Dr. Bogus, a close friend of ours, during a long session at the Dixie Chicken re

Question

Dr. Bogus, a close friend of ours, during a long session at the Dixie Chicken restaurant, was doodling on a napkin and "proved" that all numbers are equal. This came as quite a surprise to us, and we have delayed his calling the president only until you can review his proof. Here is a translation of the napkin doodles: Pick two different numbers, a and b, and a nonzero number c that is the difference between a and b, thus: a= b + c c notequalto 0 Multiply both sides by a - b a(a -b) = (b + c)(a - b) or a^2 - ab = ab - b^2 + ac - bc Subtract ac from both sides, a^2 - ab - ac = ab - b^2 - bc Factor out a from the left side of the equation, and b from the right, a(a - b - c) = b(a - b - c) Eliminate the common factor from both sides, and a = b Because a and b can be any number, c can be positive or negative; thus, all numbers are equal to each other. Exact which step(s) are wrong in the proof above, and why?

Explanation / Answer

step 5 of cancelling out common factor is incorrect

as it leads to a=b .We have already assumed in the question that

a+b = c

if we make a=b then c=0 which is not true

So, we should do the steps like as given belwo:

a( a-b -c) = b (a-b -c)

subtract both sides with b (a-b -c)

a( a-b -c) - b (a-b -c) =0

(a- b -c)(a-b) =0

So, a - b -c =0

a = b+c

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