Sophism #3: Every Number Can Sometimes Be Equal to Zero

Let a, b, and c be arbitrary numbers, such that

a = b + c

Multiply both sides of the equation by (a-b) and open the brackets:

a(a-b) = (b+c)(a-b)
a² - ab = ab + ac - b² - bc

Move ac to the left and extract:

a² - ab - ac = ab - b² - bc
a(a-b-c) = b(a-b-c)

Now divide both sides by (a-b-c) to obtain

a=b

Combining this with the first assumption a = b+c, we obtain that c, a number that was chosen arbitrarily, must necessarily be equal to zero.

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Please, avoid posting spoilers in the comments. For other sophisms check out this list.