Let a=b so a^2 = ab or a^2-b^2 = ab-b^2 or (a+b)(a-b) = b(a-b) or a+b = b

now, putting the value of a, b+b=b or 2b=b or 2=1 :P But how??

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## Comments

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TopNewestSince a-b=0, in the 4th step you have divided 0 by 0. In mathematics division by zero is meaningless.

Fallacy is an argument built on the base of poor reasoning. Your conclusion might be true or maybe not but it is still fallacious.

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yeah, i also got it on wikipedia...

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Instead of a=b, let's let a-b=k

\(a=b+k\)

\(a^2 = ab+ak\)

\(a^2-b^2 = ab+ak-b^2\)

\((a+b)(a-b) = b(a-b)+ak\)

\(a+b=b+\frac{ak}{a-b}\)

You are saying that \(lim_{x \rightarrow 0} \frac{a \times x}{x} = 0\), which causes this fallacy.

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dude dats a bit hi-fi!! limit? dats out of range :D

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We are seeing what happens if b is close to a, to determine what went wrong.

Likewise, you cannot evaluate \(\frac{x-4}{\sqrt{x}-2}\) where x=4 directly by substitution, although it is possible to evaluate it. By substitution, we get \(\frac{0}{0}\) which is kind of meaningless.

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You just can't cut (a-b) as 0/0 is indeterminate.

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a=b means a-b=0, and division by 0 is meaningless.

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