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??

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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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. – Aditya Parson · 3 years, 8 months ago

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– Shubham Sharma · 3 years, 8 months ago

yeah, i also got it on wikipedia...Log in to reply

a=b means a-b=0, and division by 0 is meaningless. – Kuldeep Guha Mazumder · 1 year, 6 months ago

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You just can't cut (a-b) as 0/0 is indeterminate. – Vivek Rao · 2 years, 3 months ago

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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. – Clarence Chew · 3 years, 8 months ago

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– Shubham Sharma · 3 years, 8 months ago

dude dats a bit hi-fi!! limit? dats out of range :DLog in to reply

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. – Clarence Chew · 3 years, 8 months ago

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– Finn Hulse · 2 years, 10 months ago

That's not meaningless. That is everything. What times 0 equals 0? is what that's asking.Log in to reply

– Clarence Chew · 2 years, 3 months ago

I agree that \(\frac{0}{0}\) can be any number, but the fact that it reduces to \(\frac{0}{0}\) says nothing more about the value an expression is unless we know that the \(0\)s mean.Log in to reply

– Shubham Sharma · 3 years, 8 months ago

yup..Log in to reply