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Topic: Basic Sequence (Read 497 times) |
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Sameer
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Basic Sequence
« on: May 28th, 2004, 8:11am » |
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I have a basic sequence. Once it is solved I have a question that I will follow up:
10, 11, 12, 13, 14, 20, 22, 101, ?, ?
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"Obvious" is the most dangerous word in mathematics.
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Proof is an idol before which the mathematician tortures himself.
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BNC
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Re: Basic Sequence
« Reply #1 on: May 28th, 2004, 8:58am » |
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I think there's a similar one floating around here.
::... 1010, 1111111111 ...
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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TenaliRaman
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Re: Basic Sequence
« Reply #2 on: May 29th, 2004, 1:02pm » |
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we may even have 11110 into consideration.
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Sameer
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Re: Basic Sequence
« Reply #3 on: Jun 1st, 2004, 6:50am » |
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Tenaliraman what was your logic?
I was looking for BNC's answer. So as a follow up can somebody (or BNC) explain the last term?
Btw BNC, :: 1111111111 is the last term and hence no "..." after that.
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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BNC
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Re: Basic Sequence
« Reply #4 on: Jun 1st, 2004, 7:27am » |
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Well, the last term is definitly an odd one. It's in base 1.
What is base 1? I'm not sure if it's mathematically correct, but basically, 1=1, 2=11, 3=111, 4=1111 etc.. A little like roman numerical without the "complicated" stuff (V, X, C ...).
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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Sameer
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Re: Basic Sequence
« Reply #5 on: Jun 1st, 2004, 10:58am » |
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Yes correct, from definition of bases, it is known that for a base b the number 0,1... b-1 are in its domain. Then for base 1, only number possible is 0. Now this totally "cooked my noodles"  . I couldn't find some good info on base 1 so I would invite everyone to discuss "base 1" and/or "base 0".
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"Obvious" is the most dangerous word in mathematics.
--Bell, Eric Temple
Proof is an idol before which the mathematician tortures himself.
Sir Arthur Eddington, quoted in Bridges to Infinity
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towr
wu::riddles Moderator
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Re: Basic Sequence
« Reply #6 on: Jun 1st, 2004, 1:02pm » |
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Well for any base b a number amam-1..a1a0 is equal to [sum]i = 0m(biai)
Clearly that holds for these supposed unary numbers, where [forall]i: bi = 1
The problem of course is the ai's. If you'd allow ai = 0 for some i < m then every integer has an infinite number of representations. But multiple representations needn't be a problem, f.i. in decimal 1 = 0.999...
For regular bases it's common to have 0 [le] ai < b, but should it really be a requirement? It wouldn't be an issue for many numbers to begin with, whereas for unary representation you can't represent any number without allowing ai = b.
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| « Last Edit: Jun 1st, 2004, 1:12pm by towr » |
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Wikipedia, Google, Mathworld, Integer sequence DB
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Noke Lieu
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Re: Basic Sequence
« Reply #7 on: Jun 1st, 2004, 11:49pm » |
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As far as I understand, (which seems to be FAR less than towr) in base 1, 1 is used by convention.
It needn't be. it could be # or @ or [ohno], it doesn't really matter.
Sort of comes down to the idea that 1 is 1 isn't 1. Its just a line with bent bit. It doesn't mean one (and for that matter "one" doesn't mean 'one' either- its just a word), we just accept it does.
ah.... semantics. That's just the word you use to highlight the difference between words. 
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BNC
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Re: Basic Sequence
« Reply #8 on: Jun 2nd, 2004, 12:11am » |
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Yes, "1" is just a graphic symbol. However, it has mathematical background. It is the binary operation "*" (another graphic symbol) "natural" element, as "0" is the "natural" element for the other principle binary operation "+". I'm not sure it's possible to define a number system with a single element.
Disclaimer: I'm writing from my (not so good) memory, and I'm not sure if I use the right terms (or concepts, for that matter).
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How about supercalifragilisticexpialidociouspuzzler [Towr, 2007]
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towr
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Re: Basic Sequence
« Reply #9 on: Jun 2nd, 2004, 1:13am » |
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on Jun 1st, 2004, 11:49pm, Noke Lieu wrote:As far as I understand, in base 1, 1 is used by convention.
It needn't be. it could be # or @ or [ohno], it doesn't really matter. |
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It should still mean 1 (to the writer and intended readers), regardless of whether you actually use that particular symbol. Otherwise how do you get from the representation to the number?
It would make sense if it was the same way as with normal bases: multiply the digit by the weight of the position and sum them. And that certainly works here as well..
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| « Last Edit: Jun 2nd, 2004, 1:13am by towr » |
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Grimbal
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Re: Basic Sequence
« Reply #10 on: Jun 2nd, 2004, 4:18am » |
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To put the question of the bases in perspective.
To represent a number in a base, you have the base, which is the value you raise to power n, and a set of symbols, having each a value, that multiply each power.
If you consider the problem of weighting any integer weight with a limited set of weights, you find that the weight 1, 3, 9, 27, ... do the job nicely by puting the weight either left, right or doing without. This is because you can write any weight as
sum(an*3n) where an is -1,0,1.
Effectively, you are writing the weight in base 3, with symbols -1, 0, 1.
Normally, to cover all integer values, you need as many symbols as the base value. 3 symbols for base 3. Else, the range of n-digit numbers would increase faster than the number of n-digit numbers you could write, so you would leave gaps.
In base 1, you can have a single symbol, say I, with value 1, if you consider that the sum(an*basen) has a finite number of terms. Then, you can write any number by just writing as many I's as necessary. 1=I, 2=II, 3=III, 10=IIIIIIIIII, etc.
The weirdest base I know is i-1, where i is [smiley=surd.gif]-1
It looks like base 2, since there are only the symbols 0 and 1. The good thing is that you also can represent negative and complex numbers. The bad thing is that the carry mechanism is much more complicated. 1+1 = 1100. That means that there are carry bits 2 and 3 positions upwards.
0 = 0
1 = 1
2 = 1100
3 = 1101
4 = 111010000
5 = 111010001
6 = 111011100
7 = 111011101
8 = 111000000
9 = 111000001
10 = 111001100
etc.
2*2 = 4
1100*1100 = 1210000 (replace 2 by 1100) = 111010000
2*4 = 8
1100 * 111010000 = 122111000000 = 111000000
because 122 = (i-1)^2 + 2(i-1) + 2 = (-1)-2i+1 + 2i-2 + 2 = 0
PS: I write 2 as a temporary notation for 1+1. 2 must be resolved as 0 with 110 as carry bits.
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| « Last Edit: Jun 2nd, 2004, 4:24am by Grimbal » |
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TenaliRaman
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Re: Basic Sequence
« Reply #11 on: Jun 5th, 2004, 7:32am » |
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on Jun 1st, 2004, 6:50am, Sameer wrote:Tenaliraman what was your logic?
I was looking for BNC's answer. So as a follow up can somebody (or BNC) explain the last term?
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ofcourse i agree with bnc's answer too.
(10)10=(11110)-2
If you are wondering what makes sense of base -2 then look at grimbals last post who made everything clearer than i ever could.
I found these bases(i.e negative and other unimaginable ones) with my friend when we were playing around with summation some four years ago in our junior college.Those were fun days.Alas i miss them now and engg student life is hell with no extra activities to work on.
(Sorry for the rant! i just finished my exams and i am just throwing these things out of my system)
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