# The product of every number

1. Oct 15, 2009

### bubbloy

1. The problem statement, all variables and given/known data

A long time ago I got asked the product of every real number except 0. The person was looking for -1 and the first mapping I tried was a -> 1/a so i got the answer. never really thought about it again. Just realized that you could map a -> 1/(2a) and the answer would seem to be an infinite product of 1/2's. can someone find why this new map wouldn't cover every real number except 0? if it was right, this would imply that the product of every real number is both -1 and 0 and infinity (a -> 2a)

2. Oct 15, 2009

### bubbloy

Multiplying every number

A long time ago I got asked the product of every real number except 0. The person was looking for -1 and the first mapping I tried was a -> 1/a so i got the answer. never really thought about it again. Just realized that you could map a -> 1/(2a) and the answer would seem to be an infinite product of 1/2's. can someone find why this new map wouldn't cover every real number except 0? if it was right, this would imply that the product of every real number is both -1 and 0 and infinity (a -> 2a)

pairing a and 1/a

(7*1/7)(8.89765798*1/8.89765798)…*-1 = -1

pairing a and 1/(2a)

(1/2*1)(2/3*3/4)…(7/8*8/14)… = 1/2*1/2*1/2 = (1/2)^∞

i don't find anything wrong with either scheme of pairing the terms

3. Oct 15, 2009

### Office_Shredder

Staff Emeritus
Re: Multiplying every number

Because this isn't well defined. First of all, how do you define an uncountable product? Second, how do you know this product even converges? It seems to be at best "conditionally convergent" (in the same sense a sum that is conditionally convergent can't be re-arranged at will).

Have you covered infinite series and sequences before? If so the conversation is a lot easier to have

4. Oct 15, 2009

### bubbloy

Re: Multiplying every number

lol yeah i'm a physicist. kind of embarrassed to have this plaguing me. go ahead with whatever level of sophistication you wish to apply

5. Oct 15, 2009

### bubbloy

Re: Multiplying every number

it seems to satisfy the convergence conditions in each case

terms in the product are of the form (a*1/(2a)) = 1/2, each term in the infinite product is then 1/2 which obviously converges.

6. Oct 15, 2009

### Mentallic

Now, for any positive a, multiplying by -a will just give you an even bigger negative number which we can't do much with. But if we multiplied by -1/a we have automatically made the number -1, and this will happen for all $a\neq 0$.
It also maps through the entire real numbers (not the extended, so don't think of infinite, and also not 0).

Now, the only ambiguity I see here is that there are an infinite number of these values for a, so we can't say if we've multiplied by an even or odd number of (a)(-1/a)

say we take two values of a=2,3
then (2)(-1/2)(3)(-1/3)=(-1)(-1)=1

but for an odd number of products, we get -1.

7. Oct 16, 2009

### Mentallic

Can I just ask how you got the answer from mapping this? I still see the same problem arising as what happened in post #2.

For all a>0, the infinite products of (a)(1/a)=1
For all a<0, the infinite product still fluctuates (I think the correct term is diverges) between 1 and -1 depending on multiplying by an odd or even number of products. So the answer could still be -1 or 1 by this method.

8. Oct 16, 2009

### bubbloy

right, i see what you are saying

if you pair each number with it's inverse

you can have every positive number a with 1/a and every negative number b with 1/b, so b and 1/b will cancel their minus signs in every pairing, the only unpairable number will be -1 so that the entire product becomes -1

in your pairing (a with -1/a) the product would seem to fluctuate even in the infinite case

in the a with 1/(2a) pairing the product seems to be 0

if you pair a and 2/a the product seems to be 2*2*2*2*2=∞

is it possible that this can have different products depending on ordering?

9. Oct 16, 2009

### Office_Shredder

Staff Emeritus

If you want to say what the 'infinite product' is you need to have a reasonable definition. For products of countably many elements this is sort of easy, you just look at finite products with more and more terms and see what it converges to.

With what you want to do it's not so easy. Even if you look at larger and larger products of a finite number of real numbers, you can never order your real numbers in a way so that every single one of them is eventually included in a product (this is because the reals are uncountable). So you need to define your product in some other way than looking at finite products.

It's not surprising that different orderings (even in countable cases) give different values. For example, the infinite series

$$\sum \frac{(-1)^n}{n}$$ converges and what it converges to changes if you change the order around

But again, in your case you cannot possibly give an ordering of the real numbers so that every real number shows up in one of your finite products, so you are guaranteed to get different values just based on which real numbers you actually decide to look at

10. Oct 16, 2009

### HallsofIvy

Staff Emeritus
Re: Multiplying every number

This was also posted under "Homework". I am merging the two threads.

11. Oct 17, 2009

### Bingk

When we consider they product of all reals except 0, we can split up the product as such: [a * (1/a)] * [b * (1/b)] * -1 * 1 = -1, where [a * (1/a)] is the product of all a>1 and [b * (1/b)] is the product of all b<-1. This is to be sure that none of the reals are repeated. This is A solution that seems pretty solid to me, so we can be relatively certain that -1 is a solution.

Your (bubbloy) question is if it would work when 1/a or 1/b are replaced with 1/(2a) or 1/(2b) respectively.

It depends on how you look at it. If you treat 1/(2a) as a number by itself, then yes the product converges to 0, and this is due to the nature of the reals.
But, if you look at that product, you're basically saying that the product of all reals is the basically the same as taking the product of all reals multiplied to an infinite number of halves. So they're not really the same ... and I guess that's where the issue of defining the product becomes a problem, because it should work, but it doesn't seem to work.
So, the product of all reals MIGHT approach 0, depending on how you set it up..

I'm not too sure if your mapping (which I'm roughly translating to as a pairing of factors) of a -> 2a works, because it seems like you get repeating numbers in this product i.e. 2*4 and 4*8
Instead of mapping a to 1/(2a), we can also map a to 2/a, and that's how we can get the convergence to infinity. Again, this comes with issues, so it MIGHT be true ...

As Office_Shredder mentioned, once you consider other cases, the problem is whether you accept that all reals are included or not ...

So now it's up to you to choose :)