\cdot 0=b\cdot 0$$ to $$a=b!$$

And then you remember "oh yeah, of course you can't. Because that would be like dividing by zero..."

Answer 4

Let's start from:

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

(Note: Here you divide by $0$, since $A =B,$ not allowed)

$(A-B)(A+B) - B(A-B) = 0;$

$(A-B)[ (A+B) -B] = 0$, or

$(A-B)(A) = 0;$

A product is zero iff one (or more) of the factors is (are) $0$.

1) $A-B= 0$; You get $A=B$.

2) $A = 0.$

Comments welcome.