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# Bernoulli's Sophism

Complex Analysis. FreeTutorial

This sophism was constructed by Swiss mathematician John Bernoulli (1667 - 1748), who was one of the eight outstanding mathematicians in the Bernoulli family.

Find a mistake in the following chain of arguments, pretending to prove that

Ln(-z) = Ln(z) for any .

 "Proof:" 1. Ln[(-z)2] = Ln(z2); 2. Ln(-z) + Ln(-z) = Ln(z) + Ln(z); 3. 2Ln(-z) = 2Ln(z); 4. Ln(-z) = Ln(z). Where is the mistake?

An explanation:

The conclusion that Ln(-z) = Ln(z) is false, because

Ln(z) = ln(|z|) + i[arg(z) +2k], k = 0, ±1, ±2, ... ,

Ln(-z) = ln(|z|) + i[arg(z) +(2k+1)], k = 0, ±1, ±2, ... ,

and none of the numbers representing the value of Ln(z) is the same as any of the numbers representing Ln(-z).

The error occurs in going from line 2 to line 3 because

Ln(-z) + Ln(-z) 2Ln(-z),

Ln(z) + Ln(z) 2Ln(z).

The following example elucidates the situation:

Let A be the set of two numbers 3 and 4.

A set B=A+A is the set of numbers 6; 7; 8 because 3+3=6, 3+4=7 and 4+4 =8.

A set C=2·A is the set of numbers 9; 12; 16 because 3·3=9; 3·4=12 and 4·4=16.

So, a set A+A2·A

Ln(-z) + Ln(-z) 2Ln(-z),

Ln(z) + Ln(z) 2Ln(z).

by Tetyana Butler

Reference

A.I. Markushevich, "Theory of functions of a complex variable" , 1–2 , Chelsea (1977) (Translated from Russian)

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