If the base of number system becomes one then we have only one digit (zero) to represent all the numbers which is a problem.
2006-09-19 14:37:58 · 6 answers · asked by Hemant 2 in Science & Mathematics ➔ Mathematics
Well my base is metastable .
It may be written as : lim(x -->0)[1+x]
2006-09-19 16:31:47 · update #1
I think you just answered your own question. There is NO possible way.
2006-09-19 14:45:58 · answer #1 · answered by PhysicsDude 7 · 0⤊ 0⤋
In base n, the number 111 = n^2 + n + 1. That still works in this theoretical base 1 system, but it looks ugly: 1 = 1, 2 = 11, 3 = 111, 4 = 1111, etc. You can still have 0 represent zero, and I suppose negative numbers are still fine: -1 = -1, -2 = -11, etc.
Bigger numbers would be a major nuisance, but maybe you can make the system manageable by inventing other symbols. For example, you could replace 11111 with V, or 1111111111 with X, or represent fifty 1s by an L. Oh wait, that's Roman numerals, isn't it...?.
2006-09-19 21:58:45 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Well, each digit should weight "base" times more than its neighbour to the right. Therefore, in a base 1 system, all positions weight the same, and you may have:
0 = 1
00 = 2
000 = 3
0000 = 4
etc...
2006-09-20 07:37:43 · answer #3 · answered by Andy D. 2 · 0⤊ 0⤋
Base one number systems are metaphysical. You achieve them by mediation and reaching a state of emptiness. Within this emptiness lies infinity. In nothingness is purity. In purity is infinity.
;-D Seek the Inner Light in nothingness and find the Universe.
2006-09-19 22:34:28 · answer #4 · answered by China Jon 6 · 0⤊ 1⤋
http://en.wikipedia.org/wiki/Unary_numeral_system
2006-09-19 22:10:04 · answer #5 · answered by Pascal 7 · 0⤊ 0⤋
dude, ever hear of binary?
2006-09-19 21:44:54 · answer #6 · answered by dave_co_78 2 · 0⤊ 0⤋