Can Two Different Base Representations of an Integer Prove It is Unique?

[sty2004]

base representation please help~

It is known that if a_sk^s+a_s-1k^s-1+...+a₀ is a representation of n to the base k, then 0<n<=k^s+1-1.

Now suppose n=a_sk^s+a_s-1k^s-1+...+a₀ and m=b_tk^t+b_t-1k^t-1+...+b₀ with a_s,b_t 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~~~

 

[telesyn]

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

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

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

We always have $$D$$ different from 0

Good luck.