The product of all positive real numbers less than R

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But we're not using a countably infinite set. We're using an uncountably infinite set. Even though the numbers are uncountably infinite, the number of operations are ALSO uncountably infinite, so everything equals out.

1/0 is undefined because the context needs to be specified. In some given case, it might be infinity, but then that infinity would need to be clarified.

For example, consider the question, what is the sine of infinity? This question is as yet meaningless because the infinity is undefined. But if I were to say, multiply 1 by 2 by 3 by 4 and continue this to infinity (integer-infinity factorial) and multiply it by pi, the sine of that infinity would be zero because this number, even though it be infinity, would be evenly divisible by 2.....because the sin of any number evenly divisbly by 2, multiplied by pi is zero.

You still have failed to explain how applying uncountably many binary operations is at all valid.

Uncountably infinite sets have countably infinite subsets. Just saying "everything equals out" is handwaving.


By the definition of division, if k = 1/0 then 0*k = 1. But 0*k = 0. Therefore 0 = 1.

It doesn't matter what we define k to be, if it exists, we get contradictions. Context is irrelevant.

Well, consider the question: What is the SUM of all real numbers less than one? It must be infinity because the resultant sum is greater than the sum of the harmonic series 1 + 1/2 + 1/3 + 1/4.....etc. Similarly for any number less than one. Only this time you're removing a finite set of the first n terms. For example: what is the sum of all real numbers in the interval (0, 1/3)? It is greater than the sum of the harmonic series with the first two terms removed. Still infinity.

This makes sense though we've used an uncountably infinite. Similarly, we can make sense out of the product as well. I've defined "making sense" as ensuring that the density of all the reals is constant throughout the entire interval.

It seems you are essentially trying to define integration. Would you call the integral of f(x) = x over (0, 1) the sum of all numbers between 0 and 1?

There is a topological definition of a dense set, but that doesn't seem to be related to the way you are using the term "density". I think that the term as you are using it is quite meaningless.

The problem is that if you are using "dense" as in "how many real numbers there are in an interval on the real line such as in (A,B)" and you want that to be the same as in every other interval (C,D) for C<D, then that is always true no matter the size of the interval. Always.

For example, there are as many real numbers in the interval (0,1) as in (1,2) which seems obvious. But there are also as many real numbers in (0,1) as in (0,2) which is not as obvious. Likewise, there are equally many number ins (.00000000001,.000000000010000000001) as in (0,10000000000000000000). Every open subinterval of the open interval has the same number of points as the interval itself.

Density, in the way I define it in this case: If x factors are being multiplied within the sub-interval (0,S),whose - that is, the sub-interval's - "magnitude" will be denoted as "S", within the interval (0, R), then x factors are also being multiplied within any arbitrary sub-interval of the magnitude S within (0,R).

This is equivalent to the function defined earlier in this thread. It is, I will concede, an arbitrary stipulation, but it's one that makes sense to me because it involves using R/2 as an axis of arithmetic symmetry. We could use an axis of geometric symmetry, if we wanted to, but then things get more difficult to define and control.

Yep. I understand you there.

But I guess, you're right: I really haven't proven much. Other than that the product of x amount of numbers whose series progress arithmetically throughout an interval approaches 0 as x approaches infinity. The series of the numbers don't have to progress arithmetically, by any means.