Number Storage

[Shadow]

Hey all,

 

this will sound like a pretty stupid questions, but given a set of N real numbers numbers, is there any operation that can be performed on them that would output a result unique only to that set of numbers, in that specific order? This question has been bugging me for some time, so I was wondering if there was some kind of proof or something that makes this impossible. If there is nothing to dispute the possibility of there being such an operation, can you think of one?

 

Cheers,

 

Gabe

[timo]

Assuming it is reals that can be written down in decimal representation then it becomes almost trivial:

 

1st line: N.

2nd line: 1st number.

3rd line: 2nd number.

.

.

.

N+1th line: Nth number.

 

Otherwise: A representation system with a finite number of digits can (I think) only represent a countable number of objects. Since the reals are non-countable, you cannot represent each real number in any such system. That would be your "it's impossible" proof, I think.

[Shadow]

You misunderstood me. I meant an operation that would return one number, or word, or anything. And yes, let's assume that they consist of a finite number of digits. The reason for this is that I was wondering whether the formula of an nth degree function can be represented by one number. Or in less than n numbers, since I could just keep applying the process over and over again. It sounds more impossible the more I'm writing about it, but I'd still like someone else's opinion.

[timo]

Add an "a" instead of a carriage return, then:

<number of numbers>a<1st number>a<2nd number>a...

 

Or represent each standard digit d with "0d" and use "10" as spacing between the entries and 11 as decimal point and 12 as *10^... (and interpret that stream of digits as a number if you want), so e.g. the tuple (1, 2, 3.14) would read:

03100110021003110104

[Shadow]

Yeah, that was also my thinking, although I only had the idea in my head for integers. Unfortunately, that would yield incredibly long numbers, which is something I was trying to avoid. Thankfully, I've found a way to bypass this step completely, so it doesn't really matter anymore. Thanks for the help.