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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 $B$:

$2=1$

MJD
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user474103
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    You can't eliminate $A-B$ because $A-B=0$, i.e., you can't divide by zero. – Carlos Jiménez Sep 05 '17 at 07:27
  • You're dividing by zero. – bloomers Sep 05 '17 at 07:28
  • DividedByzero Exception :) – GAVD Sep 05 '17 at 07:33
  • ++ Out of Cheese Error. Redo From Start. ++ – Joffan Sep 05 '17 at 07:45
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    Here's another quick mistery: \begin{eqnarray}1=\lim_{n\rightarrow \infty}1&=&\lim_{n\rightarrow \infty}(\frac{1}{n}+\frac{1}{n}+\dots +\frac{1}{n})\ &=& \lim_{n\rightarrow \infty}\frac{1}{n}+\lim_{n\rightarrow \infty}\frac{1}{n}+\dots +\lim_{n\rightarrow \infty}\frac{1}{n}\ &=& 0+0+\dots +0\ &=&0.\ \end{eqnarray} – Mathematician 42 Sep 05 '17 at 07:54
  • This false proof hides a division by zero. Division by zero as no defined result otherwise $ 4 \times 0 = 2 \times 0 \Rightarrow 4 = 2 $ by cancelling out the zeros from both sides. – Warren Hill Sep 05 '17 at 08:01
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    @Mathematician42 Nice one. Had to do a double take to find the mistake :) – 5xum Sep 05 '17 at 08:09
  • @5xum what is the mistake :( ? – Joao Noch Sep 05 '17 at 08:26
  • @JoaoNoch Basically, there's a hidden "$n$" variable in the expression - the number of sumands is $n$ in the first line, and.. well... undetermined in the second line. – 5xum Sep 05 '17 at 08:30
  • @JoaoNoch Indeed: $$\lim_{n\rightarrow \infty}(\frac{1}{n}+\frac{1}{n}+\dots +\frac{1}{n})=\lim_{n\rightarrow\infty}(n\cdot \frac{1}{n})\neq \sum_{i=1}^n\lim_{n\rightarrow \infty}\frac{1}{n}.$$ You can't pull out a variable to an undetermined sum, writing it as above makes this very clear. In fact, the last expression doesn't make any sense and even less sense if you stare longer at it. In other words, I fooled you by using extremely bad notation. This is something any mathematician should be wary of. – Mathematician 42 Sep 05 '17 at 08:42
  • @Mathematician42 Thank you. I see you performed an undefined number of sums to get a sneaky result. Nice one, I'll recognise that mistake in the future. – Joao Noch Sep 05 '17 at 08:48

4 Answers4

3

since $$A = B$$

$$A-B = 0$$

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

Ong Hai Xiang
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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

$$42+42=42$$

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..."

5xum
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0

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.

0

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.

Peter Szilas
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