# 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 🙂

Filed under geekery

### 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

Ah-hah – so that’s what they call it 🙂

• 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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