Proving the Uniqueness of Base Representation for Integers

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The discussion focuses on proving the uniqueness of base representation for integers in base k, specifically addressing the scenario where two different representations, n and m, exist. It establishes that if n is represented as asks + as-1ks-1 + ... + a0 and m as btkt + bt-1kt-1 + ... + b0, then m cannot equal n when as and bt are non-zero. The participants suggest dividing both numbers by ks+1 to demonstrate that one quotient will be less than 1 while the other will be greater than or equal to 1, confirming the uniqueness of representation.

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sty2004
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base representation please 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(existence and uniqueness of such representation of an integer), prove directly that m not equal to n.

Many many thanks~~~
 
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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.
 
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