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Proposition: Let n $in$ N. If $n = sum_i=0^p x_i10^i = sum_i=0^qy_i10^i $ where $p,q, in Z_geq0$, each $x_i$ and each $y_i$ is a digit, $x_pneq0$ and $y_qneq0$,

then $p=q$, and $x_i = y_i$ for all i

So far I have:

$n = sum_i=0^p x_i10^i = sum_i=0^qy_i10^i $

$sum_i=0^p x_i = sum_i=0^qy_i $

$sum_i=0^p x_i - sum_i=0^qy_i = 0$

I feel like I can perform some summation algebra to write this as one summation, but I'm not sure how.

How do I prove that $p=q$ and the sequence $x_i = y_i$ ?

The key issue with the base $10$ representation being unique is that the digits are restricted to being between $0$ and $9$ inclusive (note that without this limitation, the representations may not necessarily be unique). The main reason this is important is that the difference in digits is limited to being between $-9$ and $9$, inclusive. To see why this is important, consider what you've stated, i.e.,

$$n = sum_i=0^p x_i 10^i = sum_i=0^q y_i 10^i tag1labeleq1$$

where $p,q, in Z_geq 0$, each $x_i$ & $y_i$ is a digit of $0$ to $9$, $x_p neq 0$ and $y_q neq 0$. To make it a bit easier for processing in case $p neq q$, let $r = max(p,q)$ and extend $x_i$ to be $0$ for any $i gt p$ and $y_i$ to be $0$ for any $i gt q$, so eqrefeq1 becomes

$$n = sum_i=0^r x_i 10^i = sum_i=0^r y_i 10^i tag2labeleq2$$

Subtracting the middle & LHS parts of eqrefeq2 gives

$$0 = sum_i = 0^r left(x_i - y_iright) 10^i tag3labeleq3$$

If $x_i neq y_i$ for one or more $i$, let $k$ be the smallest one. Thus, for all $i lt k$, $x_i = y_i$ so $x_i - y_i = 0$, causing those terms to all become $0$. They can therefore be ignored, simplifying eqrefeq3 to

$$0 = sum_i = k^r left(x_i - y_iright) 10^i tag4labeleq4$$

Next, dividing both sides by $10^k$ gives

$$0 = sum_i = k^r left(x_i - y_iright) 10^i - k tag5labeleq5$$

Note that all of the terms on the RHS of eqrefeq5 have at least one factor of $10$ except for the first one. Since $10$ divides $0$, plus all of the terms on the RHS of eqrefeq5 after the first term, this means the first term of $left(x_k - y_kright)10^0 = x_k - y_k$ must be a multiple of $10$. Since $-9 le x_k - y_k le 9$, this means that $x_k - y_k = 0$, i.e., $x_k = y_k$, so they can't be different. This contradicts the assumption that there is at least one digit different, thus showing that all of the digits up to $r$ are the same. This also means the highest non-zero ones must be the same, i.e., $p = q$.

This same basic argument can be used to show that an integer can be uniquely represented in any base $b ge 2$ as long as the digits used are in the range of $0$ to $b - 1$, inclusive (e.g., binary using $0$ and $1$, or hexadecimal using $0$ to $F$).

Rather the proving any two representations on $N$ must be equal, try proving $N$ has exactly one unique representation.

But the division theorem, for any integer $N$ the are are unique integers $M_1$ and $x_0$ so that $N = 10M_1 + x_0; 0 le x_0 < 10$. This proves that $x_0$ must be uniquely determined.

Inductively for each $M_k$ there are unique $M_k+1$ and $x_k$ so that $M_k = 10M_k+1 + x_k; 0le x_k < 10$.

And that's it really. As $N$ is bounded above by some $10^p+1$ this will have a finite number of steps.