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Fallacy-explain the point of fallacy

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??

Note by Shubham Sharma
4 years, 2 months ago

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Since 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 · 4 years, 2 months ago

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@Aditya Parson yeah, i also got it on wikipedia... Shubham Sharma · 4 years, 2 months ago

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a=b means a-b=0, and division by 0 is meaningless. Kuldeep Guha Mazumder · 2 years ago

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You just can't cut (a-b) as 0/0 is indeterminate. Vivek Rao · 2 years, 9 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 · 4 years, 2 months ago

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@Clarence Chew dude dats a bit hi-fi!! limit? dats out of range :D Shubham Sharma · 4 years, 2 months ago

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@Shubham Sharma 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. Clarence Chew · 4 years, 2 months ago

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@Clarence Chew That's not meaningless. That is everything. What times 0 equals 0? is what that's asking. Finn Hulse · 3 years, 4 months ago

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@Finn Hulse 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. Clarence Chew · 2 years, 9 months ago

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@Clarence Chew yup.. Shubham Sharma · 4 years, 2 months ago

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