Thursday, April 1, 2010

revealing the algebraic falacy, 2 != 1

Previously I used only basic algebraic manipulation I showed that 2 = 1, which is also false, now I am going to show where it went wrong, by doing the same steps as before, using the same forms as before.

a = b

Substitute b for a in the entire rest of the article, since they are the same.

b = b

we can do whatever we want to an equation as long as we do the same thing to both sides, so first add b

b + b = b + b

partially simplify

2b = b + b

subtract 2b

2b - 2b = b + b - 2b

partially group like terms

2b - 2b = b + (1-2)b


2b - 2b = b + (-1)b

adding a negative number is the same as subtracting

2b - 2b = b - b

factor out (b - b)

2(b - b) = 1(b - b)

divide by (b - b), which is also known as zero.

\frac{2(b - b)}{b - b} = \frac{1(b - b)}{b - b}

re-factor to isolate the one, now we have zero over zero, anything can happen

2\left(\frac{b - b}{b-b}\right) = 1\left(\frac{b - b}{b-b}\right)

reduce the one, and we are just not allowed to get here

2(1) = 1(1)


2 = 1

Division by zero is not allowed, and zero over zero is just undefined.

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