Wherein 2 is shown to be equal to 1

Here’s a little maths trick I have for some reason carried around in my brain for years:

\text{Let } a = b
\Rightarrow a^2 = ab
\Rightarrow a^2 - b^2 = ab - b^2
\Rightarrow (a + b)(a - b) = b(a-b)
Dividing through by the common (a - b)
\Rightarrow (a + b) = b
\text{But } a = b
\Rightarrow (b + b) = b
\Rightarrow 2b = b
\Rightarrow 2 = 1

Why this is possible is left as an excercise for the reader 🙂

4 Comments

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4 Responses to Wherein 2 is shown to be equal to 1

  1. grecken

    Line 3: (a^2)-(b^2) = ab-(b^2)

    How do you get:
    (a+b)(a-b) = b(a-b)

    From line 3? If you divide ‘b’ from the right side of the equation, you must do the same on the left, giving you:

    ( (a^2)-(b^2) /b) = b(a-b)
    simplified: (a^2)/b) – b = a – b
    Eg replacing with #’s
    4/2 – 2 = 2 – 2

    Was the answer faulty math? XD

    • rowan

      Had to check wikipedia for this one

      http://en.wikipedia.org/wiki/Mathematical_fallacy

      very interesting though

    • kim

      Hi,

      on line 3, I rewrote both sides of the equation, but didn’t actually perform any calculations.
      On the left, (a^2)-(b^2) can be written as (a+b)(a-b), since if you multiply that out you get (a^2) -ab + ab -(b^2). -ab + ab is 0, so you get (a^2)-(b^2)
      On the right, ab-(b^2) has b common to both ab and b^2, so the b can be factored out and the line written as b(a-b). If you multiply b(a-b) out you get ab – (b^2).

      In both cases the lines are rewritten in a different form without any calculation being done.

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