×

# 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
3 years, 10 months ago

Sort by:

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. · 3 years, 10 months ago

yeah, i also got it on wikipedia... · 3 years, 10 months ago

a=b means a-b=0, and division by 0 is meaningless. · 1 year, 8 months ago

You just can't cut (a-b) as 0/0 is indeterminate. · 2 years, 5 months ago

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. · 3 years, 10 months ago

dude dats a bit hi-fi!! limit? dats out of range :D · 3 years, 10 months ago

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. · 3 years, 10 months ago

That's not meaningless. That is everything. What times 0 equals 0? is what that's asking. · 3 years 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. · 2 years, 5 months ago

yup.. · 3 years, 10 months ago

×