A question about $\prod_{x\in \mathbb{R}^{*}}{x}$
When I was an elementary school student, I encountered some puzzles like "What's product of numbers of hair of each people in the world?", which answer was zero, because there's people with no hair(bald), and 0*any number = 0. There was similar question like "What's the product of every number in a telephone?".
After know that '0' makes the whole answer 0, I made three questions and asked first two questions to my friends (to induce the last question):
"What's the product of every numbers?" (or $\prod_{x\in \mathbb{R}} x = ?$)
"What's the sum of every numbers?" (or $\sum_{x\in \mathbb{R}} x = ?$)
They answered '0' in the first question easily, and they also answered '0' in the second question, though they spent several seconds to think (that $x+(-x) = 0$).
After they answered the second question, I asked the last question:
"The answer of the first question was zero, because 'every numbers' includes zero. Then, what's the product of every numbers, except zero? (or $\prod_{x\in \mathbb{R}^{*}} x = ?$)"
My answer was -1, because except 1 and -1, every number $x$ has it's inverse, $\frac{1}{x}$, and the product of two numbers is 1. Therefore, the answer = 1*-1*1*1*.... = -1.
It was quite interesting, because some answered 'infinite?', some answered '1', and some answered '-1', which was the correct answer I thought, before one of my friend says that
"Because the set of real numbers are uncountable, your question doesn't have an answer."
I don't remember whether my friend said "doesn't have an answer", "'s answer is infinite", or other things...
Anyway, the question is, what is the answer of $\prod_{x\in \mathbb{R}^{*}} x$? is it -1? or undecidable?
If it's not -1 (undecidable or infinite or others), then what's the answer of above two questions? or $\prod_{x\in \mathbb{Q}^{*}} x$?
If it's 1, then what's the answer in hyperreal number?
Solution 1:
First of all, it is almost impossible to have zero hairs. Even bald people presumably have some amount of hair on their chest, their arms, or their legs. I shudder to think what kind of chemicals you'd need to douse on yourself to get rid of all of your hair. A much more reasonable answer is something like $(3 \times 10^5)^{7 \times 10^9}$: I happen to know that the average human head has about $1.5 \times 10^5$ hairs, so counting all of the hair on a human body should get you about twice that on average, and there are about $7$ billion people in the world. This is a very, very rough estimate, though (the logarithm of it should be within an order of magnitude). Edit: Apparently it is possible to have zero hairs; see the comments.
Second, while these are interesting answers, mathematicians have a formal definition of infinite sums and infinite products, and in those definitions
- you need to specify an order to take the sum or product in, which you haven't, and
- for the vast majority of orders, none of these expressions are well-defined.
The reason for the first is the Riemann rearrangement theorem, which says that under certain conditions it is possible to rearrange a series of numbers to have any sum, and the reason for the second is that a sum of uncountably many positive numbers always diverges. This is because for some positive integer $n$ there must be uncountably many numbers in the sum greater than $\frac{1}{n}$. Thus, mathematicians generally do not consider sums or products over an uncountable set at all.
Nevertheless, the real numbers are an example of a field, and it is possible to apply the reasoning that you've described to finite fields. In a finite field of characteristic not equal to $2$, it is true for precisely the reasons you describe that the sum of every element is $0$ and that the product of all of the nonzero elements is $-1$. More generally, in any finite abelian group, the product of all the elements is equal to the product of all the elements of order $2$, which is equal to $1$ unless there is exactly one non-identity such element $-1$, in which case it is equal to $-1$. This is a nice exercise, and the proof is very similar to the one you've already described.
Edit: It occurs to me that I don't need to go quite so far as the Riemann rearrangement theorem to convince you that you can't naively apply symmetry arguments to infinite sums and products. You claim that the sum of all real numbers is zero because a number cancels with its inverse. Similar reasoning would tell you that
$$1 - 1 + 1 - 1 \pm ... = 0$$
because you can write it as
$$(1 - 1) + (1 - 1) + (1 - 1) + ... = 0.$$
However, you can also write it as
$$1 + (-1 + 1) + (-1 + 1) + ... = 1.$$
So by applying symmetry in a different way, you get a different answer. This is a basic example known as Grandi's series which highlights the problem with thinking about infinite sums this way, although in the particular case of Grandi's series there's a funny answer: in certain senses, the "right value" of the above sum is $\frac{1}{2}$, the average of the two values above!