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Base representation ~

  1. Jan 28, 2010 #1
    base representation plz help~

    It is known that if asks+as-1ks-1+...+a0 is a representation of n to the base k, then 0<n<=ks+1-1.

    Now suppose n=asks+as-1ks-1+...+a0 and m=btkt+bt-1kt-1+...+b0 with as,bt not equal to 0, are two different representations of n and m to base k, respectively. Without loss of generality we may assume t>=s. Without using Theorem 1-3(existance and uniqueness of such representation of an integer), prove directly that m not equal to n.

    Many many thanks~~~
     
  2. jcsd
  3. Jan 28, 2010 #2
    Re: base representation plz help~

    We consider the difference :
    [tex]D = m-n = c_tk^t + c_{t-1}k^{t-1} + ... + c_0[/tex]
    Where [tex]c_i = b_i - a_i[/tex] ( [tex]i[/tex] from [tex]1[/tex] to [tex]t[/tex], and [tex]a_i = 0[/tex] if [tex] i > s[/tex])

    Due to the fact that [tex]a_sa_{s-1}...a_1[/tex] and [tex]b_tb_{t-1}...b_1[/tex] are two different representations, [tex]c_i[/tex] must not equal to 0 for every i, otherwise [tex]a_i = b_i[/tex] for all [tex]i[/tex].

    Choose [tex]j[/tex] is the largest number such that [tex]c_j[/tex] is different from
    If [tex]c_j > 0[/tex],then:
    [tex]D >= k^j -(k-1)(k^{j-1} + ... 1) = k^j -(k^j-1) = 1 >0[/tex]
    If [tex]c_j < 0[/tex], similarly [tex]-D > 0[/tex]

    We always have [tex]D[/tex] different from 0

    Good luck.
     
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