Why the hell is [math]\infty \cdot 0[/math] undefined? wolframalpha.com
If you have the set of natural numbers [math]N = \{1, 2, 3, ...\}[/math]
and the set of natural numbers that are negative [math]M = \{n: n \in N, n < 0\}[/math]
...then we know that [math]|M| = 0 \cdot |N|[/math], since [math]M[/math] is an empty subset of [math]N[/math]. But according to the WolframAlpha logic, this would imply that [math]|M|[/math] is undefined, which in turn implies that 0 is undefined.
Other urls found in this thread:
en.wikipedia.org
twitter.com
Because it's equal to 1 they just haven't proven it.
You can't multiply something by infinity since it's not a number. Yes, |N| is infinity (aleph0 rather) and |M| = 0, but doesn't mean 0*infinity is defined.
Infinity isn't a number in the standard complex number system. One good reason may be that we want the field axioms to hold
this
the term infinity is vague because it is simply defined as non-finite, this includes aleph0, aleph1, ... etc. here you defined infinity as aleph0 and yes aleph0 * 0 = 0 but that won't work with other infinites.
consider the following example:
>A point has 0 length
>There are infinite points in a line
>If we divided a line evenly into some number of pieces then the length of that line would be = (the length of each piece)*(the number of pieces)
If 0*infinity=0 then that would imply all line must have 0 length, however that is not the case, 0*infinity is not always 0.
Q.E.D.
It depends which infinity you use to get a proper answer.
Well if you multiply the size of all natural numbers times 0 you seem to get 0.
How about the size of all complex numbers times 0?
>Halmos. Measure Theory.
I don't get it either.
>If we divided a line evenly into some number of pieces then the length of that line would be = (the length of each piece)*(the number of pieces)
You cannot do that.
In measure theory the additivity is countable.
As you can see in
[math]\pm \infty \cdot 0[/math] is defined.
>and the set of natural numbers that are negative
There are no negative natural numbers. Either [math]n \in \mathbb{N}:-n
Can anyone provide a working example on why [math]\pm \infty \cdot 0[/math] is generally considered to be undefined?
Sorry, I did not read the whole message.
Because [math]0 \cdot \infty[/math] can literally be any positive real number you want depending on how you approach the limit.
[eqn]\lim_{x\to 0} \sin(x) \cdot \frac{\alpha}{x} = ``0 \cdot \infty" = \alpha[/eqn]
[eqn]\lim_{x\to \infty} \exp(-x) \cdot x = ``0 \cdot \infty" = 0[/eqn]
[eqn]\lim_{x\to \infty} \frac{1}{x} \cdot \exp(x) = ``0 \cdot \infty" = \infty[/eqn]
Thanks. How come this is not a problem in measure theory?
Why are you talking of measure theory? You obviously didn't finish high school.
Because of the two posts above.
post more gookers and i will tell you
mildly savage
The same reason the product of 1 divided by 0 is not equal to zero.
0 = (1-1)
x · (1-1) = 1x - 1x = 0
1 · (1 / (1-1) = ?
