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pikachamp
swore in chat!

 Posted: Thu Mar 07, 2002 11:28 pm    Post subject: 1 1. what do you get if you multiply every real number together? puzzle 2 will be up when puzzle 1 is solved, as puzzle 2 might give away the answer to puzzle 1
Daedalian Member

 Posted: Thu Mar 07, 2002 11:29 pm    Post subject: 2 0.
pikachamp
swore in chat!

 Posted: Thu Mar 07, 2002 11:31 pm    Post subject: 3 wow, fast puzzle 2: what if you multiply all the real numbers together except for 0
Chuck
Daedalian Member

 Posted: Thu Mar 07, 2002 11:39 pm    Post subject: 4 Answers are supposed to be invisible, not the puzzle.
pikachamp
swore in chat!

 Posted: Thu Mar 07, 2002 11:42 pm    Post subject: 5 i didn't want to give away the answer to puzzle 1
Bicho the Inhaler
Daedalian Member

 Posted: Thu Mar 07, 2002 11:48 pm    Post subject: 6 I don't think so. "Zero" is not going to work here. Did you mean that 0 is in the set of reals, and 0 times any real number is 0, so 0 times all of them is 0? To show that, you would have to show that the product of all nonzero real numbers was itself a real number (even a complex or a quaternion would cut it). That is what you meant, right?
Laramie
Daedalian Member

 Posted: Fri Mar 08, 2002 12:26 am    Post subject: 7 What do you want it to be? Consider any real number other than 1 or zero(call it x). It has a unique counterpart 1/x such that when we multiply them we get one. We choose all such pairs and multiply them to get one. We then multiply them all together to get 1. So the answer is one. Now, again consider any real number other than 1 or zero(call it x). It has a unique counterpart 2/x such that when we multiply them we get two. We choose all such pairs and multiply them to get two. We then multiply them all together to get infinity. So the answer is infinity. The product is not well-defined mathematically, and I can make the limit literally anything I want by choosing the multiplication "sequence" properly. [This message has been edited by Laramie (edited 03-07-2002 07:27 PM).]
pikachamp
swore in chat!

 Posted: Fri Mar 08, 2002 12:41 am    Post subject: 8 if you do it by the reciprocal thing, you will find that the recipricol of -1 is -1, thus -1 remains negative, causing the answer to be -1 in the 2 thing, the one for -2 is -2, thus the answer is negative infinity
quercitron
Don't trust Robinson

 Posted: Fri Mar 08, 2002 2:58 am    Post subject: 9 hate to break it, but infinity really isn't a number. q
tigg
Daedalian Member

 Posted: Fri Mar 08, 2002 3:36 am    Post subject: 10 what do you get if you multiply every real number together? very, very old
Bicho the Inhaler
Daedalian Member

Posted: Fri Mar 08, 2002 3:47 am    Post subject: 11

Well, it certainly isn't a real or complex number. People have come up with many ways to make infinity a "number" in a more general sense. Still, I don't think there's any way to define the product of a continuum of numbers in the real number system. Implicit in any product is the order and grouping of the multiplication. Remember, the multiplication operator is defined as mapping *ordered *pairs** of real numbers to individual real numbers. Because multiplication on reals is associative and commutative, we tend to ignore grouping and order because every grouping and order we could pick (on a finite set) yields the same answer, but formally, grouping and order do matter and cannot be ignored when making abstract definitions like extending multiplication beyond just 2 numbers. With this in mind, how would you define a product of a continuum? Would you like to try?

 Quote: and I can make the limit literally anything I want by choosing the multiplication "sequence" properly.
No, you can't, actually. The reals are uncountable, meaning no sequence of real numbers contains all of them.
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