Playing with mathematics

We want to demonstrate that all numbers are equal each other

Demonstration

Let’s assume that a and b are both real numbers, different from zero.

The mean value is c= (a+b) / 2

Then a + b = 2 c

I can multiply both members of the equation by (a – b)

[1]  (a + b) (a – b) = 2 c (a – b)

I can follow with further algebraic passages:

[2] (a^2 – b^2) = 2 c (a – b)

[3] a^2 – 2ac = b^2 – 2bc

I can add c^2 to both members

[4]  a^2 – 2ac + c^2 = b^2 – 2bc + c^2

[5]   (a – c)^2 = (b – c)^2

[6]  (a – c) = (b – c)

[7]   a = b

sicuti demonstrandum erat

This is a “mathematic sophisma” that I have found on the “Dizionario Enciclopedico Italiano (Treccani)” edition year 1960 vol XI page 406, similar sophismas can be found on the web.

But……where’s the weakness?

Answer (from the same source)

From equation [5] does not follow necessarily equation [6], the other solution is

[6bis] (a – c) = – (b – c)

[7bis) = a + b = 2c

 

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