The product of all positive real numbers less than R

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Is there some number R such that every real number greater than 0 but equal to or less than R has a product other than 0 or infinity? Has someone already answered this?

I found it's easy to show that if R is less than or equal to 2, the product converges to Zero.

A way to find an answer is to take increasing values of Q for the following equation: (R * 10^Q)! * 10^(-(R * 10^Q) * Q). If the result of this equation converges to some value with arbitrarily large values of Q, then that value is the aforementioned product.

It seems that the highest number R such that this infinite product is still finite is around 2.655 or so. The resultant product seems to be about that number (or possibly e). If this is true, then it would be easy to show that any value for R higher than that number would diverge to infinity. Possibly, every number lower than that would converge to Zero, but I'm not sure.

Any thoughts?

I don't think that it makes sense to try to multiple all the real numbers in an interval.

Multiplication involves a countable number of binary operations. It doesn't logically follow that trying to extend that to an uncountable number of binary operations would be at all possible.

But math has historically always found a way to define operations that don't make sense. For example, who would have thought taking a number to the power of a complex exponent makes sense, and yet Euler performed wonders with that. Actually, taking a number to the power of any irrational or transfinite number doesn't make sense to me for that matter, because taking any number N^(X/Y) should return Y possible answers where X/Y does not equal 1.



But, actually, I think what I said above was wrong. After trying that function for different numbers, I found that even though the product seemed to converge to some non-zero value, I would try an even higher value for Q and found that it suddenly ran away to zero. There seems to be some highest value for each R at which the function has a zero derivative and then the graph drastically downturns asymptotically to zero. If this is the case, then the product of every positive real number greater than 0 but less than R, for ANY value of R would equal Zero, which is wonderfully counter-intuitive.

When talking about uncountable numbers, there can be many things that are counter-intuitive.

For example, intuitively, one would expect that the line segment (0,10) would have more points in it than the line segment (0,1), but one would be wrong.

Back to the original question, consider (0,infinity). Think about it broken into three sets, (0,1), {1}, and (1,infinity).

Pair every number x in (1,infinity) with 1/x in (0,1). Multiplying the x by 1/x in each pair, you clearly have a value of 1. Thus, to multiply all the numbers in (0,infinity) by themselves, you can reduce that to multiplying 1 by itself uncountably many times. So if multiplying uncountably many numbers together were to be legitimate, it seems logical that the result would be the value 1.

Now consider the partition of (0,1), {1}, (1,2), {2}, and (2,infinity). For each x in (0,1), multiply by the point 2/x in (2,infinity). You would have uncountably many 2s. So you would have 1 * 2 * {uncountably many 2's) * (the product of the points in (1,2)) which, if legitimate, would seem to result in a product of infinity.

Let R > 1. Given a large N, we can find a finite set of real numbers r1, r2, in [1, R) such that there product is larger than N. In short, your proposed product is not well defined. for R < 0 the product of numbers in the interval (0,R) can be made arbitrarily small so the product is 0.

ruveyn

The equation (R * 10^Q)! * 10^(-Q * (R * 10^Q)) - is that what I typed before? I might have messed up on my previous post - this equation is an expression of the stipulation that the density of terms be constant throughout the entire interval (0, R).

For example, let's make R = 2.65 and Q = 2. Then this equation becomes 265! * 10^(-530). The result would then be the same as 2.65 * 2.64 * 2.63 *.....*.01. As you can see, where the terms are ordered in decreasing value in this way, any two given consecutive terms differ by .01 and therefore the density of numbers is constant throughout the entire interval.

Were we to take the product 1/(abcde....) * abcde...., each term on the right side would differ by the same value from its adjacent term if, for example, a,b,c,d,e wer consecutive integer values each differing from its adjacent value by 1, whereas each term on the left side: 1/a, 1/b, 1/c, ...etc. would differ by its adjacent term by varying values.

For example, in 1/(1 * 2 * 3 * 4 * 5) * 1 * 2 * 3 * 4 * 5, the terms on the right side each differ by one, but on the left side, the differences of any two adjacent terms vary from .5 to .05. Similar results would follow even if a, b, c, d, and e were not well-ordered: the density on the left side would not equal the density on the right side, so the density wouldn't be constant throughout the entire interval (0, e).

Where Q approaches infinity, the principle is the same, any two adjacent terms where the terms are ordered in value, differ by the same value; that value just happens to be infinitesimal. So the density is still constant.

Remember that the set of reals includes both the rational numbers and the irrational numbers.

It looks to me like you are only considering the rational numbers.

It would still follow for the irrationals/transfinites. as per the extension of the factorial via the gamma function.

I was thinking about this last night. I think I can prove that the product of all real numbers greater than 0 but less than or equal to R will be zero for any value of R. For example, consider the following product: (2^8)/((8^8)/8!). This is the same as .25 * .5 * .75 * 1 * 1.25 * 1.5 * 1.75 * 2. If we input 16 instead of 8, we get a similar sequence, but one in which each term is separated by .125. We can increase this arbitrarily to any power of 2 (by the way, the "2" in the numerator is "R" in this case - it could be any positive real number whatsoever). If we take the product in the denominator OF the denominator and put it in the top-most numerator (the same product would result), we would get, for the first example, 2 * 4 * 6 * 8 *......*16. This is the same as taking the geometric mean of (2^8) * 16! and taking it to the power of 8. But because the geometric mean of that sequence will always be less than the geometric mean of 8^8 (extrapolate to x^x) - which, of course, is 8, the denominator will increase at a dramatically faster rate than the numerator. Therefore, the product will approach 0 as x approaches infinity.

So you say that you can represent the interval from (0,N) which contains an uncountable number of points by a the product of a countable number of arbitrary rational numbers?

No matter how large an N you choose, there will always be uncountably many irrational numbers on the real line between any two rational numbers.

Woops. How did some of my 8's become cool-guy smileys?

Every time x is increased, there will be rational numbers that differ from irrationals by an arbitrarily small value. (By the way, are transfinites considered a sub-set of irrationals? That would make this discussion much more succinct.) For example, let's say we were using powers of 10 rather than powers of 2. In the interval (0,2), we could get .5, .56, .567, .5671, .567143, .5671432, etc. with increasing powers of 10. As the power of 10 approaches infinity, we would be able to include the Omega Constant as one of the real factors, which is transfinite.

That's a good question. It took me a while to figure out those were 8s.

The only use of "transfinite" that I'm familiar with is as another name for the cardinal numbers. As such, a transfinite number itself is not a number on the real number line.



Do you mean the ordinal number omega? If so, then I don't think it would be considered transfinite since the transfinites are the cardinal numbers, not the ordinal numbers.

Remember that the cardinals are used to represent the size of a set while the ordinals are used to represent the number of elements in a well ordered set.

The omega constant is the solution to the equation Ω * e ^ Ω = 1. It is clearly a real number.

Transcendentals. Sorry. Not transfinites.

That makes more sense, then, but I don't think that it matters at all as for the product of all nonzero real numbers less than some arbitrary number N.

No. Division by zero is undefined, because there is no way that makes sense to define it. We can define the square root of -1 as "i" and proceed, but we can't define 1/0.


An 8 followed by a ) with no space is a cool-guy smiley. There's a preview button you can use to catch this and other formatting errors, and if you're not using smilies in a post, there's a checkbox "Disable Smilies in this post" that does what it says it does.


This can't work. If you take any interval between 0 and 1, of any size, then you can take that interval's largest member (call it "a") and get a subset of the interval which are all < a, and multiply them together to get a number smaller than a^x for any x. Even with as few as a countably infinite subset of the interval, we get a product < any real number, but not equal to zero; in other words, not a real number. A similar argument works for any interval with every element > 1.

You might think that intervals crossing 1 might have a chance, but they don't. Try matching up numbers from the high and low end and you run into problems: the only way they can cancel out to 1 is if each number on the low end is paired up exactly with its inverse. But you can pair up the same 2 sets of numbers different ways. You can even do this with countably infinite sets, it's quite easy to pair up every integer with an even integer, leaving out all the odd ones.

The same arguments work to show this wouldn't work on the rationals either.