Crowley wrote:Why does the equation seem to work out so well? - Because You make a mistake by dividing by infinity.
I don't know how to write the infinity symbol here, so I use "i" instead. Let's say I have a number "a" which is neither zero nor infinity. Your assumption is that:
a / i = 0 (for all (a != 0 and a != i))
This is wrong. It is just as wrong as
a / 0 = i
would be. Tempting, but wrong. There is a trivial proof to it. Take the equation
a / i = 0
and multiply it with i:
a = 0 * i
0 and i are constants, but a is a variable. Remember? a stands for every number that is neither 0 nor i! So a could be 1, 2, 3 or -2.9873 . How could two constants mulitplied with each other result in a variable? So now: What's the result of zero mutliplied with infinity?
Changing a / i = 0 to a = 0 * i is like doing this:
a / i = 0
a * i / i = 0 * i
a * 1 = 0 * i
a = 0 * i
Isn't i / i an indeterminate form? So it shouldn't be treated like the number one. Zero times infinity is also an indeterminate form.
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Better ways of writing the equations a / i = 0 and a / 0 = i would be in terms of limits like this:
lim x->i a/x = 0
lim x->0 a/x = i
When I was in precalculus half a decade ago, the teacher had us write the = symbol in equations like the two above as -> so we'd understand that the limit is approached by the function as the variable, like the x in the examples, approached a number or infinity.