A problem posed by T. Hewitt, London:
Ye learned in numbers, how will you contrive
To prove that Nothing is equal to five?
Solution: Find the error's in the Hewitt's proposed answer: “Put a = 1 and n = 5+1 = 6, that is one more than the given number, then the following work will clear this matter: (1-an)/(1-a) = 1 = a + a2 + a3 + a4 + &c. to an-1 = (1-a6)/(1-a) = 1 + a + a2 + a3 + a4 = (1-16)/(1-1) + (1-1)/(1-1) = 1 only because the numerator and denominator are equal. Then will a + a2 + a3 + a4 + a5 = 5 be also = 0.” Or, is there no error in this response?
Source: The Gentleman’s Mathematical Companion, 1797