Let ABCD represent the digits of the starting number.
The four digit number would be represented in basis $X\in \mathbb{N}$ by :
$$\textrm{ABCD}=X^{3}.A+ X^{2}.B+ X^{1}.C+ X^{0}.D$$
Am I right ?
Thanks a lot.
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1Yes, though base X is more common than basis X and you usually require the digits to be less than $X$, which then precludes $X=1$ – Henry Sep 02 '14 at 11:23
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1base is more common than basis – Henry Sep 02 '14 at 11:38
1 Answers
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The common conventions for positional systems is that the least significant digit is the rightmost one. This is shared by the Latin, Arabic and Hebrew scripts, notwithstanding the fact that the writing direction is different.
Thus the best bet is that $ABCD$ in base $X$ represents the number $$ D\cdot X^0+C\cdot X^1+B\cdot X^2+A\cdot X^3 $$
Note that an integral base should be greater than $1$.
egreg
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It seems to me that your re-ordering of the OP's expression serves no purpose, but might create confusion. – TonyK Sep 02 '14 at 12:20
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@TonyK I answered when the question was formulated differently (with $A$ considered the least significant digit). If we write the number in the form $a_3a_2a_1a_0$ instead of $ABCD$, the ordering I used is more natural. – egreg Sep 02 '14 at 13:16
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1@Educ Digits should be in the range $0..X-1$ in order to achieve uniqueness of the representation. So base $1$ would just have the digit $0$ which is inadequate for representing all numbers. – egreg Sep 02 '14 at 13:18