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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~

    You can try dividing both numbers by ks+1... one quotient will be <1 and the other >=1.

    Edit: Oh, I'm sorry. I thought t>s, while you said instead t>=s. In this case I think you need something else, because if t=s, as=bs and all ai,bi are 0 for i<s, then n=m.

    Oh, but then again... you said they are different representations. I assume this means that, even if t=s, as and bs cannot be the same. Then, in this case, divide both numbers by max(as,bs) . ks, and that should give you one quotient <1 and another >=1.
    Last edited: Jan 28, 2010
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