Wednesday, December 21, 2011

Why 1 ≠ 2 (and how divide by zero can work)

So, the argument is posed that:

0 * 1 = 0
0 * 2 = 0


0 * 1 = 0 * 2

So far so good, but according to the rules of algebra, you can add any operation to 1 side as long as you do it to the other side so let's add a division by 0:

(0 * 1)/0 = (0 * 2)/0


0 * (1/0) = 0 * (2/0)


(0/0) * 1 = (0/0) * 2

and through simplification of x/x = 1:
(1) * 1 = (1) * 2

Obviously this is not correct...

As I talked about in my last blog entry, x/0 = {}, the answer to a division by zero is actually a set. Now the question is, what is returned in the set? Is it always -inf to inf? Or is there a relevance factor that has to be taken into account that returns a subset?

Let's say for a moment that 0/0 = {-inf, inf}.

the the above proof actually translates to:

{-inf, inf} * 1 = {-inf, inf} * 2

which is true. All numbers multiplied by 2 results in the same numbers as all numbers multiplied by 1.

But let's try breaking it by dividing both sides by {-inf, inf}:

({-inf, inf}/{-inf, inf}) * 1 = ({-inf, inf}/{-inf, inf}) * 2

Well, now we're back in trouble.

Or are we? If we treat these sets as matrices with an infinite number of values ranging from -inf to inf then we can almost do this! The only problem we face is that the matrices are all in the same dimension. What if we made a rule that said if the matrices are in the same dimension, that one (perhaps the bottom one) should always be transposed? From there we simply invert and multiply.

The result would simply become:
inf * 1 = inf * 2

Which is again true since infinity is as big as you can get. OK, it is said that some infinities are bigger than others, but really? Infinity literally means "boundless" or "without limit". You can get an equation that solves to 2*inf, but like any x*0=0, any x*inf really is just inf.

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