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It may sometimes be easier to notice that the coefficient is zero if we aim to multiply by its reciprocal instead of directly dividing by the coefficient. In the first false proof example from the referenced math humor page:Not really. If the coefficient is zero, then its reciprocal is undefined.

Statement '5.' is vacuously true, but given statement '1.' as true, statement '6.' cannot be true. The false proof works partly by obscuring the zero divisor in the bilateral division operation employed to justify the transition from statement '5.' to statement '6.'. I think it is harder to fail to notice that the divisor is zero if the division operation is seen as multiplying by ##1/(a-b-1)## instead of as simply canceling the ##(a-b-1)##, whatever that quantity might be, from both sides.

Given that a and b are integers such that a = b + 1, Prove: 1 = 0 1. a = b + 1 1. Given 2. (a-b)a = (a-b)(b+1) 2. Multiplication Prop. of = 3. a ^{2}- ab = ab + a - b^{2}- b3. Distributive Property 4. a ^{2}- ab -a = ab + a -a - b^{2}- b4. Subtraction Prop. of = 5. a(a - b - 1) = b(a - b - 1) 5. Distributive Property 6. a = b 6. Division Property of = 7. b + 1 = b 7. Transitive Property of = (Steps 1, 7) 8. Therefore, 1 = 0 8. Subtraction Prop. of =

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