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Q:prove $1=2$ ?

method1:-

let us consider $x=1$

then $x=x^2$

$x-1 =(x^2)-1$

$x-1=(x-1)(x+1)$

$1=x+1$

finally $1=2$

i have little confusion here, is the mistake is considering $x=1$ or else any thing other than this.

method2:-

$-2=-2$

$1-3=4-6$

$(1^2)-(\frac{2*1*3}{2})=(2^2)-(\frac{2*2*3}{2})$

$(1^2)-(\frac{2*1*3}/2)+((\frac{3}{2})^2)=(2^2)-(\frac{2*2*3}{2})+((\frac{3}{2})^2)$

$(1-\frac{3}{2})^2=(2-\frac{3}{2})^2$

$1-\frac{3}{2}=2-\frac{3}{2}$ [here the formula if $x^n=y^n$ then $x=y$ is failed]

$1=2$

point out the mistakes in above two methods exactly and please explain

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closed as not constructive by AWertheim, N. S., lab bhattacharjee, Amzoti, Micah Jun 13 '13 at 4:58

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possible duplicate of What are some classic fallacious proofs? –  lab bhattacharjee Jun 13 '13 at 4:36
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We get so many of this question... –  Sujaan Kunalan Jun 13 '13 at 4:42

5 Answers 5

up vote 3 down vote accepted

For the first: Don't divide by zero. (That's what you did going from $x-1=(x-1)(x+1)$ to $1=x+1$, since $x=1$.)

For the second: $x^2=y^2$ does not imply that $x=y$. After all, $(-1)^2=1=1^2,$ but $-1\ne 1$. The best we can conclude is that $x=\pm y$.

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thank you understood –  Gangadhar Jun 13 '13 at 4:53

The step $$x - 1 = (x - 1)(x + 1) \implies 1 = x + 1$$

is not valid since $x - 1 = 0$, and division by zero is not allowed. Also,

$$ -\frac{1}{2} = 1 - \frac{3}{2} \ne 2 - \frac{3}{2} = \frac{1}{2}$$

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In the first one you divided by $x-1$, which is illegal cuz $x-1=0$. I don't know what you're doing in the second one but I suspect you also divided by 0 there too lol.

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You have divided by $0$, which is not allowed. (During the step when you divided $x-1$ from both sides.)

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thank you understood –  Gangadhar Jun 13 '13 at 5:54

$(1-3/2)^2 = (2-3/2)^2$ does not imply $(1-3/2) = (2-3/2)$. It only implies $(1-3/2) = \pm (2-3/2).$

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